<%BANNER%>
Concepts in Calculus I, Second Edition
Buy This Book ( Related URL )
ALL VOLUMES CITATION PDF VIEWER
Full Citation
STANDARD VIEW MARC VIEW
Permanent Link: http://ufdc.ufl.edu/AA00011694/00001
 Material Information
Title: Concepts in Calculus I, Second Edition
Physical Description: Book
Language: en-US
Creator: Bona, Miklos
Shabanov, Sergei
University of Florida Department of Mathematics
University Press of Florida
Publisher: University Press of Florida
Place of Publication: Gainesville, FL
Publication Date: 8/2/2011
 Subjects
Subjects / Keywords: Calculus I
MAC 221
MAC 223
MAC 230
MAC 241
MAC 300
Mathematics
 Notes
Abstract: From the University of Florida Department of Mathematics, this is the first volume in a three volume presentation of calculus from a concepts perspective. The emphasis is on learning the concepts behind the theories, not the rote completion of problems. WebAssign online ancillaries are available. Most questions from this textbook are available in WebAssign. The online questions are identical to the textbook questions except for minor wording changes necessary for Web use. Whenever possible, variables, numbers, or words have been randomized so that each student receives a unique version of the question. This list is updated nightly. Downloads are freely available (no cost). Click the shopping cart button to purchase a low cost printed textbook online or you may also phone the University Press of Florida to order a textbook on their toll free number: 800-226-3822.
General Note: MAC2230 - CALCULUS FOR BUSINESS & SOC. SCI. I, MAC2253 - CALCULUS FOR ENGINEERING TECHNOLOGY I, MAC2241 - LIFE SCIENCE CALCULUS I, MAC2233 - CALCULUS FOR BUSINESS & SOC. SCI. I, MAC2234 - CALCULUS FOR BUSINESS & SOC. SCIENCE II, MAC2223 - CALCULUS FOR ARCHITECTURE I
 Record Information
Source Institution: University Press of Florida
Holding Location: University Press of Florida
Rights Management: This work is licensed under a modified Creative Commons Attribution-Noncommercial-No Derivative Works 3.0 Unported License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/3.0/. You are free to electronically copy, distribute, and transmit this work if you attribute authorship. However, all printing rights are reserved …
Resource Identifier: isbn - 978-1-6161-0170-1
System ID: AA00011694:00001

Downloads

This item is only available as the following downloads:

Calc-I_final ( PDF )


Full Text

PAGE 1

ConceptsinCalculusI SecondEditionUNIVERSITYPRESSOFFLORIDAFloridaA&MUniversity,Tallahassee FloridaAtlanticUniversity,BocaRaton FloridaGulfCoastUniversity,Ft.Myers FloridaInternationalUniversity,Miami FloridaStateUniversity,Tallahassee NewCollegeofFlorida,Sarasota UniversityofCentralFlorida,Orlando UniversityofFlorida,Gainesville UniversityofNorthFlorida,Jacksonville UniversityofSouthFlorida,Tampa UniversityofWestFlorida,Pensacola OrangeGroveTexts Plus

PAGE 3

ConceptsinCalculusI SecondEdition Mikl osB onaandSergeiShabanov UniversityofFloridaDepartmentof MathematicsU niversity P ressof F lorida Gainesville  Tallahassee  Tampa  BocaRaton Pensacola  Orlando  Miami  Jacksonville  Ft.Myers  Sarasota

PAGE 4

Copyright2012bytheUniversityofFloridaBoardofTrusteesonbehalfoftheUniversityof FloridaDepartmentofMathematics ThisworkislicensedunderamodiedCreativeCommonsAttribution-Noncommercial-No DerivativeWorks3.0UnportedLicense.Toviewacopyofthislicense,visithttp:// creativecommons.org/licenses/by-nc-nd/3.0/.Youarefreetoelectronicallycopy,distribute,and transmitthisworkifyouattributeauthorship. However,allprintingrightsarereservedbythe UniversityPressofFlorida(http://www.upf.com).PleasecontactUPFforinformationabout howtoobtaincopiesoftheworkforprintdistribution .Youmustattributetheworkinthe mannerspeciedbytheauthororlicensor(butnotinanywaythatsuggeststhattheyendorse youoryouruseofthework).Foranyreuseordistribution,youmustmakecleartoothersthe licensetermsofthiswork.Anyoftheaboveconditionscanbewaivedifyougetpermissionfrom theUniversityPressofFlorida.Nothinginthislicenseimpairsorrestrictstheauthor'smoral rights. ISBN978-1-61610-170-1 OrangeGroveTexts Plus isanimprintoftheUniversityPressofFlorida,whichisthescholarly publishingagencyfortheStateUniversitySystemofFlorida,comprisingFloridaA&M University,FloridaAtlanticUniversity,FloridaGulfCoastUniversity,FloridaInternational University,FloridaStateUniversity,NewCollegeofFlorida,UniversityofCentralFlorida, UniversityofFlorida,UniversityofNorthFlorida,UniversityofSouthFlorida,andUniversityof WestFlorida. UniversityPressofFlorida 15Northwest15thStreet Gainesville,FL32611-2079 http://www.upf.com

PAGE 5

Contents Chapter1.Functions1 1 .Functions1 2 .ClassesofFunctions4 3 .OperationsonFunctions8 4 .ViewingtheGraphsofFunctions12 5 .InverseFunctions15 6 .TheVelocityProblemandtheTangentProblem21 Chapter2.LimitsandDerivatives25 7 .TheLimitofaFunction25 8 .LimitLaws33 9 .ContinuousFunctions39 10 .LimitsatIn“nity43 11 .Derivatives48 12 .TheDerivativeasaFunction51 Chapter3.RulesofDifferentiation57 13 .DerivativesofPolynomialandExponentialFunctions57 14 .TheProductandQuotientRules61 15 .DerivativesofTrigonometricFunctions64 16 .TheChainRule67 17 .ImplicitDierentiation71 18 .DerivativesofLogarithmicFunctions74 19 .ApplicationsofRatesofChange77 20 .RelatedRates82 21 .LinearApproximationsandDierentials89 Chapter4.ApplicationsofDifferentiation99 22 .MinimumandMaximumValues99 23 .TheMeanValueTheorem106 24 .TheFirstandSecondDerivativeTests116 25 .TaylorPolynomialsandtheLocalBehaviorofaFunction125 26 .LHospitalsRule133 27 .AnalyzingtheShapeofaGraph140 28 .OptimizationProblems146v

PAGE 6

viCONTENTS 29 .NewtonsMethod153 30 .Antiderivatives161 Chapter5.Integration167 31 .AreasandDistances167 32 .TheDe“niteIntegral176 33 .TheFundamentalTheoremofCalculus188 34 .Inde“niteIntegralsandtheNetChange194 35 .TheSubstitutionRule200

PAGE 7

CHAPTER1 Functions 1.Functions A function f isarulethatassociatestoeachelement x inaset D a unique element f ( x )ofanotherset R .Heretheset D iscalledthe domain of f ,whiletheset R iscalledthe range of f .Thefactthat f associatestoeachelementof D anelementof R isrepresentedbythe symbol f : D R .Insteadofsayingthat f associates f ( x )to x ,we oftensaythat f sends x to f ( x ),whichisshorter.SeeFigure1.1for anillustration. DomainRangex a b y z f Figure1.1. Domainandrange. Ifthesetsmentionedinthepreviousde“nitionaresetsofnumbers, thenitisofteneasiertodescribe f byanalgebraicexpression.Let N bethesetofallnaturalnumbers(whicharethenonnegativeintegers). Thenthefunction f : N N givenbytherule f ( x )=2 x +3is thefunctionthatsendseachnonnegativeinteger n tothenonnegative integer2 n +3.Forinstance,itsends0to3,1to5,17to37,andsoon. Inthiscase,thealgebraicdescriptionissimplerthanactuallysaying  f isthefunctionthatsends n to2 n +3.Ž Therulethatdescribes f maybesimpleorcomplicated.Itcould bethatafunctionisde“nedbycasessuchas f ( x )= 0 1 x if0 x 40 4+0 15( x Š 40)if40 80. Thisexamplecoulddescribeanincometaxcode.The“rst$40,000 ofincomeistaxedatarateof10%,incomeabove$40,000butbelow$60,000istaxedatarateof15%,andincomeabove$80,000is1

PAGE 8

21.FUNCTIONS taxedatarateof20%.Thevalueof f ( x )istheamountoftaxto bepaidafteranincomeof x thousanddollarsforanypositivereal number x Therearetimeswhentherulesthatapplyinvariouscasesare closelyconnectedtoeachother.Aclassicexampleisthe absolutevalue function ,thatis, f ( x )= | x | = x if0 x, Š x if x< 0. 3 2 10123 x 0.5 1.0 1.5 2.0 2.5 3.0 y Figure1.2. Graphof | x | Inthiscase, f ( x )= f ( Š x )forall x .Whenthathappens,wesaythat f isan evenfunction .Forinstance, g ( x )=cos x and h ( x )= x2areeven functions.Therearealsofunctionsforwhich Š f ( x )= f ( Š x )holdsfor all x .Thenwesaythat f isan oddfunction .Examplesofoddfunctions include g ( x )=sin x and h ( x )= x3. TherearetimeswhenaplainEnglishdescriptionofafunctionis simplerthananalgebraicone.Forinstance,let g bethefunction thatsendseachintegerthatisatleast2intoitslargestprimedivisorŽ issimplerthandescribingthatfunctionwithalgebraicsymbols(and symbolsofformallogic).Ifthesets D and R arenotsetsofnumbers, analgebraicdescriptionmaynotevenbepossible.Anexampleofthis iswhen D and R arebothsetsofpeopleand f ( x )isthebiological fatherofperson x .Notethatitisnotbyaccidentthatwesaidthat f ( x )isthefather(andnottheson)of x .Indeed,afunctionmustsend x toa unique f ( x ).Whileapersonhasonlyonebiologicalfather,he orshemayhaveseveralsons. Sometimestherulethatsends x to f ( x )canonlybegivenbylisting thevalueof f ( x )foreach x ,asopposedtoageneralrule.Forinstance, let D bethesetof200speci“ccitiesintheUnitedStates,let R bethe setofallnonnegativerealnumbers,andforacity x ,let f ( x )bethe

PAGE 9

1.FUNCTIONS3 amountofprecipitationthat x hadin2011.Then f isafunctionsince itsendseach x D intoanelementof R .Thisfunctionisgivenbyits listofvalues,notbyarulethatwouldspecifyhowto compute f ( x )if given x Finally,functionscanalsoberepresentedbytheir graphs .If f : D R isafunction,thenletusconsideratwo-dimensionalcoordinatesystemsuchthatthehorizontalaxiscorrespondstoelements of D ,andtheverticalaxiscorrespondstoelementsof R .Thegraphof f isthesetofallpointswithcoordinates( x,f ( x ))suchthat x D .The requirementthat f ( x )isuniqueforeach x willensurethatnovertical lineintersectsthegraphof f morethanonce.Thisiscalledthe vertical linetest .1.1.Exercises.(1)Foreachperson x ,let f ( x )denotethebirthday(day,month, andyear)of x .Is f afunction? (2)Foreachperson y ,let g ( y )denotethebiologicalmotherof y Is g afunction?Ifyes,whatisthedomainof g andwhatis therangeof g ? (3)Fortwopeople x and y ,letussaythat f ( x )= y if y isachild of x .Is f afunction? (4)Howmanyfunctionsaretherewithdomain { A,B,C,D } and range { 0 1 } ? Fortheremainingexercisesinthissection,allfunctionsare de“nedonsome realnumbers (5)Let f ( x )= x + | x | .Findthedomainandtherangeof f (6)Let f ( x )=( x +1) / ( x Š 2).Findthedomainandtherange of f (7)Let g ( x )= x/ | x | .Findthedomainandtherangeoff (8)Let f ( x )=1 sin x.Findthedomainandtherangeof f (9)Let f ( x )= sin2x +cos2x .Findthedomainandtherange of f (10)Let h ( x )=x x +3+x +3 x.Findthedomainandtherangeof f (11)Let x bethesmallestinteger y suchthat x y .Whatis thedomainandtherangeofthefunction x x ? (12)Let x bede“nedasinthepreviousexerciseandlet f ( x )= x Š x .Findthedomainandtherangeof x (13)Canthegraphofafunctionintersectaverticallinetwice? (14)Canthegraphofafunctioncontainacircle?

PAGE 10

41.FUNCTIONS (15)Canthegraphofafunctionintersectahorizontalline twice? (16)Canthegraphofafunctionintersectahorizontallinein“nitely manytimes? (17)Anin“nite sequence isanin“nitearrayofnumbers a1,a2,... Explainwhyin“nitesequencesare,infact,functions.What isthedomainofthesefunctions? (18)Let f ( x )=3 x +2.Findfourpointsthatareonthegraph of f .Whatcanbesaidaboutthecurvedeterminedbythose fourpoints? (19)Let f and g betwofunctionsandletusassumethatthereis exactlyonepoint( x,y )thatisonthegraphofboth f and g Whatisthealgebraicmeaningofthatfact? (20)Let f g ,and h bethreefunctionsandletusassumethatthere isnopointthatisonthegraphofallthreeofthem.Whatis thealgebraicmeaningofthatfact? 2.ClassesofFunctions2.1.PowerFunctions.A powerfunction isafunction f givenbythe rule f ( x )= xa,where a isa“xedrealnumber.Notethat xŠ a=1 /xa, so,forinstance, xŠ 3=1 /x3.Thespecialcaseof a = Š 1,thatis,the function f ( x )=1 /x ,iscalledthe reciprocalfunction .Notethatthe rule g ( x )=1forallrealnumbers x alsode“nesapowerfunction,one inwhich a =0.If a =1 /n ,where n isapositiveinteger,thenthe powerfunction f givenbytherule f ( x )= xa= x1 /n=n x isalsocalleda rootfunction .2.2.Polynomials.A polynomialfunction isthesumofa“nitenumber ofconstantmultiplesofpowerfunctionswithnonnegativeintegerexponents,suchasthefunction f givenbytherule f ( x )=3 x4+2 x2+ 7 x Š 5.Thedomainofthesefunctionsisthesetofallrealnumbers. Thelargestexponentthatispresentinapolynomialfunctioniscalled the degree ofthepolynomial.Sothedegreeof f inthelastexampleis 4.Therealnumbersthatmultiplythepowerfunctionsinapolynomial arecalledthe coecients ofthepolynomial.Inthelastexample,they are3,2,7,and Š 5. Somesubclassesofpolynomialfunctionshavetheirownnamesas follows:

PAGE 11

2.CLASSESOFFUNCTIONS5 € Polynomialsofdegree0,suchas f ( x )=6,arecalled constant functions. € Polynomialsofdegree1,suchas g ( x )=3 x Š 2,arecalled linearfunctions. € Polynomialsofdegree2,suchas h ( x )= x2Š 4 x Š 21,are called quadraticfunctions. € Polynomialsofdegree3,suchas p ( x )= x3Š x2+6 x Š 2,are called cubicfunctions.2.3.RationalFunctions.A rationalfunction istheratiooftwopolynomialfunctionssuchas R ( x )= 3 x2+4 x Š 7 x3Š 8 Thedomainofarationalfunctionisthesetofallrealnumbers,except forthenumbersthatmakethepolynomialinthedenominator0.In theprecedingexample,theonlysuchrealnumberis x =2.2.4.TrigonometricFunctions.Periodicity.Thereaderhassurelyencounteredthetrigonometricfunctionssin,cos,tan,cot,sec,andcscin earliercourses.Wewilldiscussthesefunctions,andtheirinverses,later inthetext.Fornow,wementiononeoftheirinterestingproperties, their periodicity .Afunction f iscalled periodic withperiod T> 0if f ( x )= f ( x + T )forall x and T isthesmallestpositiverealnumber withthisproperty. Forexample,sinandcosarebothperiodicwithperiod2 ,andtan andcotareperiodicwithperiod .SeeFigure1.3foranillustration. ThereaderwillbeaskedinExercise2.7.1abouttheperiodicityofsec andcsc.2.5.AlgebraicFunctions.An algebraicfunction isafunctionthatcontainsonlyaddition,subtraction,multiplication,division,andtaking roots.Forinstance,powerfunctionswithintegerexponentsarealgebraicfunctions,sincetheyonlyusemultiplication,thoughpossibly manytimes.Therefore,polynomialsarealgebraicfunctionsaswellsince theyaresumsofconstantmultiplesofpowerfunctions.Thisimplies thatrationalfunctionsarealsoalgebraicsincetheyareobtainedby dividingapolynomial(alsoanalgebraicfunction)byanotherone. Theprecedinglistdidnotcontainallalgebraicfunctionssinceit didnotcontainanyfunctionsinwhichrootswereinvolved.Soweget additionalexamplesifweincluderoots,suchasthefunctionsgivenby therules f ( x )= x +3, g ( x )=3 x h ( x )= ( x +1) / ( x Š 1).

PAGE 12

61.FUNCTIONS 23222322 1 0.5 0.5 1sin x 2322232210.5 0.5 1cos x 232223221 csc x 232223221 sec x 23222322 1 1 tan x 23222322 1 1 cot x Figure1.3. Trigonometricfunctions.2.6.TranscendentalFunctions.Functionsthatarenotalgebraicare called transcendentalfunctions .Theseincludetrigonometricfunctions andtheirinverses,exponentialfunctions,whicharefunctionsthatcontainavariableintheexponent,suchas f ( x )=2x,andtheirinverses, whicharecalled logarithmicfunctions .SeeFigure1.4foranillustration.Wewilldiscussthesefunctionsinlatersectionsofthischapter. Therearemanyadditionalexamples,whichdonothavetheirown names.2.7.Exercises.(1)Aresecantandcosecantperiodicfunctions?Ifyes,whatis theirperiod? (2)Canapolynomialbeaperiodicfunction? (3)Are f ( x )=3 x5+7 x Š 31and g ( x )=(2 x +7) / (3 x Š 1) polynomialfunctions? (4)Are f ( x )=2xand g ( x )=sin2x powerfunctions?

PAGE 13

2.CLASSESOFFUNCTIONS7 1 12 23 34 45 5 x 1 1 2 2 3 3 4 4 5 5 y x log2( x ) 2x Figure1.4. Logarithmicfunctions. (5)Are1 / ( x +3), g ( x )=( x2+3 x +9) / ( x3+1),and h ( x )= (sin x ) / ( x +2)rationalfunctions? (6)Showanexampleofarationalfunctionthatisde“nedforall realnumbers. (7)Showanexampleofarationalfunctionthatisde“nedforall realnumbers except 1,2,and3. (8)Let f ( x )= x2 / 3.Is f analgebraicfunction? (9)Issin(3 x )aperiodicfunction?Ifyes,whatisitsperiod? (10)Iscos(1 /x )aperiodicfunction?Ifyes,whatisitsperiod? (11)Issin( | x | )aperiodicfunction?Ifyes,whatisitsperiod? (12)Showanexampleofaperiodicfunctionthathasperiod1. (13)Let f ( x )= xŠ 2 / 7.Is f analgebraicfunction? (14)Is g ( x )=(2 / 3)xanalgebraicfunction? (15)Showanexampleofaperiodicfunctionwithperiod (16)Issin x +tan x aperiodicfunction?Ifyes,whatisitsperiod? (17)Issin x tan x aperiodicfunction?Ifyes,whatisitsperiod? (18)Issin2x aperiodicfunction?Ifyes,whatisitsperiod? (19)Let x beequaltothelargestinteger y suchthat y x .Now set f ( x )= x Š x .Is f aperiodicfunction?Ifyes,whatis itsperiod? (20)Let f and g beperiodicfunctionsde“nedforallrealnumbers. Let f haveperiod5andlet g haveperiod7.Is f + g aperiodic function?Ifyes,whatisitsperiod?

PAGE 14

81.FUNCTIONS 3.OperationsonFunctions3.1.TransformationsofaFunction.Wehaveseenthebasicmathematicalfunctionsandtheirgraphsinthelastsection.Inthissection,we willlookattheirtransformations. Itiseasytoseewhathappenstothegraphofafunctionifwe increaseordecreaseeachvalueofafunctionbyaconstant.Indeed,the graphofthefunction g givenby g ( x )= f ( x )+5forall x issimplythe graphofthefunction f translatedby“veunitstothenorth.Similarly, thegraphofthefunction h givenby h ( x )= f ( x ) Š 7isthegraphof f translatedbysevenunitstothesouth. Horizontaltranslationsarealittlebittrickier.Thereaderisinvited toverifythatif g isthefunctiongivenby g ( x )= f ( x Š 2),thenthe graphof g isthegraphof f translatedbytwounitstotheeast,thatis, inthepositivedirection.Indeed,wemustsubstitutea larger number into g togetthesamevalueasfrom f .Forinstance, g (8)= f (6). SeeFigure1.5foranillustration.Similarly,if h isthefunctiongiven 2 1 0 1 2 3 4 x 1 2 3 4 5 6 7 8 9 1 0 y f x 2 f x 5 f x Figure1.5. Horizontalandverticaltranslationsof f ( x ).

PAGE 15

3.OPERATIONSONFUNCTIONS9 by h ( x )= f ( x +3)forall x ,thenthegraphof h isthegraphof f translatedbythreeunitstothewest,thatis,inthenegativedirection. Theinteractivewebsitehttp://www.math.u”.edu/ mathguy/ufcalc book/translations.htmlprovidesfurthertoolstovisualizetransformationsoffunctions. Theeectofmultiplicationanddivisiononfunctionscanbedescribedsimilarly.If f isafunctionand g isthefunctiongivenby g ( x )= c f ( x ),where c> 1isarealnumber,thenthegraphof g issimplythegraphof f stretchedŽverticallybyafactorof c .That is,eachpointonthegraphof g is c timesasfarawayfromthehorizontalaxisasthecorrespondingpointonthegraphof f .Itgoes withoutsayingthatdividingby c> 1hastheoppositeeect.Thatis, if h ( x )= f ( x ) /c ,thenthegraphof h isaverticallycompressedversion ofthegraphof f .Inotherwords,eachpointonthegraphof h is c timesasclosetothehorizontalaxisasthecorrespondingpointonthe graphof f .SeeFigure1.6foranillustration. Atthispoint,thereadershouldstopandthinkaboutwhathappensif c< Š 1isanegativeconstant.Asthereaderprobably“gured out,thestretchingorcompressingeectwillnotchange(itwillonly dependon | c | ),buteachpointonthegraphwillbere”ectedthrough thehorizontalaxis. 12345 x 2 1 1 2 3 y f ( x ) 3 f ( x ) 1 2 f ( x ) f ( x ) Figure1.6. Eectsofmultiplyingafunctionbyaconstant. Thereaderisencouragedtoconsulttheinteractivewebsitehttp:// www.math.u”.edu/ mathguy/ufcalcbook/squeeze.htmlforfurther illustrations. Horizontaltransformationsinvolvingmultiplicationanddivisionare similartotheircounterpartsinvolvingadditionandsubtractioninthat theireectistheoppositeofwhatonemightthinkat“rst.If c> 1

PAGE 16

101.FUNCTIONS and g isthefunctionobtainedfrom f bytherule g ( x )= f ( cx ),then thegraphof g isthegraphof f compressedhorizontallybyafactor of c .Thatis,eachpointonthegraphof g is c timesasclosetothe verticallineasthecorrespondingpointonthegraphof f .Inother words,if( x,y )isapointonthegraphof f ,then( x/c,y )isapointon thegraphof g .Ontheotherhand,if h isobtainedfrom g bytherule h ( x )= f ( x/c ),thenthegraphof h isahorizontallystretchedversion ofthegraphof f .Thatis,eachpointonthegraphof h is c timesas farfromtheverticalaxisasthecorrespondingpointonthegraphof f .Soif( x,y )isapointonthegraphof f ,then( cx,y )isapointon thegraphof h .Again,thereadershouldstopforaminuteandthink aboutthegraphsofthefunctions f ( cx )and f ( x/c )when c< Š 1isa negativeconstant.3.2.CombiningTwoFunctions.If f and g aretwofunctions,thentheir sum,dierence,andproductarede“nedwhereverboth f and g are de“ned.Thatis,thedomainof f + g f Š g ,and fg istheintersection ofthedomainsof f and g .Furthermore,( f + g )( x )= f ( x )+ g ( x ), ( f Š g )( x )= f ( x ) Š g ( x ),and( fg )( x )= f ( x ) g ( x ).Figure1.7illustrates thesumoftwofunctions.Wehavetobejustalittlebitmorecareful with f/g ,sincethisfunctionisnotde“nedwhen g ( x )=0,evenif x isinthedomainofboth f and g .Sothedomainof f/g isthe intersectionofthedomainof f andthedomainof g ,withtheexception ofthepoints x satisfying g ( x )=0.Foreachpointofthisdomain, ( f/g )( x )= f( x ) /g ( x ). 12345 x 1 1 2 3 y f ( x ) g ( x ) g ( x ) f ( x ) Figure1.7. Addingtwofunctionstogether. Iftherangeof f ispartofthedomainof g ,thenwecan compose f and g by“rstapplying f andthen g .Thefunctionweobtaininthis

PAGE 17

3.OPERATIONSONFUNCTIONS11 waysends x to g ( f ( x ))andiscalledthe composition of f and g .Itis denotedby f g .Notethatin f g ,“rst f ,andthen g isapplied. Example 1.1 Let R bethesetofallrealnumbers.If f and g are bothfunctionsfrom R to R and f ( x )= x2and g ( x )= x +1 ,then ( f g )( x )= g ( f ( x ))= x2+1 while ( g f )( x )= f ( g ( x ))=( x +1)2= x2+2 x +1 Notethat f g and g f are,ingeneral,dierentfunctions.3.3.Exercises.(1)Sketchthegraphof f ( x )= x2, g ( x )=( x Š 3)2,and h ( x )= (2 x +5)2. (2)Sketchthegraphof f ( x )=( x +4)2and g ( x )= x2+4. (3)Sketchthegraphof f ( x )= | x +5 | and g ( x )= | x | +5. (4)Sketchthegraphof f ( x )= | x | +1and g ( x )= | x +1 | (5)Sketchthegraphof f ( x )=sin( x/ 2)and g ( x )=(sin x ) / 2. (6)Sketchthegraphof f ( x )= | sin x | (7)Sketchthegraphof f ( x )= x and g ( x )=1 / x (8)Sketchthegraphof f ( x )= ex Š 3and g ( x )= exŠ 3. (9)Sketchthegraphof f ( x )=ln( x2)and g ( x )=(ln x )2. (10)Sketchthegraphof f ( x )=cos( x + )and g ( x )= +cos x (11)Sketchthegraphof f ( x )= x +10and g ( x )= x +10. (12)Sketchthegraphof f ( x )=cos(2 x ), g ( x )=sin( x Š 2),and h ( x )=3tan x (13)Sketchthegraphof f ( x )=2cos x g ( x )=2(sin x ) Š 2,and h ( x )=tan(3 x ). (14)Showexamplesfor f and g when g f isde“nedforallreal numbers,but f g isnot. (15)Showexampleswhen f g = g f (16)Showexampleswhen( f g ) h = f ( g h ). (17)Showexampleswhen( f g ) h = f ( g h ). (18)Sketchthegraphof g ( x )=sin x Š 4 (19)Let f ( x )=sin x and g ( x )= x2.Determine f g and g f and sketchtheirgraph. (20)Let f ( x )=cos x ,let g ( x )= ex,andlet h ( x )=ln x .Determine f ( g h )and( f g ) h .

PAGE 18

121.FUNCTIONS 4.ViewingtheGraphsofFunctions4.1.TheGraphofFunction.The graph ofafunction f istheset { ( x,f ( x )) | x D ( f ) } Thegraphofafunctionisagoodwayofvisuallydescribingwhata functiondoes.Today,wehaveplentyofadvancedtools,suchascomputersoftwarepackagesandgraphingcalculators,tostudythegraphof functions.Inthissection,wepointoutafewofthecommonmistakes inusingthesetools. Inordertofacilitatethediscussion,letusagreeonsometerminology.Ifthedomainof f containsaninterval I andforallrealnumbers x and xin I ,itistruethat xf ( x),thenwesaythat f is decreasing on I .Intermsofthe graphof f ,thismeansthatthegraphgoesroughlyfromthenorthwest tothesoutheast. Ifwesimplyaskacomputerorgraphingcalculatortoplotthegraph ofafunctionwithoutspecifyingtheinterval[ x1,x2]inwhichthevalue of x canrange,wemaygetanerrormessage,orthecomputermay simplysubstitutedefaultvaluesfor x1and x2.Forexample,thesoftwarepackageMaple13usesthedefaultvalues x1= Š 10and x2=10. Theinterval[ x1,x2]isoftencalledthe viewingwindow .SeeFigure1.8 foranillustration. 10 5510 2000 2000 4000 0.2 0.2 0.4 0.6 0.8 1.0 0.2 0.4 0.6 Figure1.8. Viewing g ( x )=4 x3+9 x2+6 x +1with viewingwindow[ Š 10,10]and[ Š 1,0]. Wehavetobecareful,however,sincenotallviewingwindowsare appropriateforallfunctions,andchoosinganinappropriateviewing windowmaycausemisleadingresults.

PAGE 19

4.VIEWINGTHEGRAPHSOFFUNCTIONS13 Forfunctionslike f ( x )= x g ( x )= | x | ,or h ( x )= x2+3,theviewingwindow[ Š 10 10]isappropriateasthebehaviorofthesefunctions outsidethatwindowissimilartotheirbehaviorinsidethewindow. Nowlet f ( x )=( x +10)2.Inthiscase,usingtheviewingwindow [ Š 10 10],wegetthegraphofanincreasingfunction.Thatismisleading since f isdecreasingontheinterval( Š Š 10].So,inthiscase,a viewingwindowthatstartsatapoint x1< Š 10isnecessary. Thisproblembecomesmoredicultifwearedealingwithfunctions thatchangefromincreasingtodecreasingmanytimes,perhapsinan irregularfashionandperhapsfarawayfromtheorigin.Forthisreason, itisworthnotingthatif f isa polynomialfunction ofdegree n ,thenit cannotchangedirectionsmorethan n Š 1times.Ifwefoundall n Š 1 directionchanges,thenwecanbesurethatwedidnotmissanyof them.Wewillreturntothistopicinalaterchapter,whenwediscuss the derivative ofafunction. Theprecedingexampleshowedwhyselectingaviewingwindow thatistoosmallcanbemisleading.Thenextexampleshowswhya viewingwindowthatistoolargecanalsomisleadus.Plotthegraph ofthefunction g ( x )=4 x3+9 x2+6 x +1.Usingthedefaultviewing window[ Š 10 10],orsomewindowcontainingthatone,manysoftware packageswillshowagraphthatincreaseseverywhereanddisappearsin asmallintervaltotheleftof0.Thisshouldraiseoursuspicionthatthe programdoesnotproperlydisplaythegraphof g around0.Indeed, g isde“nedforallrealnumbers,soitsgraphshouldnotdisappear anywhere.Takingacloserlook,thatis,changingtheviewingwindow to[ Š 1 1],weseeafunctionthatisactually decreasing between x = Š 1 and x = Š 1 / 2. Trigonometricfunctions,withtheirperiodicity,areparticularly goodexamplestodemonstratewhatsoftwarepackagescanandcannotdo.Thereaderisencouragedtoplotthegraphofthefunctions sin x ,cos(2 x ),tan( x/ 4),and,“nally,sin(1 /x )andexplaintheobtained graphs.Inparticular,thereadershouldtrytoexplainwhy,forsin(1 /x ), thechoiceoftheviewingwindowisnotimportantaslongasitcontains x =0. Applicationsofgraphicalrepresentationsoffunctionsincludecountingthesolutionsofcertainequationsevenwhenwecannotexplicitly solvethoseequations,and“ndingasymptotes.(Notethatasolution obtainedbysimplyviewingthegraphicalrepresentationofafunction isnotmathematicallyrigorous,butitcanprovidea“rststepfora morerigoroussolution.)A horizontalasymptote ofafunction f isa horizontalline y = a sothatthevaluesof f ( x )areneverequalto a ,

PAGE 20

141.FUNCTIONS butgetcloserandclosertoitas x getscloserandclosertopositive in“nityornegativein“nity.A verticalasymptote of f isaline x = b sothatthefunction f isnotde“nedat x = b ,butas x getscloserand closerto b ,thevaluesof f ( x )getcloserandclosertoin“nity,ornegativein“nity.Forinstance,thefunction f ( x )=1 /x hasahorizontal asymptoteat y =0,andaverticalasymptoteat x =0.Wewillmake thesenotionsmorepreciseinthenextchapter,whenweintroducethe conceptof limits .Fornow,wecanuseagraphingsoftwarepackageto “ndasymptotes,asyouwillbeaskedtodointheexercises.4.2.Exercises.Inthefollowingexercises,useagraphingsoftware packagewiththeappropriateviewingwindowto“ndthenumberof solutions(amongrealnumbers)forthegivenequation.Also“ndthe intervalsonwhichtheleft-handsideisincreasingandonwhichthelefthandsideisdecreasing.Approximatetheendpointsoftheseintervals toonedecimal. (1) x4Š x +1=0. (2) x4Š 1=0. (3) x3Š 6 x +1=0. (4) x3+ x2Š 1=0. (5) x4Š 4 x2+1. Inthefollowingexercises,useagraphingsoftwarepackagewiththe appropriateviewingwindowto“ndthenumberofsolutions(among realnumbers)forthegivenequation. (6) x3Š x2Š 1=0. (7) x2Š x Š 7= x3Š 1. (8) x =sin x (9) x2=sin x (10) x/ 2=cos x (11) x +2=2x. Inthefollowingexercises,useagraphingsoftwarepackagetodecideif thegivenfunctionhasaverticalorhorizontalasymptote. (12) f ( x )=( x +3) / ( x +2). (13) f ( x )=( x Š 7) / ( x +5). (14) g ( x )=1 / (2 Š x ). (15) h ( x )= x +(1 /x ). (16) h ( x )= x2Š (1 /x2). (17) s ( x )= ( x Š 4) / ( x Š 3). (18) s ( x )= ( x +5) / ( x Š 7). (19) z ( x )=( x2+1) / (2 x2Š 3). (20) z ( x )=( x2+4 x +5) / ( x +3).

PAGE 21

5.INVERSEFUNCTIONS15 5.InverseFunctions Theinverse fŠ 1ofafunction f : A B undoesŽwhat f did. Thatis,if f ( x )= y ,then fŠ 1( y )= x ,so f sends x to y ,while fŠ 1sends y backto x .Itgoeswithoutsayingthatthis fŠ 1willonlybe afunctionif fŠ 1( y )isunambiguous,thatis,whenthereisonlyone x A sothat f ( x )= y .Inthatcase,andonlyinthatcase,itisclear that fŠ 1( y )= x Letusnowformalizetheseconcepts. Definition 1.1 Afunction f : A B iscalled one-to-one if itsendsdierentelementsintodierentelements,thatis,if x = ximpliesthat f ( x ) = f ( x) One-to-onefunctionsarealsocalled injectivefunctions or injections Visually,nohorizontallinecanintersectthegraphofaone-to-one functionmorethanonce. Forinstance,if A and B areboththesetofrealnumbers,then f ( x )= x and g ( x )= x3arebothone-to-one,but h ( x )= x2isnot. Definition 1.2 Let f beaone-to-onefunctionwithdomain A and range B .Thenthe inverse of f isthefunction fŠ 1: B A givenby fŠ 1( y )= x if f ( x )= y Example 1.2 Let A and B bothbethesetofallrealnumbers.Let f : A B begivenby f ( x )=2 x +7 .Then fŠ 1( y )=( y Š 7) / 2 Solution: If f ( x )= y ,then y =2 x +7,so y Š 7=2 x andso ( y Š 7) / 2= x .As x = fŠ 1( y ),itfollowsthat fŠ 1( y )=( y Š 7) / 2. Theprecedingexampleshowsageneralstrategyfor“ndingtheinverse ofafunction.Writetheequation f ( x )= y ,withtheappropriate algebraicexpressionreplacing f ( x ).Thensolvefor x .Ifthereismore thanonesolution,then f isnotone-to-one,andsoithasnoinverse function.Ifthereisonesolution,thenthatexpressionisthevalueof fŠ 1( y ). Example 1.3 If A isthesetofpositiverealnumbers, B isthe setofrealnumbersthatarelargerthan 1 ,and f : A B isgivenby f ( x )= x2+1 ,then fŠ 1( y )= y Š 1 Solution: Wehave f ( x )= x2+1= y .So x2= y Š 1,andbecausewe knowthat x ispositiveand y> 1,wecantakethesquarerootofboth sides,leadingto x = y Š 1.Hence, fŠ 1( y )= y Š 1. Notethatthegraphsof f and fŠ 1arere”ectedimagesofeachother throughtheline y = x asillustratedinFigure1.9.

PAGE 22

161.FUNCTIONS Finally,wepointoutthatif f isaone-to-onefunctionwithdomain A andrange B ,then f fŠ 1istheidentityfunctionof A and fŠ 1 f istheidentityfunctionof B 0123 x 1 2 3 y x f 1( x ) f ( x ) Figure1.9. f ( x )and fŠ 1( x )aresymmetricaboutthe identityfunction x Forinstance,usingthefunctionsofExample1.3,forallpositive realnumbers x ,theidentity( f fŠ 1)( x )= ( x2+1) Š 1= x2= x holds,andforall y> 1,theidentity( fŠ 1 f )( y )=( y Š 1)2+1= y Š 1+1= y holds.5.1.LogarithmicFunctions.Ifafunctioncontainsonlyadditions,subtractions,multiplications,anddivisions,thenitsinverseisofteneasy tocompute.Powerfunctions,thatis,functionsoftheform f ( x )= x, where isarealnumber,arenotmuchmoredicult.However,what istheinverseofan exponentialfunction ? Let f ( x )=2x.Itiseasytosee,byplottingthegraphof f or otherwise,that f isaone-to-onefunctionwhosedomainisthesetof allrealnumbersandwhoserangeisthesetofallpositiverealnumbers. Sotheinverseof f isafunctionfromthesetofpositiverealstotheset ofallreals.But what isthatinversefunction fŠ 1?Bythede“nition ofinversefunctionsingeneral,thisisthefunctionthatsends2xto x forallpositiverealnumbers2x.Inparticular, fŠ 1(2)=1, fŠ 1(4)=2, fŠ 1(32)=5,and fŠ 1(1 / 2)= Š 1.Thatis, fŠ 1( y )tellsus towhatpower wehavetoraise2 iftheresultistobe y .Thisimportantconcepthas itsownname. Definition 1.3 Let m beapositiverealnumber.Thentheinverse ofthefunction f ( x )= mxiscalledthe logarithmicfunctionwithbase m ,andisdenotedby logm.

PAGE 23

5.INVERSEFUNCTIONS17 Soif f ( x )= xm= y ,thenlogm( y )= x .Forinstance,log2(64)=6, log3(81)=4,log5(1 / 25)= Š 2,andlog0 5(16)= Š 4. Logarithmicfunctionssatisfycertainrulesthatareverysimilarto thosesatis“edbyexponentialfunctionsandcan,infact,bededuced fromthem.Theseare (I)log( xy )=log x +log y (II)log( x/y )=log x Š log y (III)log( xa)= a log x (IV)logb x =log x b. (V) alogax= x (VI)loga( ax)= x Thelasttworulessimplyexpressthefactthatthefunctions f ( x )= axand fŠ 1( y )=loga( y )areinversesofeachother,sotheircomposition isanidentityfunction. Ifweknowthelogarithmofanumberinabaseandwanttocompute itinanotherbase,wecandosousingthefollowingtheorem. Theorem 1.1 Forpositiverealnumbers a b ,and x ,wehave logax = logbx logba Proof. Startwiththeidentity x = alogax. Nowtakethelogarithmofbase b ofbothsidestoget logbx =logax logba. Nowdividebothsidesbylogba togettheidentityofthetheorem. Example 1.4 WecanuseTheorem1.1tocompute log16(256) from log2(256) asfollows: log16(256)= log2(256) log2(16) = 8 4 =2 Soifacalculatororcomputercanprovidethelogarithmofall positiverealnumbersin one base,itcancomputethelogarithmofany positiverealnumberinanybase.Forthisreason,manycalculatorsand computersareprogrammedtoworkprimarilywithlogarithmsofone givenbase,namelyofbase e ,where e 2 718isanirrationalnumber thatwillbeformallyde“nedinChapter2. Thelogarithmofbase e issoimportantthatithasitsownname, naturallogarithm ,anditsownnotation,ln.Soln x =logex .

PAGE 24

181.FUNCTIONS 5.2.InversesofTrigonometricFunctions.Basictrigonometricfunctions, suchassin,cos,andtan,areveryimportantincalculus,soitisnosurprisethattheirinversefunctionsareimportantaswell.However,we havetobeprecisewhenwede“nethemsincetrigonometricfunctions are not one-to-one.Infact,theyareperiodical,ofperiod2 or ,and sotheytakeeveryvalueintheirrangein“nitelyoften. Inordertogetaroundthisdiculty,wewillrestrictourtrigonometricfunctionstojustashortinterval,inwhichtheyareone-to-one, andde“netheirinversesbasedonthatrestriction. Forinstance,considersinasafunctionwhosedomainis[ Š / 2 ,/ 2]. Inthatinterval,sinisaone-to-onefunction(sinceitisincreasing), anditsrangeistheinterval[ Š 1 1].SeeFigure1.10foranillustration. Sotheinverseofsin:[ Š / 2 ,/ 2] [ Š 1 1]isthefunctionsinŠ 1: [ Š 1 1] [ Š / 2 ,/ 2].Thatis,if y [ Š 1 1],thensinŠ 1y isthe(only) x [ Š / 2 ,/ 2]forwhichsin x = y .Forinstance,sinŠ 1(1 / 2)= / 6, whilesinŠ 1(0)=0andsinŠ 1( 2 / 2)= / 4.Figure1.11showsthe graphofsinŠ 1x Theinversesoftheothertrigonometricfunctionsarede“nedsimilarly,justtheintervalstowhichwerestrictthefunctions(inorderto makethemone-to-one)canchange. Thatis,cosŠ 1istheinversefunctionofthecosfunctionthatis restrictedtotheinterval[0 ].SocosŠ 1isafunctionwithdomain [ Š 1 1]andrange[0 ].Similarly,tanŠ 1istheinversefunctionofthe tanfunctionthatisrestrictedtotheinterval( Š / 2 ,/ 2).Itsdomain isthesetofallrealnumbers,anditsrangeistheinterval( Š / 2 ,/ 2). SeeFigure1.12forillustrations. Theinversefunctionsofcot,sec,andcsc,whilenotusedoften,can alsobede“nedanalogously.5.3.Exercises.(1)Isthereafunction f de“nedonallpositiverealnumbersfor which fŠ 1= f ? 2 32 22322 x 1 1 y Figure1.10. sin x isone-to-oneontheinterval[ Š / 2 ,/ 2].

PAGE 25

5.INVERSEFUNCTIONS19 1 1 x 2 2 y Figure1.11. GraphofsinŠ 1x 2 2 x 1 1 ycos x 1 1 x 2 ycos1x 22 x 5 10 5 10 ytan x 5 5 x 22ytan1x Figure1.12. Graphsofcos x andtan x withtheirinverses.

PAGE 26

201.FUNCTIONS (2)Ifwearegivenlogax ,howcanwecomputelog1 /ax ? (3)Forwhichvaluesof a islogaanincreasingfunction,andfor whichvaluesof a isitadecreasingfunction? (4)Whatisthegeometricconnectionbetweenthegraphsof f and fŠ 1? (5)WhatisthevalueoftanŠ 11? (6)Findallrealnumbers y suchthattanŠ 1y =cotŠ 1y (7)Isittruethatif g istheinversefunctionoftheone-to-one function f ,then g isone-to-one? (8)Let f : R R bede“nedby f ( x )= | x | .Is f aone-to-one function? (9)Let f : R R bede“nedby f ( x )= x5.Is f aone-to-one function? (10)Let f : R+ R+bede“nedby f ( x )= x2.Is f aone-to-one function? (11)Is f ( x )=logax aone-to-onefunctiononthesetofallpositive realnumbers? (12)Express x intermsof y ifloga(logax )= y (13)Letusassumethat f : R R isastrictlyincreasingfunction, thatis,if x
PAGE 27

6.THEVELOCITYPROBLEMANDTHETANGENTPROBLEM21 6.TheVelocityProblemandtheTangentProblem6.1.TheVelocityProblem.Letusassumethatacarwasontheroad from3:00p.m.to5:00p.m.onagivenafternoon,andittraveleda distanceof100miles,allduewest.Fromthedata,itiseasytocompute theaveragespeedofthecarbytheformula (1.1) v = s t where t isthetimepassed, s isthedistancecoveredintime t ,and v istheaveragespeedforthegiventimeperiod.Inphysics,whenthe directioninwhichanobjectismovingistakenintoaccount,wetalk about velocity insteadof speed ,hencetheabbreviation v .Inthegiven example,alltravelwasinonedirection(west),sothereisnodanger ofconfusion,andwecanuseeitherword.Letusassumethattimeis measuredinhoursanddistanceismeasureinmiles. ThenEquation(1.1)yields v = 100mi 2hr =50 mi hr sotheaveragevelocityofthecarforthegiven2-hourperiodis50miles perhour. Thecarprobablydidnotcovertheentiredistanceatitsaverage velocity.Forvarioustrac-relatedorotherreasons,itsometimesmay havegonefasterorslower.Ifwewanttoknowitsaveragevelocity forthetimeperiodbetween4:00p.m.and4:10p.m.,thenweneed knowthedistanceitcoveredinthattimeperiod.Ifthatdistanceis 10miles,thenweconcludethatinthat10-minutetimeperiod,the averagevelocityofthecarwas v = 10mi 1 / 6hr =60 mi hr Ifwewantmorepreciseinformation,liketheaveragevelocityof thecarbetween4:02p.m.and4:05p.m.,wecanproceedsimilarly, decreasingthevalueofboththenumeratorandthedenominatorof thefraction s/t .However,whatifwewanttoknowthe instantaneous velocity ofthecarinagiven moment ,suchasexactlyat4:02:23p.m. (andnotinthesecondthatpassedbetween4:02:23p.m.and4:02:24 p.m.)?Inthatcase,adirectapplicationofEquation(1.1)isimpossible, becausethedenominator t isequalto0.Thenumerator s isalsoequal to0,sincethecarneedstimetocoveranydistance;ifitisgivenno time,itwillcovernodistance. Inthissection,wewillnotgiveacompletelyformalanswertothe problemofde“ninginstantaneousvelocity;wewillleavethattaskto

PAGE 28

221.FUNCTIONS anupcomingsection.However,wewillsaythefollowing.Theinstantaneousvelocityofacarinagivenmoment m canbeapproximatedby choosingsmallerandsmallertimeperiodscontaining m andcomputing theaveragespeedofthecarforthosetimeperiods.Theseaverageswill approximatetheinstantaneousvelocity.6.2.TheTangentProblem.Theproblemof“ndingtheinstantaneous velocityofamovingobjectissimplyaspecialcaseofamuchmore generalproblem,thatof“ndingtheslopeofatangentlinetoacurve atagivenpoint. Inthepreviousproblem,thedistancethecarcoveredcanbeviewed asafunctionofthetimethatpassedsincethecarstartedmoving.So s ( t )isthedistancecoveredfromthemomentwhenthecarstarted movingtothemoment t hourslater.Inordertocomputetheaverage velocityforthetimeperiodfrom t1to t2,wesimplycomputethevalue ofthefraction s ( t2) Š s ( t1) t2Š t1. Thisfractionispreciselytheslopeofthelinethatintersectsthegraph ofthefunction s atpoints( t1,s ( t1))and( t2,s ( t2)).Ifwechoose t1and t2closerandclosertogether,thenthesepointswillgetcloserand closertogetheraswell.Finally,ifweset t1= t2,thenwewillnot immediatelyknowtheslopeofthelinethattouchesthegraphof s at thepoint( t1,s ( t1))sincewewillknowonlyone,nottwo,pointofthis line.However,andthiswillbemademorepreciseinthenextsection, theslopewearelookingforwillbeapproximatedbythesequenceof slopesofthelinesthatwegotwhenwechose t1and t2closerandcloser together. Finally,wepointoutthatthereisnothingmagicalaboutthefunction s ( t )here.Wecouldconsideranyfunction f : R R ,andask whattheslopeofthetangentlinetothiscurveisatthepoint( x,f ( x )).6.3.Exercises.(1)Acartravels1hourataspeedof60milesperhour,then2 hoursataspeedof45milesperhour.Whatistheaverage speedofthecarduringthis3-hourperiod? (2)Considerthecarofthepreviousexercise.Whatisitsaverage speedduringthe“rst2hoursofitstrip? (3)Idroveat40milesperhourfor2hours.HowfastdoIhave todriveinmythirdhourifIwanttoreachanaveragespeed of45milesperhourformy3-hourdrive?

PAGE 29

6.THEVELOCITYPROBLEMANDTHETANGENTPROBLEM23 (4)Acartravels300milesonagivenday.Duringthe“rst100 miles,thecartravelsataspeedof40milesperhour,during thesecond100miles,ittravelsataspeedof50milesperhour, andduringthethird100miles,ittravelsataspeedof60miles perhour.Whatistheaveragespeedofthecarfortheentire 300-miletrip? (5)Timhasriddenhisbicycletoschool,coveringa5-miledistance inhalfanhour.Canweconcludethattherewasasegmentof hisrideforwhichhisaveragespeedwasmorethan10miles perhour? (6)Jimhasdrivenhiscarfor3daysindierentconditions.On the“rstday,hewasabletodrive20milespergallonoffuel used,foratotalof200miles.Onthesecondday,hedrove25 milespergallonoffuelused,foratotalof275miles.Onthe thirdday,hedrove24milespergallonoffuelused,foratotal of240miles.Whatwashisaveragenumberofmilesdriven pergallonoffuelfortheentire3-daytrip? (7)Aballisthrownverticallyintheair.In t seconds,itsheight (inmeters)isgivenbythefunction h ( t )=50 t Š 20 t2.What istheaveragevelocityoftheballduringits“rst2secondsof motion? (8)Considertheballofthepreviousexercise.Whatistheaverage velocityoftheballbetween t =1and t =2? (9)Considertheballofthepreviousexercise.Whatistheaverage velocityoftheballbetween t =1 4and t =1 6? (10)Giveareasonableestimateofthevelocityoftheballofthe previousexerciseinthemoment t =1 5. (11)Considerthefunction f ( x )= x2.Canyou“ndtwopoints P and Q onthegraphof f suchthattheslopeoftheline PQ is between0and0.01? (12)Let f ( x )= x andlet P =(1 1).Findtheslopeofthe threelinesthatconnect P tothepoints(4 2),(2 25 1 5),and (1 44 1 2). (13)Let f beasinthepreviousexercise.Findtheslopeofthethree linesconnecting P =(1 1)tothepoints(0 25 0 5),(0 64 0 8), and(0 81 0 9). (14)Considertheresultsofthetwoprecedingexercises.Doyou seeatrend? (15)Let g ( x )= exandlet P =(0 1).Findtheslopeofthethree linesconnecting P tothepoints( Š 1 ,eŠ 1),(1 ,e ),and(ln2 2).

PAGE 30

241.FUNCTIONS (16)Considerthefunction f ( x )= x2.Let P =(1 1).Canyou “ndapoint Q onthegraphof g suchthattheslopeoftheline PQ is2? (17)Considerthefunction g ( x )= x3.Let P =(1 1).Canyou“nd apoint Q onthegraphof g suchthattheslopeoftheline PQ isbetween1and1.01? (18)Considerthefunction f ( x )=1 /x .Choosetwopoints P and Q ofthegraphof f suchthat P = Q andthe x coordinatesof P and Q aresmallandpositive.Whatcanbesaidaboutthe slopeoftheline PQ ? (19)Considerthefunction f ( x )= ex.Canyou“ndtwopointson thegraphof f suchthattheslopeoftheline PQ is negative ? Explainyouranswer. (20)Considerfunction f ( x )=1 Š ln x .Canyou“ndtwopointson thegraphof f suchthattheslopeoftheline PQ is positive ? Explainyouranswer.

PAGE 31

CHAPTER2 LimitsandDerivatives 7.TheLimitofaFunction7.1.Two-SidedLimits.Considerthefunctiongivenbytherule f ( x )= 1 / (1+ x ).Letuscomputethevaluesof f ( x )forvariousrealnumbers x thatarecloseto0.We“ndthat € f (1)=1 / 2, € f (1 / 2)=2 / 3, € f (1 / 3)=3 / 4,and,ingeneral, € f (1 /n )= n/ ( n +1). Similarly,fornegativevaluesof x ,weget € f ( Š 1 / 2)=2, € f ( Š 1 / 3)=3 / 2, € f ( Š 1 / 4)=4 / 3,and,ingeneral, € f ( Š 1 /n )= n/ ( n Š 1). Whatweseeisthatif x getscloseto0(fromeitherside),then f ( x )getscloseto f (0)=1.Infact,wecanget f ( x )tobeasclose to f (0)=1aswewant;allweneedtodoistochoose xsuciently closeto0.Indeed,lookingatthepreviousexamples,weconcludethat if0
PAGE 32

262.LIMITSANDDERIVATIVES So,if f isthestartingexampleofthissection,thenlimx 0f ( x )=1. Notethatthede“nitionoflimx af ( x )requiresthat f ( x )stayclose to L when x iscloseto a regardlessofwhichof x or a islarger .That is, f ( x )hastobecloseto L if x isalittlebitlessthan a ,and f ( x )has tobecloseto L if x isalittlebitmorethan a ,though f ( x )doesnot havetobecloseto L if x = a Severalcommentsareinorder.First,limx ag ( x )doesnotalways exist. Example 2.1 Let g ( x )= 1 if 0 x 0 if x< 0 Thenthelimitof g at a =0 doesnotexist.Indeed,nomatterhow smallaninterval I wetakearoundthepoint a =0 ,thatinterval I willcontainsomepositiveandsomenegativerealnumbers.Hence,the valuesof g ( x ) willsometimesequal 1 andsometimesequal 0 for x I nomatterhowsmall I is .Thereisnonumber L suchthat 0 and 1 are arbitrarilyclosetoit„infactthereisnonumbersuchthatboth 0 and 1 arebothcloserthan 0 5 toit.So limx 0g ( x ) doesnotexist. Second,iflimx 0f ( x )exists,itis unique ;thatis, f cannothave twodierentlimitsatanygivenpoint a .Letusillustratethisusing theintroductoryexampleofthissection,thefunction f ( x )=1 / (1+ x ). Wehaveseenthatlimx 0f ( x )=1.Indeed,wesawthatthevalues of f ( x )cangetarbitrarilycloseto1iftherealnumbers x arechosen fromasuitablysmallintervalaround0.Atthispoint,onecouldaskthe followingquestion. If1satis“estherequirementstobethe limx 0f ( x ) whydoes1.0001not?Afterall,whatiscloseto1isalsocloseto1.0001. Inordertoanswerthisquestion,wemusthaveagoodunderstandingofthede“nitionoflimits.Thatde“nitionsaysthatiflimx 0f ( x )= L ,thenthevaluesof f ( x )willget arbitrarily closeto f (0)if x ischosen fromasuitablysmallintervalaround0.Thekeywordintheprevious sentenceis arbitrarily. While1.0001iscloseto1,itisnot arbitrarily closeto1;itisexactly0.0001 away.Andthatisaproblem,sincewe haveseenatthebeginningofthischapterthat,as x approaches0,the valuesof f ( x )willgetarbitrarilycloseto1.Inparticular,if x isclose enoughto0,then f ( x )willbecloserthan1 106to1,butthenitcannot alsobecloserthan1 106to1.0001. Ananalogousargumentshowsthatnofunctioncanhavetwodifferentlimitsatanyonepoint. Sometimesitcanhappenthat h isnotevende“nedin a ,but limx 0h ( x )stillexists.Notethatthefactthat h ( a )isnotde“ned

PAGE 33

7.THELIMITOFAFUNCTION27 321012345 1 2 3 4 5 6 7 8 Figure2.1. h ( x )=x2Š 9 x Š 3. isnotaproblemsincethede“nitionoflimitsspeci“callystatesthat x shouldnotbeequalto a anyway. Example 2.2 Let h ( x )=( x2Š 9) / ( x Š 3) .Then h isde“nedfor allrealnumbersexcept x =3 .Still, limx 3h ( x )=6 .Inparticular, limx 3h ( x ) exists. SeeFigure2.1foranillustration. Solution: If x =3,then f ( x )= x2Š 9 x Š 3 = ( x +3)( x Š 3) x Š 3 = x +3 Soifwewant f ( x )= x +3tobecloserto6thanagivendistance a thenallwehavetodoistochoose x suchthat | x Š 3 |
PAGE 34

282.LIMITSANDDERIVATIVES However,wehavenotyetlearnedthetechniquestorigorously provethis.Plottingthegraphofthefunctionorproducingmore numericaldatashouldnotbeconsideredasacompleteanswer,since, as x approaches0,eventually x andsin x willgetsosmallthatthe computerwillnolongermanipulatethem,ortheirratio,accurately. Figure2.2. Viewing(sin x ) /x ontheTI-89graphing calculatorwithviewingwindow[ Š 5 5] [ Š 0 5 1 5]. Finally,wepointoutthatinthede“nitionofthelimit,therequirementthat f ( x )getcloseto L andstay closeto L isimportant.Consider thefunction f ( x )=sin(1 /x )around x =0.As x approaches0,the valueof1 /x willincreaseveryfast,andsoitwillequalamultipleof manytimes.Allthosetimes, f (0)=0willhold,so f ( x )willbeas closeto0aspossible.However,limx 0f ( x )doesnotexist,since f ( x ) willtakeallothervaluesintheinterval[ Š 1 1]in“nitelyoftenaswell as x approaches0.Sothevalueof f ( x )will notstay arbitrarilyclose to0,nomatterhowclose x isto0.SeeFigure2.3foranillustration. 1.0 0.50.51.0 x 1.0 0.5 0.5 1.0 y Figure2.3. f ( x )=sin(1 /x ).

PAGE 35

7.THELIMITOFAFUNCTION29 7.2.ThePreciseDenitionofLimits.Itistimeforustogiveaprecise mathematicalde“nitionoflimits.Theadvantageofthisformalde“nitionisthatwecan“nallydoawaywiththewords arbitrarilyclose and sucientlyclose .Thepricetopayforthatisthatwehavetousemore notation. Definition 2.2 Let f beafunctionde“nedonsomeopeninterval thatcontainstherealnumber a ,withthepossibleexceptionof a itself. Thenwesaythatthelimitof f at a is L ,denotedby limx af ( x )= L if,forall > 0 ,thereexists > 0 suchthatif | x Š a | < ,then | f ( x ) Š L | < SeeFigure2.4foranillustration. a L 112345 5 5 10 15 Figure2.4. As x approaches a f ( x )approaches L Example 2.3 Wehave limx 02 x sin x =0 Solution: Let beanypositiverealnumber.Thenlet = / 2.We knowthat | sin x | 1forall x .Soif | x Š 0 | = | x | < = / 2,then | f ( x ) Š 0 | = | f ( x ) | = | 2 x sin x || 2 x | < 2 = ,asrequired. 7.3.One-SidedLimits.Therearefunctionsthatbehaveinacertain wayuptoapoint a ,andthenbehaveverydierentlyafterthat.We haveseensuchafunctioninExample2.1.Thefunction g ofthat examplesatis“ed g ( x )=0fornegativevaluesof x ,and g ( x )=1for positivevaluesof x .Wehaveseenthatlimx 0g ( x )doesnotexist, sincenorealnumber L isarbitrarilyclosetoboth0and1. Nevertheless,thereareweaker, one-sided notionsoflimitsthatare relevantinthisexample.

PAGE 36

302.LIMITSANDDERIVATIVES Definition 2.3 Let f : R R beafunctionandlet a beareal number.Wesaythatthe left-handlimit of f in a istherealnumber L ifthevaluesof f ( x ) getarbitrarilycloseto L andstayarbitrarilyclose to L when x issuitablycloseto a and x
a Thefactthat L istheright-handlimitof f in a isdenotedby limx a+f ( x )= L. Forinstance,if g isthefunctionde“nedinExample2.1,then limx 0+g ( x )=1 Indeed,ifwechoose x closeto0 butmorethan0 ,then g ( x )=1,so g ( x )isarbitrarilyclose(infact,equal)to1. Atthispoint,thereadershouldcomparethede“nitionsoflimit, left-handlimit,andright-handlimit.Thede“nitionoflimit(De“nition2.1)imposesthestrongestrequirementsonthevaluesof f .Indeed, thevaluesof f ( x )havetobecloseto L when x iscloseto a and xa .Thede“nitionsofthelefthandandright-handlimitsimposeweakerrequirementsinthateach de“nitiononlyrequiresthat f ( x )becloseto L when x isona given side of a andcloseto a Itthenfollows„andthereadershouldspendaminuteverifyingit„ thatiflimx af ( x )= L ,thenlimx aŠf ( x )= L andlimx a+f ( x )= L Conversely,ifboththeleft-handlimitandtheright-handlimitof f in a isequalto L ,thenthelimitof f in a existsandisequalto L Atthispoint,thereadershouldcheckhisorherunderstandingof thematerialbyconsideringthefunction h ( x )= x | x |

PAGE 37

7.THELIMITOFAFUNCTION31 as x approaches0anddecidingifthelimitslimx 0h ( x ),limx 0Šh ( x ), andlimx 0+h ( x ),exist.ItmayhelptoconsultFigure2.5. 2 112 2 1 1 2 Figure2.5. Graphof h ( x )= x/ | x | .7.4.InniteLimits.Inourde“nitionsoflimitsinthissection,thelimit L wasalwaysarealnumber.Inthissection,weextendthosede“nitions tothecasesof in“nite limits.If L = ,thenthevaluesof f haveto getarbitrarilycloseto ;thatis,theyhavetogetaslargeaswewant. Thisisthecontentofthefollowingde“nition. Definition 2.5 Let f : R R beafunction.Wesaythatthe limitof f in a is ifwecanget f ( x ) arbitrarilylargeandkeepit arbitrarilylargeifwechoose x suitablycloseto a withoutbeingequal to a Similarly,if g : R R isafunction,wesaythatthelimitof g in a is Š ifwecanmake g ( x ) anegativenumberwithanarbitrarilylarge absolutevalueandkeep g ( x ) thatwayifwechoose x suitablycloseto a withoutbeingequalto a Thefactthatthelimitof f in a is isdenotedby limx af ( x )= Example 2.4 Let f ( x )=1 /x2.Then limx 0f ( x )= Solution: Ifwewant f ( x )tobelargerthananarbitrarypositive realnumber N ,allweneedtodoistochoose x fromtheinterval ( Š 1 /N, 1 /N ).Then x2< 1 /N willhold,implyingthat f ( x )= 1 /x2>N Similarly,if g ( x )= Š 1 /x4,thenlimx 0g ( x )= Š .Notethat ifthelimitofafunctionatagivenpoint a is or Š ,then,as x approaches a ,thegraphofthefunctionwillapproachaverticalline intersectingthehorizontalaxisat x = a .Thisphenomenonisreferred tobysayingthat f hasa verticalasymptote at a .

PAGE 38

322.LIMITSANDDERIVATIVES 7.4.1.ThePreciseDenitionofInniteLimits.Theformalde“nitionof in“nitelimitsissimilartothatof“nitelimits.Thedierenceliesin thefactthatitisnotthesametobecloseto ortobeclosetoareal number. Definition 2.6 Let f : R R beafunction.Wesaythatthe limitof f in a is if,forallpositiverealnumbers N ,thereexists > 0 suchthatif | x Š a | < ,then f ( x ) >N Similarly,let g : R R beafunction.Wesaythatthelimitof g in a is Š ifforallnegativerealnumbers M ,thereexists > 0 such thatif | x Š a | < ,then g ( x ) a Example 2.5 Let f ( x )=1 /x .Then f isnotde“nedin0.Furthermore, limx 0Š= Š and limx 0+= .Asthetwoone-sided limitsaredierent, limx 0doesnotexist. Solution: Wecanmake f ( x )=1 xsmallerthananygivennegative number M bychoosing x fromtheinterval(1 /M, 0).Wecanmake x largerthananypositivenumber P bychoosing x fromtheinterval (0 ,P ). 7.5.Exercises.(1)Findlimx 3x +7. (2)Findlimx 13x2Š 10 x +7. (3)Findlimx 3 x2Š 4 x +3 x Š 3. (4)Findlimx 1 x2+ x Š 2 x Š 1. (5)Doeslimx 3 x2Š 4 x +7 x Š 3exist? (6)Findlimx 0cos x (7)Findlimx 0 x2 | x |. (8)Findlimx 1 x Š 1 x Š 1. (9)Findlimx 2 x3Š 8 x Š 2.

PAGE 39

8.LIMITLAWS33 (10)Findlimx Š 2 x3+8 x +2. (11)Let f ( x )= x beequaltothelargestintegerthatisatmost aslargeas x .So f (3 7)=3.Notethat f isoftencalledthe ”oorfunction or integerpartfunction .Findthevalues a for whichlimx af ( x )exists.If a issuchthat f hasnotwo-sided limitat a ,decideif f hasone-sidedlimitsat a (12)Let g ( x )= x beequaltothesmallestintegerthatisat leastaslargeas x .So g (3 7)=4.Notethat g isoftencalled the ceilingfunction .Findthevalues a forwhichlimx ag ( x ) exists.If a issuchthat g hasnotwo-sidedlimitat a ,decide if g hasone-sidedlimitsat a (13)Doeslimx / 2tan x exist? (14)Doeslimx 0Šcot x exist? (15)Doeslimx 0+cot x exist? (16)Doeslimx 0 1 | x |exist? (17)Giveanexampleofafunction f suchthatlimx 0Šf ( x )=0, andlimx 0+f ( x )= (18)Doeslimx 01 x3+1 x2 exist? (19)Doeslimx 01 x4+1 x2 exist? (20)Giveanexampleofafunction f suchthatlimx 1Šf ( x )= limx 1+f ( x )= Š ,and f (1)isarealnumber. 8.LimitLaws8.1.BasicLimitLaws.If f and g aretwofunctionsandweknowthe limitofeachofthematagivenpoint a ,thenwecaneasilycompute thelimitat a oftheirsum,dierence,product,constantmultiple,and quotient.Therulesthatprovidethislimitaregivenbelow,andthey areverysimilartothewaysinwhichthesum,dierence,product, constantmultiple,andquotientoftwofunctionsarede“ned.Indeed, (I) limx a( f + g )( x )=limx af ( x )+limx ag ( x ) (II) limx a( f Š g )( x )=limx af ( x ) Š limx ag ( x ) (III) limx a( f g )( x )=limx af ( x ) limx ag ( x ) (IV) limx a( c f )( x )= c limx af ( x ) where c isarealnumber,and

PAGE 40

342.LIMITSANDDERIVATIVES (V) limx a f g ( x )= limx af ( x ) limx ag ( x ) iflimx ag ( x ) =0. Itisnotdiculttobelievethattheserulesarevalid.Forinstance, if f ( x )getsarbitrarilycloseto L as x approaches a and g ( x )gets arbitrarilycloseto Las x approaches a ,then,as x approaches a ,the valueof f ( x )+ g ( x ),thatis,thevalueof( f + g )( x ),willgetarbitrarily closeto L + L.Thisintuitiveargumentcanbemadeformalusingthe precisede“nitionoflimits. Example 2.6 Let f ( x )= | x | andlet g ( x )= x2.Findthelimitsof f + g f Š g fg 3 f +2 g ,and f/g at a =2 Solution: Basedonthe“velimitlawsgivenearlier,itmakessenseto “rstcomputethelimitsof f and g at2.Thereaderisinvitedtoverify that limx 2f ( x )=limx 2| x | =limx 2x =2 and limx 2g ( x )=limx 2x2=limx 2x limx 2x =2 2=4 whereweusedthefactthat g ( x )= x2= x x ,solawIIIcanbeapplied tocomputethelimitof g at2. Nowitissimplyamatterofbasicalgebratocomputethe“velimits thatwehavebeenaskedto“nd.Indeed,applyingthe“velimitlaws, wegetthat (I)limx 2( f + g )( x )=limx 2f ( x )+limx 2g ( x )=2+4=6, (II)limx 2( f Š g )( x )=limx 2f ( x ) Š limx 2g ( x )=2 Š 4= Š 2, (III)limx 2( f g )( x )=limx 2f ( x ) limx 2g ( x )=2 4=8, (IV)limx 2(3 f +2 g )( x )=3limx 2f ( x )+2limx 2g ( x )=3 2+ 2 4=14(notethathereweappliedlimitlawIVto“rst f thento g ,andthenweappliedlawIto3 f and2 g ),and (V) limx 2 f g ( x )= limx 2f ( x ) limx 2g ( x ) = 2 4 = 1 2 8.2.FrequentlyUsedSpecialCasesofLimitLaws.Afewspecialcases oflimitlawsI…Vareusedsofrequentlythatitisworthmentioning themseparately.First,ifwerepeatedlymultiplyafunctionbyitself,

PAGE 41

8.LIMITLAWS35 wegetapowerofthatfunction.ApplyinglawIIIeachtime,weget thatforallpositiveintegers n (2.1)limx a( f ( x ))n= limx af ( x ) n. Notethatwehaveessentiallyappliedthisruleinthespecialcase of n =2whenwecomputedlimx 2x2inExample2.6. Thereaderisinvitedtoverifythatthelimitsoftheconstantfunction f ( x )= c andtheidentityfunction f ( x )= x aregivenbylimx ac = c forall a andlimx ax = a .Formalproofswillbegiveninthe nextsection. ApplyingEquation(2.1)totheidentityfunction f ( x )= x yields theequation (2.2)limx axn= an. Itturnsout(thoughitisnotobvious)thatinEquation(2.1)the exponent n canbereplacedby1 /n ;inotherwords,powerscanbe replacedbyroots,yielding (2.3)limx an f ( x )=n limx af ( x ) (Here f ( x )hastobenonnegativeif n iseven.)So,inparticular,if f ( x )= x ,then limx an x =n a.8.3.OtherUsefulFactsAboutLimits.Inthissection,wediscussafew factsaboutlimitsthatareoftenusedtocomputelimits,butareslightly dierentinnaturefromthelimitlawswediscussedsofar. First,letusrecallthatthede“nitionof L =limx af ( x )requires that f ( x )getarbitrarilycloseto L if x issucientlycloseto a but notequaltoa .Thatis,thevalueof f ( a )doesnothavetosatisfyany requirements.Infact,wecanchange f ( a )toanythingwewant,and L =limx af ( x )willnotchange.Whatmattersiswhathappensat pointsotherthan a .Hence,wecanconcludethatif f ( x )= g ( x )forall points x = a ,thenlimx af ( x )=limx ag ( x )aslongastheselimits exist.Forinstance,let f ( x )=( x2Š 4) / ( x Š 2)forallrealnumbers x =2andlet f (2)=2014.Let g ( x )= x +2forallrealnumbers.Then f ( x )= g ( x )unless x =2,andhencelimx af ( x )=limx ag ( x )=4. Thestatementthatif f ( x )= g ( x )forallpoints x = a ,then limx af ( x )=limx ag ( x )aslongastheselimitsexistcanbesigni“cantlystrengthened.SeeExercise8.4.1forapossibledirectionfor that.

PAGE 42

362.LIMITSANDDERIVATIVES Second,Equation(2.2)canbeinterpretedbysayingthatthelimit ofa powerfunction f ( x )= xnatanypoint a issimplythe value of f ( a ).Nownotethat polynomials arenothingelsebutsumsofconstantmultiplesofpowerfunctionswithnonnegativeintegerexponents. Hence,usinglimitlawsIandIV,wegetthefollowingtheorem. Theorem 2.1 Let p beapolynomialfunction.Then,foranyreal number a ,wehave limx ap ( x )= p ( a ) Nowrecallthata rationalfunction isjusttheratiooftwopolynomials.Hence,usinglimitlawV,wegetthefollowingstatementfrom Theorem2.1. Corollary 2.1 Let R ( x ) bearationalfunctionandlet a beareal numbersuchthat R ( a ) isde“ned.Then limx aR ( x )= R ( a ) Proof. If R ( x )= p ( x ) /q ( x ),where p and q arepolynomials,then by“rstapplyinglimitlawV,andthenTheorem2.1,weget limx aR ( x )=limx ap ( x ) q ( x ) = limx ap ( x ) limx aq ( x ) = p ( a ) q ( a ) = R ( a ) Sofaralltherelationshipsthatwediscussedforlimitsinvolved equations .Wewillnowdiscusstworulesthat,involve inequalities Theorem 2.2 Let f and g betwofunctionsandassumethat,for allrealnumbers x ,theinequality f ( x ) g ( x ) holds.Then (2.4)limx af ( x ) limx ag ( x ) foranyrealnumber a aslongasbothlimitsexist. Proof. If(2.4)didnothold,then Lf=limx af ( x )= D +limx ag ( x )= D + Lgwouldhold,forsomepositiverealnumber D .Thatwouldleadtoa contradiction,sinceif x issocloseto a that | f ( x ) Š Lf| < ( D/ 3),then, inparticular, f ( x ) >LfŠ ( D/ 3),so g ( x ) >LfŠ D 3 = Lg+ 2 D 3 Thisinequalitysaysthatnomatterhowclose x isto a ,thedistance between g ( x )at Lgismorethan2 D/ 3.Thiscontradictsthede“nition

PAGE 43

8.LIMITLAWS37 of Lg,sinceif Lgexists,thenthevaluesof g ( x )shouldgetarbitrarily closetoit,providedthat x issucientlycloseto a NotethatinTheorem2.2,thefactthattheinequalitiesarenot strictisimportant.SeeExercise8.4.7forarelevantquestion. Corollary 2.2(SqueezePrinciple) If f g ,and h arefunctions suchthat,forallrealnumbers x ,theinequality f ( x ) g ( x ) h ( x ) holdsand limx af ( x )=limx ah ( x )= L, then limx ag ( x ) existsand limx ag ( x )= L SeeFigure2.6foranillustrationofthisimportantprinciple. Proof. Iflimx ag ( x )exists,thenbyapplyingTheorem2.2to f and g ,itfollowsthat L limx ag ( x ),andbyapplyingTheorem2.2 to g and h ,itfollowsthatlimx ag ( x ) L .Soiflimx ag ( x )exists,it isequalto L .InExercise8.4.3youareaskedtoprovethatthislimit exists. Thesqueezeprincipleisveryusefulsinceitallowsustocompute thelimitsofrathercomplicatedfunctionsaslongaswecan squeeze thembetweentwofunctionswithidenticallimits. f ( x ) g ( x ) h ( x ) Figure2.6. Conceptofsqueezetheoremwhere f ( x ) g ( x ) h ( x ). Example 2.7 Let g ( x )= x cos(log x ) .Then limx 0g ( x )=0 .

PAGE 44

382.LIMITSANDDERIVATIVES Solution: Indeed,let f ( x )= Š x and h ( x )= x .Then,sincecos(log z ) isalwaysarealnumberintheinterval[ Š 1 1],theinequality f ( x ) g ( x ) h ( x )holdsforallrealnumbers x .Furthermore,limx 0f ( x )= limx 0h ( x )=0,sowecanapplyCorollary2.2toproveourclaim. WecouldnothaveusedlimitlawIIItocomputelimx 0g ( x )since limx 0cos(log x )doesnotexist.Youareaskedtoprovethisin Exercise8.4.4.8.4.Exercises.(1)Findlimx 23 x2+4 x +9. (2)Findlimx 3 3 x2+5 x Š 2 x +1. (3)Findlimx 2 x Š 4 x Š 2. (4)Findlimx 4 x2+2 x +5 x3+1. (5)Findlimx 2 x2Š 4 x3+8. (6)Let f ( x )and g ( x )betwofunctionsthatonlydierfora “nitenumberofvaluesofthevariable x .Isittruethat limx af ( x )=limx ag ( x )aslongastheselimitsexist?Why orwhynot? (7)Findanexampleoftwofunctions f and g suchthat f ( x ) < g ( x )forallrealnumbers x ,butthereexistsarealnumber a suchthatlimx af ( x )=limx ag ( x ). (8)Explainwhylimx ag ( x )existsiftheconditionsofCorollary2.2hold. (9)Provethatlimx 0cos(log x )doesnotexist. (10)Provethatlimx 0| x sin( x ) | =0. (11)Computelimx 0x3sin(1 /x ). (12)Computelimx 0 x4+ x5sin(ln x ). (13)Computelimx 0 1 1+ x. (14)Computelimx 1 x7+ x5 x8+ x3. (15)Computelimx 15 x x +1. (16)Computelimx 103 18 Š x (17)Let a beapositiverealnumber.Provethatlimx = a x Š a x Š a=1 2 a. (18)Computelimx 3+x Š 3 x +1. (19)Computelimx 4( x +4)2 5.Explainwhichlawsyouareusing. (20)Computelimx 30( x +2)0 4.Explainwhichlawsyouareusing.

PAGE 45

9.CONTINUOUSFUNCTIONS39 9.ContinuousFunctions Intuitivelyspeaking,afunctioniscalled continuous atapoint x = a ifitsgraphinaneighborhoodof x = a canbedrawnwithoutlifting thepencilfromthepaper,thatis,byacontinuousŽline.Theformal de“nitionofcontinuityisasfollows. Definition 2.8 Afunction f iscalled continuous at a ifthe equality limx af ( x )= f ( a ) holds. NotethatDe“nition2.8reallyrequiresthreethings.Thelimitof f at a mustexist,thefunction f mustbede“nedin a suchthat f ( a ) exists,andthevalueof f ( a )mustagreewiththelimitof f at a Ifalltheseconditionshold,thenthebehaviorof f at a isvery similartothebehaviorof f around a ;inparticular,thegraphof f can bedrawnwithoutliftingthepencilfromthepaper.Ifwehadtolift thepencilfromthepaper,thatwouldmeanthatsomekindofgapŽ wouldexistinthegraphof f ,sotherequirementsofDe“nition2.8 wouldnotbesatis“ed. Ifafunction f : R R iscontinuousatall a R ,thenitiscalled continuous .If f iscontinuousateachpointoftheopeninterval( c,d ), thenwesaythat f is continuouson ( c,d ).Finally,ifyoureallywant aformalde“nition,the neighborhood of a isaset S thatcontainsan openinterval( c,d )containing a .9.0.1.ThePreciseDenitionofContinuity.Astheinformalde“nition ofcontinuityisveryclosetothatoflimits,itisnotsurprisingthat theirprecisede“nitionsarealsosimilar. Definition 2.9 Let f bede“nedinanopenintervalcontaining a Wesaythat f iscontinuousin a if,forall > 0 ,thereexists > 0 suchthatif | x Š a | < ,then | f ( x ) Š f ( a ) | < .9.1.ExamplesofContinuousFunctions.Letusconsidersomeofthe mostfrequentlyusedcontinuousfunctions. Example 2.8 Polynomialfunctionsarecontinuous. Solution: ThisisadirectconsequenceofTheorem2.1,whichwe discussedinthelastsection.Theorem2.1statedthatthelimitof

PAGE 46

402.LIMITSANDDERIVATIVES apolynomialfunctionat a isequaltothe value ofthepolynomialat a ,whichispreciselywhatthede“nitionofcontinuityrequires. Therearemanyclassesoffunctionsthatarecontinuousatevery pointwheretheyare de“ned .Iftheyarenotde“nedsomewhere,then, ofcourse,theycannotbecontinuousthere. Example 2.9 Thefollowingareexamplesoffunctionsthatare continuousineverypointwheretheyarede“ned. (I) Rationalfunctions (II) Exponentialfunctions (III) Trigonometricfunctions (IV) Logarithmicfunctions (V) Inversetrigonometricfunctions Thereaderisinvitedtorecallthegraphsofeachofthesefunctions andverifythattheyconsistofcontinuouslinesaslongastheyare de“ned.9.2.FunctionsThatAreNotContinuous.Itistimetostopforamoment andthinkaboutfunctionsthatarenotcontinuousatagivenpoint a Therecanbethreereasonsforthis.First,itcouldbethat f ( a )isnot de“ned,forinstance,when f isarationalfunctionwhosedenominator becomes0when x = a .Oritcouldbethat g isde“nedat a ,but limx ag ( x )doesnotexistat a .Anexampleofthisisthefunction de“nedby g ( x )=1if x 0and g ( x )=0if x< 0.Aswehaveseen before,thelimitofthisfunctiondoesnotexistin a =0,evenif g (0)is de“ned.So g isnotcontinuousat0.Finally,itcouldhappenthat h is de“nedin a andthelimitof h at a exists,but h ( a )isnotequaltothis limit.Thathappens,forexample,if h ( x )=( x +3) / ( x2Š 9)if | x | =3 and h ( x )=1if | x | =3.Let a = Š 3.Then h ( a )=1 =limx ah ( x )= Š 1 6 Theinterestedreaderisinvitedtothinkaboutthefollowing example. Excursion 2.1 Thefollowingfunctionisnotcontinuousanywhere.Let f ( x )=1 if x isrationalandlet f ( x )=0 if x isirrational.9.3.NewContinuousFunctionsfromOld.Itfollowsfromthelimitlaws thatseveraltransformationspreservethecontinuouspropertyoffunctions.

PAGE 47

9.CONTINUOUSFUNCTIONS41 Theorem 2.3 Let f and g betwofunctionsthatarecontinuous at a andlet c bearealnumber.Thenallofthefollowingarealso continuousfunctionsat a : (I) f + g (II) f Š g (III) f g (IV) cf ,and (V) f/g aslongas g ( a ) =0 Example 2.10 Itfollowsfromsuccessiveapplicationsoftheprevioustheoremthat h ( x )= ex sin x +3ln x Š x iscontinuousatall positiverealnumbers a Thefollowingimportanttheoremalsoholds,thoughitisnota directconsequenceofourlimitlaws. Theorem 2.4 Let f and g betwofunctionssuchthat f iscontinuousat a and g iscontinuousat f ( a ) .Thenthecompositionfunction f g iscontinuousat a Thistheoremisimportantsinceitenablesustoprovethecontinuity offunctionsthatwouldotherwisebecumbersometohandle. Example 2.11 Thefunction h ( x )= 2+sin x iscontinuousat allrealnumbers a Solution: Let f ( x )=2+sin x andlet g ( x )= x .Then f iscontinuouseverywhere,and g iscontinuousatallpositiverealnumbers.As f ( x )isalwaysapositiverealnumber,thestatementfollows. 9.4.One-SidedContinuity.Afunctionmayhappentobecontinuousin onlyonedirection,eitherfromtheleftŽorfromtheright.ŽFormally, thismeansthefollowing. Definition 2.10 Wesaythatthefunction f is left-continuous at a if f ( a )=limx aŠf ( x ) .Similarly,wesaythat f is right-continuous at a if f ( a )=limx a+f ( x ) Example 2.12 Let g bethefunctionde“nedby g ( x )=1 if x 0 and g ( x )=0 if x< 0 .Then limx 0Šg ( x )=0 =1= g (0) ,so g isnot left-continuousat 0 .Ontheotherhand, limx 0+g ( x )=1= g (0) ,so g isright-continuousat 0 Thereaderisinvitedtoverifythat f iscontinuousat a ifandonly if f isbothleft-continuousandright-continuousat a .

PAGE 48

422.LIMITSANDDERIVATIVES Wesaythatafunctioniscontinuousonaninterval[ a,b ]ifitiscontinuousatallpointsof( a,b ),left-continuousat a ,andright-continuous at b .9.5.IntermediateValueTheorem.Perhapsthemostimportantproperty ofcontinuousfunctionsisthattheydonot skip anyvaluesbetweentwo valuesthattheyactuallytake.Forinstance,ifatreegrowsfrom3feet to6feet,thenthereisatimeinbetweenwhenthetreeisexactly4.47 feettall.Theintuitivereasonforthisisthatiftherewereavaluein betweenthatisnottakenbythefunction,thentherewouldbeagap inthegraphofthefunction,contradictingtherequirementthatthe functionbecontinuous.Thisisthecontentofthenexttheorem. Theorem 2.5(IntermediateValueTheorem) Let f beafunction thatiscontinuousontheinterval [ a,b ] .Then,if f ( a )= y1and f ( b )= y2and y isarealnumberthatisbetween y1and y2,thenthereexists x [ a,b ] suchthat f ( x )= y Inotherwords, f takesallvaluesbetween y1and y2ontheinterval [ a,b ] Example 2.13 Thereisarealnumber x intheinterval [0 1] such that x + ex=2 Solution: Let f ( x )= x + ex.Then f iscontinuouseverywhere, f (0)= 1,and f (1)=1+ e> 3 71.So,bytheintermediatevaluetheorem, wegetthat f takesallvaluesbetween1and1+ e onthatinterval, including y =2. 9.6.Exercises.(1)Is e3 x +7sin x continuouseverywhere? (2)Is( x2+1)ln( x +1)continuouseverywhere? (3)Isx3+2 x2+3 x +4 x2+4continuouseverywhere? (4)Inwhatpointis x right-continuous,butnotcontinuous? (5)Whereis | ln x | continuous? (6)Whereisx2+2 x +37 e2 xcontinuous? (7)Whereistan x continuous? (8)Whereis1 /x notcontinuous? (9)Isthereapointinwhich xŠ 2isleft-continuous,butnotcontinuous? (10)Whereis1 1 Š ln(1 Š x )continuous? (11)Whereis3 x +2 5 x2Š 6 x +1continuous? (12)Whereissin( x2)continuous?

PAGE 49

10.LIMITSATINFINITY43 (13)Let f ( x )= x2+3if x =2.Whatshould f (2)beif f istobe acontinuousfunction? (14)Let f ( x )=( x2Š 16) / ( x Š 4)if x =4.Whatshould f (4)beif f istobeacontinuousfunction? (15)Let f ( x )= x .Determinethesetofpoints a forwhich f iscontinuousat a .Whatcanbesaidabout f atthepoints where f isnotcontinuous? (16)Let g ( x )= x .Determinethesetofpoints a forwhich g iscontinuousat a .Whatcanbesaidabout g atthepoints where g isnotcontinuous? (17)Provethattheequation x5Š x Š 1=0hasarootintheinterval ( Š 1 2). (18)Provethattheequation x3Š 3 x Š 1=0hasatleasttworoots intheinterval( Š 1 2). (19)Provethattheequation x4+ x = 2hasatleastonesolution intheinterval(0 1). (20)De“neafunction f : R R thatisnotcontinuousin any point a ,and f ( x ) x holdsforall x R 10.LimitsatInnity10.1.FiniteLimitsatInnity.InSection7,wede“nedwhatitmeant forafunctiontohavealimit L atarealnumber a .Inthissection,we extendthatde“nitionandde“newhatitmeansforafunctiontohave alimit L at orat Š Definition 2.11 Let f : R R beafunctionthatisde“nedon someinterval ( b, ) .Wesaythatthe limit of f at istherealnumber L ifthevaluesof f ( x ) getarbitrarilycloseto L andstayarbitrarilyclose to L when x issuitablylarge. Thefactthatthelimitof f at is L isexpressedbythenotation limx f ( x )= L. Thisde“nitionfollowstheideaofthede“nitionoflimitsat“nite points.Indeed,inorderforlimx f ( x )= L tohold,werequirethat thevaluesof f ( x )getarbitrarilycloseto L andstayarbitrarilyclose to L if x islargeenough.Here x islargeenoughŽmeansthat x is inasuitablyselectedneighborhoodof ,inotherwords,inanopen interval( c, ).Recallthatthisisanalogoustowhatwerequiredinthe “nitecase.Therewesaidthatlimx af ( x )= L if f ( x )gotarbitrarily closeto L andstayedarbitrarilycloseto L once x wassuitablyclose to a ,thatis,when x wasinasuitablyselectedneighborhoodof a .

PAGE 50

442.LIMITSANDDERIVATIVES Example 2.14 Let f ( x )=1 /x .Then limx f ( x )=0 Solution: Ifwewantthevalueof f ( x )tobecloserthan to0,allwe havetodoistoselect x suchthat x> 1 / holds.Once x getspast1 / thevaluesof f ( x )willstaybetween0and Thede“nitionoflimitsat Š iswhatthereaderprobablyexpects. Definition 2.12 Let f : R R beafunctionde“nedonsome interval ( Š ,b ) .Wesaythatthe limit of f at Š istherealnumber L ifthevaluesof f ( x ) getarbitrarilycloseto L andstayarbitrarily closeto L when x isanegativenumberwithasuitablylargeabsolute value. Thefactthatthelimitof f at Š is L isexpressedbythenotation limx Šf ( x )= L. Example 2.15 Let f ( x )=1 /x2.Then limx Šf ( x )=0 Solution: Ifwewanttoget f ( x )closerthan to0andkeepitthere, itsucestochoose x suchthat x< Š 1 / .Then x2> 1 / ,andhence f ( x )=1 /x2< 10.1.1.TheFormalDenitionofLimitsatInnity.Theformalde“nition oflimitsatin“nityisverysimilartothatoflimitsat“nitepoints.The onlydierenceisintheformaldescriptionofwhatitmeanstobeina neighborhoodofin“nityversuswhatitmeanstobeinaneighborhood ofarealnumber. Definition 2.13 Let f : R R beafunctionde“nedonsome interval ( b, ) .Wesaythat limx f ( x )= L if,forallpositivereal numbers ,thereexistsapositiverealnumber N suchthatif x>N then | f ( x ) Š L | < Theformalde“nitionoflimitsatnegativein“nityisanalogous.The onlydierenceisagainintheformaldescriptionofwhatitmeansfor x tobeinaneighborhoodof Š .Itmeanstobeinaninterval( Š ,c ). Definition 2.14 Let f : R R beafunctionde“nedonsome interval ( Š ,b ) .Wesaythat limx Šf ( x )= L if,forallpositivereal numbers ,thereexistsanegativerealnumber N suchthatif x
PAGE 51

10.LIMITSATINFINITY45 10.1.2.TheGraphicalMeaningofaFiniteLimitatInnity.Ifafunction f haslimit L at or Š ,thenthegraphofthefunctionwillapproach thehorizontalline y = L atthatin“nity.Thegraphmayormay notactuallytouchthatlineorevenbecomethatline.Theline y = L iscalleda horizontalasymptote ofthegraphof y = f ( x )when limx f ( x )= L orlimx Šf ( x )= L holds.10.2.InniteLimitsatInnity.Itcanhappenthatthelimitofafunctionat isnotarealnumberbutrather or Š Definition 2.15 Let f : R R beafunctionde“nedonsome interval ( b, ) .Wesaythatthelimitof f at is ,denotedby limx f ( x )= if f ( x ) getsarbitrarilylargeandstaysarbitrarilylargeif x getssucientlylarge. Example 2.16 Let f ( x )= ex.Then limx f ( x )= Solution: Inordertoget f ( x )tobelargerthansomegivenpositive realnumber M ,itsucestochoose x> ln M Thefollowingnotationisde“nedinananalogousway: (I)limx f ( x )= Š (II)limx Šg ( x )= (III)limx Šh ( x )= Š Eachofthesede“nitionsreferstoafactthatthevaluesofafunction getarbitrarilyfarawayfrom0andstayarbitrarilyfarawayfrom0(in theappropriatedirection)if x getssucientlyfarawayfrom0(inthe appropriatedirection).Thereadershouldtesthisorherunderstandingoftheseconceptsbyverifyingthatlimx 1 Š x = Š ,while limx Šx2= ,andlimx Šx3= Š .10.2.1.TheFormalDenitionofInniteLimitsatInnity.Bynow,the formalde“nitionofin“nitelimitsatin“nityprobablydoesnotcome asasurprise.Weareprovidingaformalde“nitionforoneofthefour possiblescenariosthatcanoccurduetochangesinsign.Theother threecasesareanalogous. Definition 2.16 Let f : R R beafunctionde“nedonsome interval ( b, ) .Wesaythat limx f ( x )= if,forallpositivereal numbers M ,thereexistsapositiverealnumber N suchthatif x>N then f ( x ) >M .

PAGE 52

462.LIMITSANDDERIVATIVES 10.3.ComputingLimitsatInnity.Thelimitlawsthatwelearnedfor limitsat“nitepointsstaytrueforlimitsatin“nityaswell,provided, ofcourse,thattheymakesense.Hereareafewexamples. Example 2.17 Wehave limx x +3 x Š 4 =1 Itwouldbe wrong toargueasfollows:Thenumeratoristhefunction f ( x )= x +3,andthedenominatoristhefunction g ( x )= x Š 4. At ,theybothhavelimit ,so,bythelimitlawforquotients,the limitoftheirquotientis1.Ž Theproblemwiththisargumentisthat is notanumber .So / isnotde“ned.Itispossiblefor f and g bothtohavelimit at ,andfor f/g tohavelimits c at ,foranygivenrealnumber c Indeed,let f ( x )= cx andlet g ( x )= x Instead,wecansolveExample2.17asfollows. Solution: limx x +3 x Š 4 =limx ( x Š 4)+7 x Š 4 =limx 1+ 7 x Š 4 =1+limx 7 x Š 4 =1+0 =1 Wewouldliketopointoutotherpitfallswhendealingwiththe applicationoflimitlawsandin“nitelimits.Thefollowingexpressions arenotde“ned: (I) +( Š ) (II) 0and Š 0 (III)1and1ŠThefollowingtheoremisveryusefulwhendealingwithlimitsat Theorem 2.6 Let r beapositiverationalnumber.Then limx 1 xr=0 If r isaninteger,thenthisstatementfollowsfromthefactthat limx 1 /x =0byapplyinglimitlawIII(forproducts) r times.If r = p/q ,where p and q arepositiveintegers,thenwecan“rstprove thetheoremfor xp,andthen,usingtherootlaw,for xp/q=q xp.

PAGE 53

10.LIMITSATINFINITY47 Manylimitscanbecomputedwiththehelpofthistheorem. Example 2.18 Wehave limx x2+3 x +1 x3=0 Solution: Wehave x2+3 x +1 x3= x2 x3+ 3 x x3+ 1 x3, andeachofthethreesummandshaslimit0at bythepreceding theorem.Hence,bythelimitlawforsums,sodoestheirsum. Notethatthelimitwouldnotchangeifwechangedthedenominator from x3to x3+3 x2+4 x +5.Thiswouldhavedecreasedthevalueofour function,butwouldhavestillkeptitpositive.Hence,bythesqueeze principle,wecanthenconcludethat limx x2+3 x +1 x3+3 x2+4 x +5 =0 .10.4.Exercises.(1)Findlimx x +1 x2+4. (2)Findlimx 3 x2+4 x +1 x2+5. (3)Findlimx x3+2 x x2+4 x +6. (4)Findlimx Š 3 x2+4 x +1 x2Š 4. (5)Computelimx Š x5Š 3 x x3Š 14. (6)Let R ( x )= p ( x ) /q ( x )bearationalfunction.Explainhow limx R ( x )dependson p ( x )and q ( x ). (7)Computelimx x +2 x Š 3+2 x2Š 3 x +1 x2Š 2 x +1. (8)Computelimx Š sin x x. (9)Doeslimx Š x sin xexist? (10)Computelimx sin x +cos2x x0 1. (11)Computelimx x2+1 x. (12)Computelimx x2+1 x2Š 1. (13)Computelimx x4+1 2 x2+11. (14)Computelimx Š 100 x +4 x2Š 4 x +5. (15)Istherearealnumber L suchthat L =limx x3+1 1000 x2+9 x +35holds?

PAGE 54

482.LIMITSANDDERIVATIVES (16)Computelimx xŠ 0 1+ xŠ 0 9. (17)Computelimx x2+ x3+ x4 x4 1. (18)Computelimx Š x +sin x x. (19)Doeslimx x sin x exist? (20)Giveanexampleofafunction f suchthatlimx f ( x )=0, butlimx (1 /f ( x ))doesnotexist. 11.Derivatives11.1.TangentLines.Letusconsiderafunction,suchas f ( x )= x2,and itsgraph.Letuschooseapointonthegraph,saythepoint P =(3 9). Nowletuslookfortheslopeofthe tangentline tothegraphatthat point. Thatis,considerasequenceofpoints P1,P2,... thatareallonthe graphof f andarecloserandcloserto P .Foreachofthesepoints, drawtheline PiP .Theslopeoftheselineswillapproachacertain slope,andsothelines PiP willapproachacertainline. That lineis calledthetangentlineof f at P .SeeFigure2.7foranillustration. Definition 2.17 Let f beafunctionandlet P =( a,f ( a )) bea pointonthegraphof f .Thenthe tangentline to f at P istheline thatcontains P andhasslope (2.5)limx af ( x ) Š f ( a ) x Š a providedthatthislimitexists. ( a f ( a ))( a f ( a )) ( x f ( x )) ( x f ( x )) 2 112 3 2 1 1 2 112 3 2 1 1 Figure2.7. Noticethatas x approaches a thesecant lineapproachesthetangentline.

PAGE 55

11.DERIVATIVES49 Theinteractivewebsitehttp://www.math.u”.edu/ mathguy/ufcalc book/derivative def.htmlprovidesfurtherexamplesofthis phenomenon. Notethatintheprecedingde“nition,( f ( x ) Š f ( a )) / ( x Š a )issimply theslopeofthelineconnectingthepoints P and( x,f ( x )). Example 2.19 Inourrunningexample,thatis,when f ( x )= x2and P =(3 9) ,thetangentlineisthelinethatgoesthrough P andhas slope limx 3f ( x ) Š f (3) x Š 3 =limx 3x2Š 9 x Š 3 =limx 3( x +3)=6 .11.2.Velocities.RecallthatinSection6,wementionedthattheaveragevelocityofamovingobject,suchasacar,canbecomputedby therule v = s/t .Thatis,theaveragevelocityisequaltothedistance covereddividedbythetimeneededtocoverthatdistance.However, whatcanbesaidaboutthe instantaneousvelocity ,thatis,thevelocity inagivenmoment? WecouldnotanswerthatquestioninSection6sincewedidnot havethetoolstohandlethefactthatwhenonlyagivenmomentis considered,boththenumeratorandthedenominatoroftheformula v = s/t are0.Nowthatwehavelearnedaboutlimits,wecanovercome thatdicultyasfollows. Definition 2.18 Let f ( t ) beafunctionsuchthat f ( t ) isthedistancecoveredbyamovingobjectin t unitsoftime.Thenthe instantaneousvelocity oftheobject a unitsoftimeafteritstartsmovingis v ( t )=limt af ( t ) Š f ( a ) t Š a providedthatthislimitexists. Example 2.20 Acarstartsoutbyacceleratingfor10secondsso thatthedistancecoveredinthe“rst t secondsisobtained(inmeters) bythefunction f ( t )=1 2t2if t 10 .Whatistheinstantaneousvelocity ofthecarafter4seconds? Solution: Bythede“nitionofinstantaneousvelocity,wemust compute v (4)=limt 4f ( t ) Š f (4) t Š 4 =limt 4t2Š 16 2( t Š 4) =limt 4t +4 2 =4 So,attheendofthefourthsecond(exactly4secondsafterstarting out),thecarwillmoveatarateof4meterspersecond.

PAGE 56

502.LIMITSANDDERIVATIVES 11.3.TheDerivativeofaFunction.Thefactthatthelasttwoconcepts, thetangentlineandtheinstantaneousvelocity,ledtoverysimilar de“nitionssuggeststhatthereisaverygeneralprincipleatworkand wehaveseentwospecialcasesofthatprinciple. Thisisindeedthecase. Definition 2.19 Let f beafunction.The derivative of f at a is thelimit f( a )=limx af ( x ) Š f ( a ) x Š a ifthislimitexistsandis“nite. So,inparticular, f( a )istheslopeofthetangentlineof f at a (unlessthattangentlineisvertical).Furthermore,theinstantaneous velocityattime a isthederivativeofthedistancecovered(asafunction ofthetime t neededtocoverthatdistance)at t = a Inotherwords,thederivativeisacommongeneralizationofthe conceptsoftangentlineandinstantaneousvelocity.11.4.Exercises.(1)Findtheslopeofthetangentlinetothecurve f ( x )=3 x2Š 7 atthepoint(2 5). (2)Findtheslopeofthetangentlinetothecurve f ( x )= x3at x =0. (3)Findtheslopeofthetangentlinetothecurve f ( x )= x (1 Š x ) at x =1 / 2. (4)Findtheslopeofthetangentlinetothecurve f ( x )= x2at threedierentpoints.Doyouseeapattern? (5)Findtheslopeofthetangentlinetothecurve f ( x )= x2+ x atthreedierentpoints.Doyouseeapattern? (6)Showanexampleofacurvethatdoesnothaveatangentline atsomepoint a becausethelimitde“nedin(2.5)doesnot existorisin“nite. (7)Thedistancecoveredbyacarinacertaintimeperiodisdescribedbythefunction f ( t )= tm + t2( b Š m ) 2 where b and m arepositiveconstants.Letusassumethat t [0 1].Findtheinstantaneousvelocityofthecarata givenmoment t = a (8)Aballisrollingdownahill.Thedistanceitcoversintime t isgivenbythefunction s ( t )=3 t +0 5 t2,where t [0 5]

PAGE 57

12.THEDERIVATIVEASAFUNCTION51 andtimeismeasuredinseconds.Whatistheinstantaneous velocityoftheballatthemomentoftime t =3? (9)Atthebeginningofadailytrainingsession,thedistancecoveredbyarunnerisdescribedbythefunction s ( t )=0 5 t2, where t [0 5],timeismeasuredinseconds,anddistanceis measuredinmeters.Atwhatmomentwilltherunnerhavean instantaneousvelocityof6m / s? (10)Acarismovingataspeedof20meterspersecondwhenits driverappliesthebrakesandthecarstartsslowingdown.The carstops10secondslater.Thedistancecoveredbythecarin t seconds,startingatthemomentwhenthedriverstepsonthe brakes,isgivenbythefunction f ( t )=20 t Š t2for t [0 10]. Whatisthevelocityofthecar t secondsafterthebrakesare applied? (11)Provethat,foranyconstant c ,thederivativeofthefunction f ( x )= c is0at any point a (12)Findthederivativeofthefunction f ( x )= x +5at a =7. (13)Findthederivativeofthefunction f ( x )=3 x +2at a =4. Whathappensifwechangethevalueof a ? (14)Findthederivativeofthefunction f ( x )=3 x Š 11at a =4. Compareyourresultwiththeresultofthepreviousexercise. (15)Findthederivativeofthefunction f ( x )=2 x2at a =2. (16)Findthederivativeofthefunctions g ( x )=2 x2+1at a =2 and h ( x )=2( x Š 1)2at a =3. (17)Findthederivativeofthefunction f ( x )= x3at x =1. (18)Findthederivativeofthefunction f ( x )= x at x =4. (19)Let g ( x )= 2 x if0 x x if x< 0 Does f(0)exist? (20)Let f bede“nedasinthepreviousexercise.Does f( a )exist if a =0? 12.TheDerivativeasaFunction12.1.RatesofChange.Inthelastsection,wesawthatthederivative ofafunctionatagivenpointwasacommongeneralizationofthe conceptsoftangentlinesandinstantaneousvelocities.Wewillnow furtherelaborateonthat,inordertounderstandhowfar-reachingthe conceptofderivativesis. If f isafunctionand f ( x )= y ,thenthequantitydenotedby y dependsonthequantitydenotedby x .Thisissometimesexpressed

PAGE 58

522.LIMITSANDDERIVATIVES bysayingthat x istheindependentvariableand y isthedependent variable.If x changes,thenthechangein y canbedescribedinterms ofthechangein x Inparticular,if x changesfrom x1to x2,then y = f ( x )changes from y1= f ( x1)to y2= f ( x2).Theaveragerateofchangeforthe interval( x1,x2)isthentheratio y2Š y1 x2Š x1= y x where x isthechange(orincrement)of x .Wehavetousetheword averageŽsinceweonlyhaveinformationaboutthevaluesof y atthe endpointsoftheinterval( x1,x2);wedonotknowhow f ( x )= y behaves intherestoftheinterval.Ifwewantmorepreciseinformation,such asthe instantaneousrateofchange of f ( x )= y atagivenpoint,then wehavetousethenotionoflimitsagain,justaswehavedonetwice inthelastsection.Thatis,atagivenpoint x = a ,wede“nethe instantaneousrateofchangeof f ( x )= y as limx2 af ( x2) Š f ( a ) x2Š a =lim x 0 y x .12.2.TheDerivativeoftheFunctionf.Recallthat,atagivenpoint a thederivativeofthefunction f isde“nedasthelimit f( a )=limx af ( x ) Š f ( a ) x Š a Notethatthisde“nitionassociatestherealnumber f( a )tothe realnumber a .Thatis, f: R R isafunction.Thefunction fis calledthe derivative of f .Theoperationthattakes f into fiscalled dierentiation .Thisexplainsthefollowingde“nition. Definition 2.20 Afunction f iscalled dierentiable at a if f( a ) exists. Wesaythat f isdierentiableontheinterval( a,b )if f isdierentiableat d forall d ( a,b ). Example 2.21 Thefunction f ( x )= x3isdierentiableinevery realnumber a ,and f( a )=3 a2.

PAGE 59

12.THEDERIVATIVEASAFUNCTION53 Solution: Wehave limx af ( x ) Š f ( a ) x Š a =limx ax3Š a3 x Š a =limx a( x Š a )( x2+ xa + a2) x Š a =limx a x2+ xa + a2 =3 a2. Thefunctionswehaveconsideredsofarhadonlyoneindependent variable,usuallythevariable x .Thedependentvariablewasusually denotedby y ,so y = f ( x )held.Soitwasalwaysclearthatthederivativewastaken withrespectto x .However,therearecircumstances whenthisisnotsoclear,usuallywhen f dependsonmorethanone variable.Therefore,thereareadditionalwaystodenotethefunction fsuchas €dy dx, €df dx, €d dxf ( x ), € Dxf ( x ),or € Df ( x ).12.3.DifferentiabilityVersusContinuity.Thede“nitionsofdierentiabilityandcontinuityaresimilar.Whichoneimposesstrongerrequirementsonafunctionatagivenpoint?Thefollowingtheoremshows thatdierentiabilityisthestrongerrequirement. Theorem 2.7 If f isdierentiableat a ,then f iscontinuousat a Proof. If f isdierentiableat a ,then f( a )=limx af ( x ) Š f ( a ) x Š a ; inparticular,thelimitshownontheright-handsideexists.Multiplying bothsidesbythefunction g ( x )= x Š a ,weget f( a )( x Š a )=( x Š a )limx af ( x ) Š f ( a ) x Š a ; Now,takinglimitsat a onbothsides,weobtain (2.6) f( a ) limx a( x Š a )=limx a( f ( x ) Š f ( a )) ,

PAGE 60

542.LIMITSANDDERIVATIVES sincewecanapplythelimitlawforproductsontheright-handsideto getthat limx a( x Š a )limx af ( x ) Š f ( a ) x Š a =limx a ( x Š a ) f ( x ) Š f ( a ) x Š a =limx a( f ( x ) Š f ( a )) Finally,notethattheleft-handsideof(2.6)isequalto0since f( a ) ( x Š a )isapolynomialthattakesvalue0when x = a .Hence,the right-handsideof(2.6)isequalto0aswell,thatis, 0=limx a( f ( x ) Š f ( a ))=(limx af ( x )) Š f ( a ) Adding f ( a )toboththefarleftandfarrightsides,wegetthat f ( a )=limx af ( x ) whichmeansthat f iscontinuousat a TheconverseofTheorem2.7isnottrue.Indeed,thefunction f ( x )= | x | iscontinuousat a =0,butitisnotdierentiable.The readerisinvitedtoprovethisbyshowingthat limx 0Š| x |Š 0 x Š 0 =limx 0Š| x | x =limx 0+| x | x =limx 0+| x |Š 0 x Š 0 andhence f(0)=limx 0| x |Š 0 x Š 0 doesnotexist. Ingeneral,thereareseveralreasonsacontinuousfunctionmayfail tobedierentiableatagivenpoint.Itcouldbethatthegraphofthe functionhasacorner,Žlikethatof | x | at0,andhencetheslopeof thetangentlinecannotbede“nedbecausetheleft-handlimitandthe right-handlimitofthelinesapproachingthepurportedtangentline arenotequal.Or,itcouldbethatthefunctionhasaverticaltangent lineatthegivenpoint.SeeExercise12.5.6foranexampleofthis.12.4.Higher-OrderDerivatives.Inupcomingchapters,itwilloftenbe usefultoconsidernotonlythederivativeofafunctionbutalsothe derivativeofthederivativeandeventhederivativeofthederivative ofthederivative.Thesefunctionsappearsooftenthattheyhavetheir ownnames. If f isadierentiablefunctiononaninterval( a,b )anditsderivative fisalsodierentiableon( a,b ),thenthederivativeof fiscalled the secondderivative of f andisdenotedby f.Similarly,if fis

PAGE 61

12.THEDERIVATIVEASAFUNCTION55 dierentiableon( a,b ),thenitsderivativeiscalledthe thirdderivative of f andisdenotedby f.Higher-orderderivativesarede“nedin ananalogousway,butdenotedslightlydierently.Forinstance,the seventhderivativeof f isdenotedby f(7),and,ingeneral,the n th derivativeisdenotedby f( n ). Example 2.22 WehaveseeninExample2.21thatif f ( x )= x3, then f( x )=3 x2.Therefore, f( a )=limx af( x ) Š f( a ) x Š a =limx a3( x2Š a2) x Š a =limx a3( x + a ) =6 a. So f( x )=6 x InExercise12.5.2,youareaskedtoprovethat f( x )=6forall x andinExercise12.5.3,youareaskedtocomputehigher-orderderivativesof f .12.5.Exercises.(1)Let f ( x )= cx + d ,where c and d are“xedrealnumbers. Compute f( x ),and f( x ). (2)Let f ( x )= px2+ qx + r ,where p q ,and r are“xedreal numbers.Compute f( x ), f( x ),and f( x ). (3)Let f ( x )= x .Compute f( a )atsomepoint a> 0. (4)Compute f( a )if f ( x )= 4 x +1. (5)Let f ( x )= x3.Provethat f( x )=6forallrealnumbers x (6)Let f ( x )= x3.Compute f(4)( x ).Whatcanbesaidabout higher-orderderivativesof f ? (7)Let f ( x )= x4.Compute f( a )atsomepoint a (8)Let f ( x )=1 /x .Compute f( a )atsomepoint a =0. (9)Let f ( x )=1 /x2.Compute f( a )atsomepoint a =0. (10)Let f and g betwofunctionssuchthat f ( x ) Š g ( x )= c for all x ,where c isaconstant.Isittruethat,ateverypoint a where f( a )exists, g( a )alsoexists,and f( a )= g( a )holds? (11)Let f bede“nedontheinterval[0 2]by f ( x )= 1 Š x2if 0 x 1,and f ( x )= Š 1 Š ( x Š 2)2if1
PAGE 62

562.LIMITSANDDERIVATIVES line,thatis, limx 1 f ( x ) Š f (1) x Š 1 = (12)Findanexampleofafunction f andarealnumber a suchthat f( a )exists,but f( a )doesnotexist. (13)Findanexampleofafunction f andapoint a suchthat f( a ) and f( a )exist,but f( a )doesnotexist. Intheremainingexercisesofthissection,decidewhetherthe derivativeofthegivenfunctioninthegivenpointexistsor not. (14) f ( x )= | x | x at a =0. (15) f ( x )= | x |2at a =0. (16) f ( x )= x at a =1. (17) f ( x )= x at a =2. (18) f ( x )=ln x at a =0. (19) f ( x )=cot x at a =0. (20) f ( x )=2 x2+ x3atanyrealnumber a .

PAGE 63

CHAPTER3 RulesofDifferentiation 13.DerivativesofPolynomialandExponentialFunctions13.1.Polynomials.Letusrecallthatpolynomialsaresumsofpower functionswithnonnegativeintegerexponents,suchasthefunction f ( x )=3 x2+4 x +6.Inthissection,wewilldeducegeneralrulesfor thederivativesofpolynomialfunctions.Westartbytheirbuilding blocks,Žpowerfunctions.Thesimplestoftheseistheclassofconstant functions. Theorem 3.1 Let c bearealnumberandlet f ( x )= c forall x Then f( a )=0 forallrealnumbers a Beforeweprovethetheorem,wepointoutthat,intuitively,itmakes perfectsense.Thederivativeofafunction f describestherateof changeof f ,butif f isaconstantfunction,then f neverchanges(it haszerochange). ProofofTheorem3.1. Wehave f( a )=limx af ( x ) Š f ( a ) x Š a =limx ac Š c x Š a =0 Notethatlimx a( c Š c ) / ( x Š a )=0since( c Š c ) / ( x Š a )=0forall values x = a Wenowturnourattentiontoamoregeneralclassofpowerfunctions,thoseoftheform f ( x )= xn,where n isapositiveinteger.Let usrecallthealgebraicidentity xnŠ an=( x Š a ) ( xn Š 1+ xn Š 2a + + xan Š 2+ an Š 1) Theorem 3.2 Let n beapositiveintegerandlet f ( x )= xn.Then f( a )= an Š 1.57

PAGE 64

583.RULESOFDIFFERENTIATION Proof. Wehave f( a )=limx af ( x ) Š f ( a ) x Š a =limx axnŠ an x Š a =limx a( x Š a ) ( xn Š 1+ xn Š 2a + + xan Š 2+ an Š 1) x Š a =limx a( xn Š 1+ xn Š 2a + + xan Š 2+ an Š 1) = nan Š 1. Notethatthisagreeswithourresultfromthelastsectionthat showedthatif f ( x )= x3,then f( x )=3 x2. ItturnsoutthatTheorem3.2holdsevenif n isnotapositive integer.Thatis,forallrealnumbers ,if f ( x )= x,then f( x )= x Š 1.Wewillseeaformalproofofthisfactlater.Intheexercises, youareaskedtoprovetwospecialcasesofthisgeneralresult.13.1.1.ThreeSimpleRules.Derivativesarelimitsofcertainfunctions, soitisnotsurprisingthat some ofthelawsgoverningtheircomputation areverysimilartolimitlaws.Thatis,ifweknowthederivativeof f and g ,thenwecaneasilycomputethederivativeof f + g f Š g ,and cf ,where c isagivenrealnumber.Therulesareasfollows. Theorem 3.3 Let f and g betwofunctionsthataredierentiable at a .Then f + g isdierentiableat a ,and ( f + g )( a )= f( a )+ g( a ) Proof. Wehave ( f + g )( a )=limx a( f + g )( x ) Š ( f + g )( a ) x Š a =limx a f ( x ) Š f ( a ) x Š a + g ( x ) Š g ( a ) x Š a = f( a )+ g( a ) Theothertworulesandtheirproofsaresosimilarthattheyare leftasexercises. Theorem 3.4 Let f and g betwofunctionsthataredierentiable at a .Then f Š g isdierentiableat a ,and ( f Š g )( a )= f( a ) Š g( a ) .

PAGE 65

13.DERIVATIVESOFPOLYNOMIALANDEXPONENTIALFUNCTIONS59 Theorem 3.5 Let f beafunctionthatisdierentiableat a and let c bearealnumber.Then cf isdierentiableat a and ( cf )( a )= cf( a ) Itisveryimportanttopointoutthatthe otherlimitlawsdonot carryovertoderivativesinthesamefashion. Thatis,ingeneral, ( fg ) = fg,and( f/g ) = f/g.Wewilllearnsomemorecomplicated rulestocomputethederivativesof fg and f/g inthenextsection. Theorems3.3to3.5enableustocomputethederivativeofany polynomialfunction. Example 3.1 Let p ( x )=3 x3+5 x2Š 6 x +8 .Find p( x ) Solution: Notethat p ( x )isjustasum(anddierence)ofconstant multiplesofpowerfunctions.Thederivativesofpowerfunctionsare computedinTheorem3.2.ThenwecanapplyTheorems3.3to3.5 toget p( x )=(3 x3)+(5 x2)Š (6 x )+(8)=3( x3)+5( x2)Š 6( x )+(8)=9 x2+10 x Š 6 13.2.ExponentialFunctions.Letusnowcomputethederivativeofthe exponentialfunction f ( x )= bx,where b issomepositiveconstant.By thede“nitionofderivatives,weget f( a )=limx af ( x ) Š f ( a ) x Š a =limx abxŠ ba x Š a =limz 0ba + zŠ ba z = balimz 0bzŠ 1 z = baf(0) Severalcommentsareinorder.First,notethesubstitution z = x Š a inthethirdline.Second,notethat baisaconstantthatdoesnot dependon z ;hence,thelimitlawforconstantmultipleswasusedin thefourthline.Third,inthespecialcasewhen a =0,thede“nition ofthederivativeyields f(0)=limz 0( bzŠ 1) /z .Weusedthisfactin thelastline.

PAGE 66

603.RULESOFDIFFERENTIATION Inotherwords, (3.1) f( x )= f(0) bx= f(0) f ( x ) Thatis,thederivativeofthefunction f isa constantmultiple of f Theconstantinquestionis f(0),thatis,limz 0( bzŠ 1) /z .Numerical experimentationsuggeststhatthelarger b is,thelargerthislimitis. Graphicalexperimentationsuggeststhisaswell.Indeed, f(0)isthe slopeofthetangentlinetothecurveof f ( x )= bxatthepoint x =0, andplotting f forvariousvaluesof b suggeststhatthelarger b is,the largerthisslopeis. Inparticular,itcanbeprovedthatthereexistsarealnumber e closeto2.71,suchthat limz 0ezŠ 1 z =1 Thisrealnumber e isthebasisofthenaturallogarithmthatwedenote byln. Thereadermaywishtoconsulttheinteractivewebsitehttp://www. math.u”.edu/ mathguy/ufcalcbook/exponent.htmlforfurtherillustrations. Definition 3.1 Let e betherealnumbersuchthat limz 0ezŠ 1 z =1 So,inthespecialcaseof b = e ,Equation(3.1)takestheform ( ex)= ex, since f(0)=limz 0ezŠ 1 z =1 Thatis,thederivativeof f ( x )= exis f ( x )= exitself.InSection 16,wewillseewhatthatimpliesforthederivativesofexponential functionswithbasesdierentfrom e .13.3.Exercises.(1)Let f ( x )= x3+2 x2+3 x +4.Compute f( x )and f( x ). (2)Let f ( x )= x4Š 3 x +9.Compute f( x )and f( x ). (3)Let f ( x )= x8Š 2 x4+1.Compute f( x ), f( x ),and f( x ). (4)Let f ( x )= Š x Š x3.Compute all derivatives(“rst,second, third,etc.)of f ( x ). (5)Provethatif f isapolynomialfunction,then f( x )isalsoa polynomialfunction.

PAGE 67

14.THEPRODUCTANDQUOTIENTRULES61 (6)Provethatif f isapolynomialfunctionofdegree d ,then f( d +1)( x )=0forallrealnumbers x (7)Provethatif f isapolynomialfunctionofdegree d ,then f( d )isa linear function. (8)Let p beapolynomialfunctionofdegree d andlet k d bea nonnegativeinteger.Whatkindoffunctionis f( k )? (9)Provethatif f ( x )= x1 / 2and a> 0,then f( a )=1 2 a. (10)Provethatif f ( x )=1 /x and a =0,then f( a )= Š1 a2. (11)ProveTheorem3.4. (12)ProveTheorem3.5. (13)Let f ( x )=3 x3Š 4 x2+ x Š 2+4 ex.Compute f( x ). (14)Let f ( x )=1 / x .UsetheremarkaftertheproofofTheorem 3.2tocompute f( x ). (15)Let f ( x )= x2Š 2 x +7,andlet g ( x )= ex.Compute( f + g )( x ) and( f Š g )( x ). (16)Isthereafunction f thatisnotidenticallyzerosuchthat f( k )( x )= f ( x )forall x ? (17)Couldithappenthat f and g aretwodierentfunctions,but f( x )= g( x )forall x ? (18)Couldithappenthat f and g aretwodierentfunctions, fand garetwodierentfunctions,but f( x )= g( x )forall x ? (19)Couldithappenthat f and g aretwopolynomialfunctionsof dierentdegree,and f( x )= g( x )forall x ? (20)Isthereapolynomialfunction f ( x )thatisnotidenticallyzero suchthatthereisarealnumber x forwhich f( k )( x )doesnot dependon k ? 14.TheProductandQuotientRules14.1.TheProductRule.Wementionedinthelastsectionthat,ingeneral,( fg ) = fg.Forinstance,if f ( x )=2 x +1and g ( x )= x +2,then ( fg )( x )=2 x2+5 x +2,so( fg )( x )=(2 x2+5 x +2)=4 x +5,while f( x )=2and g( x )=1,so f( x ) g( x )=2. Itturnsoutthatthereisaruletocomputethederivativeofa product;itisjustalittlebitmorecomplicatedthanthelimitlawfor products.Thisisthefocusofour“rsttheoreminthissection. Theorem 3.6 Let f and g betwofunctionsthataredierentiable at a .Then fg isdierentiableat a ,and ( fg )( a )= f ( a ) g( a )+ f( a ) g ( a ) .

PAGE 68

623.RULESOFDIFFERENTIATION Proof. Byde“nition,wehave (3.2)( fg )( a )=limx af ( x ) g ( x ) Š f ( a ) g ( a ) x Š a Thecrucialideaistodecomposethedierence f ( x ) g ( x ) Š f ( a ) g ( a )as ( f ( x ) g ( x ) Š f ( x ) g ( a ))+( f ( x ) g ( a ) Š f ( a ) g ( a ))inthenumeratorofthe right-handsideof(3.2). Usingthisidea,weobtainfromEquation(3.2) ( fg )( a )=limx a f ( x ) g ( x ) Š f ( x ) g ( a ) x Š a + f ( x ) g ( a ) Š f ( a ) g ( a ) x Š a =limx af ( x ) g ( x ) Š f ( x ) g ( a ) x Š a +limx af ( x ) g ( a ) Š f ( a ) g ( a ) x Š a =limx af ( x ) g ( x ) Š g ( a ) x Š a +limx ag ( x ) f ( x ) Š f ( a ) x Š a = f ( a ) g( a )+ g ( a ) f( a ) Example 3.2 Thederivativeof h ( x )= x2excanbecomputedas follows.Let f ( x )= x2and g ( x )= ex.Then h( x )=( fg )( x ) = f ( x ) g( x )+ f( x ) g ( x ) = x2( ex)+( x2)ex= x2ex+2 xex= ex( x2+2 x ) .14.2.TheQuotientRule.Theruleforthederivativeofthequotientof twofunctionsisalittlebitmorecomplicatedthanthatforthederivativeoftheproductoftwofunctions.Thoughmorecomplex,boththe ruleanditsproofbearsomesimilaritytotherulegiveninTheorem3.6. Theorem 3.7 Let f and g betwofunctionsthataredierentiable at a andletusassumethat g ( a ) =0 .Then f/g isdierentiableat a andwehave f g ( a )= g ( a ) f( a ) Š f ( a ) g( a ) g ( a )2. Proof. Byde“nition,wehave (3.3) f g ( a )=limx a f ( x ) g ( x )Šf ( a ) g ( a ) x Š a .

PAGE 69

14.THEPRODUCTANDQUOTIENTRULES63 Letusmultiplyboththenumeratorandthedenominatoroftherighthandsideby g ( x ) g ( a )toget f g ( a )=limx af ( x ) g ( a ) Š f ( a ) g ( x ) ( x Š a ) g ( x ) g ( a ) Nowtransformthenumeratoroftheright-handsidebysubtracting andthenadding g ( a ) f ( a )toget f g ( a )=limx af ( x ) g ( a ) Š g ( a ) f ( a )+ g ( a ) f ( a ) Š f ( a ) g ( x ) ( x Š a ) g ( x ) g ( a ) =limx ag ( a ) g ( x ) g ( a ) f ( x ) Š f ( a ) x Š a Š limx af ( a ) g ( x ) g ( a ) g ( x ) Š g ( a ) x Š a = g ( a ) f( a ) Š f ( a ) g( a ) g ( a )2. Theorem3.7nowenablesustocomputethederivativeofrational functions. Example 3.3 Let h ( x )=( x +3) / ( x2+1) .Find h( x ) Solution: Let f ( x )= x +3andlet g ( x )= x2+1.Then f( x )=1 and g( x )=2 x .So,byTheorem3.7,wehave h( x )= g ( x ) f( x ) Š f ( x ) g( x ) g ( x )2= x2+1 Š ( x +3)2 x x4+2 x2+1 = Š x2Š 6 x +1 x4+2 x2+1 14.3.Exercises.(1)Let h ( x )= exx3.Find h( x )and h( x ). (2)Let f ( x )=(2 x +7) ex.Compute f( x ). (3)Findaruletocompute( f2)( x ). (4)Findaruletocompute( f3)( x ). (5)Findaruletocompute(1 /f )( x ). (6)Usetheresultofthepreviousexercisetoproveaformulafor g( x )if g ( x )= xnfora negative integer n (7)Let g ( x )= eŠ x.Find g( x ). (8)Let h ( x )= x/ex.Find h( x ). (9)Let f ( x )=1 Š 2 x 1+3 x.Compute f( x ). (10)Let f ( x )=2 Š 3 x 1 Š 5 x.Compute f( x ). (11)Let f ( x )= ex/ ( x +2).Compute f( x ). (12)Let g ( x )= e2 x.Usetheproductruletocompute g( x ). (13)Let g ( x )= eŠ 2 x.Compute g( x ).Tryto“ndthreedierent waystoobtainyourresult.

PAGE 70

643.RULESOFDIFFERENTIATION (14)Let h ( x )=( ex+1)( ex+2).Compute h( x ). (15)Let g ( x )=( x Š 3) / ( ex+1).Compute g( x ). (16)Let f ( x )=(2 x +3) / (4 x +7).Compute f( x ).Tryto“ndtwo dierentwaysofgettingthesameanswer. (17)Let f ( x )=1 / (1 Š x ).Find f( x ). (18)Let f ( x )=1 / (1 Š x )n.Find f( x ). (19)Let f ( x )= g ( x ) h ( x ),where g isapolynomialfunctionof x and h ( x )= ex.Provethat f( x )and f( x )areeachequalto theproductofapolynomialfunctionandthefunction h ( x )= ex. (20)Provethatif f ( x )isarationalfunction,then f( x )isalsoa rationalfunction. 15.DerivativesofTrigonometricFunctions Inthissection,weshowhowtocomputethederivativesoftrigonometricfunctions.First,wecompute(sin x ).Thiswillbeasomewhat lengthyprocedure,duetothefactthatthisisthe “rst trigonometric functionwewilldierentiateandwewillhavetoapplynewmethods. However,onceweknowthederivativesofsin x andcos x ,itwillbe muchsimplertodeducethederivativesofothertrigonometricfunctions,sincethosefunctionscanbeobtainedfromsinandcos,andthen thevariousdierentiationrulescanbeused. Theorem 3.8 Wehave (sin x )=cos x Proof. Recalltheidentitysin( a + b )=sin a cos b +sin b cos a .We have (sin x )=limh 0sin( x + h ) Š sin x h =limh 0sin x cos h +sin h cos x Š sin x h =limh 0 sin x cos h Š 1 h + sin h cos x h =sin x limh 0cos h Š 1 h +cos x limh 0sin h h Notethatas, h approaches0,wecertainlyhavelimh 0sin x =sin x and limh 0cos x =cos x ,sincethesefunctionsdonotevendependon h Thereremainsthetaskofcomputingthetwonontriviallimits limh 0cos h Š 1 h andlimh 0sin h h Wewillcarryoutthistaskintwolemmas.

PAGE 71

15.DERIVATIVESOFTRIGONOMETRICFUNCTIONS65 Lemma 3.1 Wehave limh 0sin h h =1 Proof. Letusconsideracirclewithunitradiusandaregular n gonwhosecenterisatthecenter O ofthecircleandwhose n vertices areallontheunitcircle.Thentheareaofthecircleis ,andthearea ofthe n -gonis n 1 2 sin ,where =2 /n istheangle AOB ,with A and B beingadjacentverticesofour n -gon. Consideringjust1 /n ofboththecircleandthe n -gon,weseethat theareaofthetriangle AOB is(sin ) / 2,andtheareaof1 /n ofthe circleborderedbythelines AO BO ,andthearc AB is / (2 )= / 2.Sotheratioofthetwoareasis (sin ) / 2 / 2 = sin Ontheotherhand,as n getslargerandlarger, getssmallerand smaller,whiletheareaofthe n -gongetscloserandclosertothearea ofthecircle.Hence,theirratio,sin / ,willgetarbitrarilycloseto1 andstayarbitrarilycloseto1. Lemma 3.2 Theequality (3.4)limh 0cos h Š 1 h =0 holds. Proof. Wewillmanipulatetheexpression(cos h Š 1) /h sothatwe canusetheresultofLemma3.1.First,wemultiplyboththenumerator andthedenominatorbycos h +1toget cos h Š 1 h = cos2h Š 1 h (1+cos h ) = Š sin2h h (1+cos h ) Therefore,wehave limh 0cos h Š 1 h = Š limh 0sin2h h (1+cos h ) = Š limh 0sin h h sin h 1+cos h = Š limh 0sin h h limh 0sin h 1+cos h =( Š 1) 0=0

PAGE 72

663.RULESOFDIFFERENTIATION Wecannow“nishtheproofofTheorem3.8.Attheendofthe“rst displayedchainofequationsinthatproof,wesawthat (sin x )=sin x limh 0cos h Š 1 h +cos x limh 0sin h h Theprevioustwolemmasshowedthat,ontheright-handside,the“rst limitis0andthesecondlimitis1,so(sin x )=cos x asclaimed. Thefollowingtheoremcanbeprovedbyverysimilarmethods. Theorem 3.9 Theequality (cos x )= Š sin x holds. YouareaskedtoprovethistheoreminExercise15.1.1. Nowthatwehavethederivativesofsinandcos,thederivatives ofothertrigonometricfunctionscanbeobtainedbysimplyusingthe quotientrule.Thenexttheoremshowsanexampleofthis. Theorem 3.10 Wehave (tan x )=sec2x Proof. Notethattan x =sin x/ cos x ,sowecanapplythequotient rule.Thisleadsto (tan x )= sin x cos x = cos x (sin x )Š sin x (cos x ) cos2x = cosx+sin2x cos2x = 1 cos2x =sec2x. Thederivativesoftheotherthreetrigonometricfunctionsaregiven intheexercises.15.1.Exercises.(1)Provethat(cos x )= Š sin x (2)Provethat(cot x )= Š csc2x. (3)Provethat(csc x )= Š csc x cot x. (4)Provethat(sec x )=sec x tan x. (5)Let h ( x )= x sin x .Find h( x ). (6)Let h ( x )=( x2Š 2 x +3)cos x .Find h( x ). (7)Let h ( x )=tan x x +1.Find h( x ). (8)Let h ( x )= excos x .Find h( x ). (9)Let h ( x )= ex/ sin x .Find h( x ).

PAGE 73

16.THECHAINRULE67 (10)Let h ( x )= e2 xtan x .Compute h( x ). (11)Let h ( x )= eŠ 2 xcot x .Compute h( x ). (12)Let h ( x )=ex+sin x x2+1.Find h( x ). (13)Compute(sin2x ). (14)Compute(cos2x ).Trytogetthesameanswerintwodierent ways. (15)Compute(sin x tan x ). (16)Compute(cot2x ). (17)Compute(tan2x ). (18)Compute(sec2x ).Trytogetthesameanswerintwodierent ways. (19)Compute(sin3x ).Youmaywanttousetheresultofexercise13. (20)Compute(cos3x ).Youmaywanttousetheresultofexercise14. 16.TheChainRule16.1.TheDerivativeoftheCompositionofTwoFunctions.Inprevious sections,welearnedhowtocomputethederivativeofthesum,dierence,product,andquotientoftwofunctions.Westilldonotknow howtocomputethederivativeofthe composition offunctions,suchas h ( x )=sin(3 x ), t ( x )= x2+1,or r ( x )= esin x.Inthissection,we willlearnarule,calledthe chainrule ,thatappliesinthesesituations. Theorem 3.11(ChainRule) Let h ( x )= f ( g ( x )) ,where g isdifferentiableat x and f isdierentiableat g ( x ) .Then h isdierentiable at x ,andwehave h( x )= f( g ( x )) g( x ) Inotherwords,we“rstdierentiatetheoutsidefunctionatapoint givenbytheinsidefunction,thenmultiplytheresultbythederivative oftheinsidefunction. Theproofofthechainruleissomewhattechnical,sowewillpostponeituntiltheendofthissection.Nowwewilldiscusssomeexamples oftheapplicationsofthechainrule. Example 3.4 Findthederivativeof h ( x )=sin(3 x ) Solution: Let f ( x )=sin x andlet g ( x )=3 x .Then h ( x )= f ( g ( x )), so,bythechainrule,wehave h( x )= f( g ( x )) g( x )=(cos(3 x )) 3=3cos(3 x )

PAGE 74

683.RULESOFDIFFERENTIATION Example 3.5 Let h ( x )= x2+1 .Find h( x ) Solution: RecallthatinSection13,wementionedthattheidentity ( xn)= nxn Š 1holdsforanynonzero realnumber .Therefore,selecting n =1 / 2,wegetthat( x )=( x1 / 2)=1 2xŠ 1 / 2=1 2 x. Nowwecanprovethestatementoftheexample.Let f ( x )= x andlet g ( x )= x2+1.Then h ( x )= f ( g ( x )),so,bythechainrule,we have h( x )= f( g ( x )) g( x )= 1 2 x2+1 2 x = x x2+1 SometimesthechainruleiswrittenintheLeibniznotation,that is,as dh dx = dh dg dg dx .16.2.TwoApplicationsoftheChainRule. 16.2.1.ASimpleWayofObtaining(cos x ).Recallthatinthelastsection,ittookconsiderabletimeandeorttoprovethat(sin x )=cos x Finding(cos x )withsimilarmethodsisjustastime-consuming.On theotherhand,thechainruleenablesustocompute(cos x )faster. Recallthatcos x =sin x + 2 .Sowecanwritecos x asthecompositionoftwofunctions,namelycos x = f ( g ( x )),with f ( x )=sin x and g ( x )= x + 2.Sothechainruleapplies,andweget (cos x )= f( g ( x )) g( x ) =cos x + 2 1 =cos x cos 2 Š sin x sin 2 =0 Š sin x = Š sin x.16.2.2.TheDerivativesofExponentialFunctions.Recallthatwede“nedthenumber e suchthatthederivativeoftheexponentialfunction f ( x )= exwas f ( x )itself.Nowthechainruleenablesustocompute thederivativesofexponentialfunctionswith any base. Theorem 3.12 Let a beapositiverealnumberandlet h ( x )= ax. Thenwehave h( x )= axln a.

PAGE 75

16.THECHAINRULE69 Proof. Notethat h ( x )= ax=( eln a)x= ex ln a. Sowehavesucceededinwriting h asthecompositionoftwofunctions, namely h ( x )= f ( g ( x )),where f ( x )= exand g ( x )= x ln a .Therefore, thechainruleapplies,andweget h( x )= f( g ( x )) g( x )= ex ln a ln a = axln a. 16.3.ProofoftheChainRule.Itistimethatweprovedthechainrule. ProofofTheorem3.11. As g isdierentiableat x ,weknow that (3.5)limr 0 g ( x + r ) Š g ( x ) r Š g( x ) =0 Set t = g ( x + r ) Š g ( x ) r Š g( x ) Notethat t dependson r ,andas r approaches0, t approaches0. Similarly,let y = g ( x ).As f isdierentiableat y ,wehave (3.6)lims 0 f ( y + s ) Š f ( y ) s Š f( y ) =0 Set u = f ( y + s ) Š f ( y ) s Š f( y ) Again,notethat u dependson s andthat u approaches0as s approaches0. Nowweundertakeaseriesofmanipulationsoftheprecedingtwo equations.Ourgoalistoexpress f ( g ( x ))=limr 0f ( g ( x + r )) Š f ( g ( x )) r intermsof f( g ( x ))and g( x ). Rearrangingtheequationthatde“nesthevariable t thatwejust introduced,weget (3.7) g ( x + r )= g ( x )+( g( x )+ t ) r. Similarly,rearrangingtheequationthatde“nesthevariable u ,weget (3.8) f ( y + s )= f ( y )+( f( y )+ u ) s. Nowapplythefunction f tobothsidesof(3.7)toget (3.9) f ( g ( x + r ))= f ( g ( x )+( g( x )+ t ) r ) .

PAGE 76

703.RULESOFDIFFERENTIATION Observethat(3.8)holdsforall y and s ,so,inparticular,itholds when y = g ( x )and s =( g( x )+ t ) r .Makingthesesubstitutionsin (3.8),Equation(3.9)yields f ( g ( x + r ))= f ( g ( x )+( g( x )+ t ) r ) (3.10) = f ( g ( x ))+( f( g ( x ))+ u ) ( g( x )+ u ) r. (3.11) Wecannowexpressthequotient( f ( g ( x + r )) Š f ( g ( x ))) /r from theequalityoftheleft-handsideof(3.10)andtheexpressionin(3.11)as f ( g ( x + r )) Š f ( g ( x )) r = ( f( g ( x ))+ u )( g( x )+ t ) r r =( f( g ( x ))+ u )( g( x )+ t ) Finally,weareinapositiontocomputethederivativewewere lookingforasthelimitoftheleft-handsideas r approaches0.Weget limr 0f ( g ( x + r )) Š f ( g ( x )) r =limr 0( f( g ( x ))+ u )( g( x )+ t ) = limr 0f( g ( x ))+limr 0u limr 0g( x )+limr 0t = f( g ( x )) g( x ) sinceboth t and u approach0as r approaches0. 16.4.Exercises.(1)Let h ( x )=( x2+1)5.Find h( x ). (2)Let h ( x )=cot(2 x ).Find h( x ). (3)Let h ( x )=cos(2 x +8).Find h( x ). (4)Let h ( x )=sin( x2).Find h( x ). (5)Let h ( x )=sin3x .Find h( x ).Compareyourresulttothe resultofexercise19oftheprevioussection.Whichotherexercisesoftheprevioussectioncanbesolvedbythechainrule? (6)Let f ( x )=2x+3x.Find f( x ). (7)Let h ( x )= esin x.Find h( x ). (8)Let h ( x )=2cos x.Find h( x ). (9)Let h ( x )= ex2sin x.Find h( x ). (10)Let h ( x )= x2+2 x +7.Find h( x ). (11)Let h ( x )=sin( ex+5 x +6).Find h( x ). (12)Let h ( x )= e x +1.Find h( x ). (13)Let h ( x )= esin( x2).Find h( x ). (14)Let h ( x )=sin(2 x ).Find h( x ).Howcouldyougetthesame resultwithoutusingthechainrule?

PAGE 77

17.IMPLICITDIFFERENTIATION71 (15)Let h ( x )=cos(2 x ).Find h( x ).Howcouldyougetthesame resultwithoutusingthechainrule? (16)Let h ( x )=2x2.Find h( x ). (17)Let h ( x )=1 / (1 Š x ).Find h( x ).Trytogetthesameanswer intwodierentways. (18)Let h ( x )=(2+tan x )3.Find h( x ). (19)Let h ( x )= 1 Š x2.Find h( x ). (20)Let h ( x )= x + x +1.Find h( x ). 17.ImplicitDifferentiation Inthelastseveralsections,wecomputedthederivativesofmany dierentfunctions.Althoughthesefunctionsweredierent,theyhad oneimportantfeatureincommon.Theywere explicitly given.That is,theyweregivenbyarulethatdirectlydescribedhow f ( x )= y is obtainedfrom x .17.1.TangentLinestoImplicitlyDenedCurves.Sometimeswehaveto dealwithcurvesthataregivenbyadierentkindofrule.Consider thecurvegivenbytheequation (3.12) x3+ y3=4 xy. Letussaythatwewanttocomputetheslopeofthetangentline tothiscurveatthepoint(2 2).Ifwecouldexpress y asafunction of x ,wecouldsimplytakethederivativeofthatfunctionat x =2. However,itisnotclearhowtowrite y explicitly intermsof x ,evenif (3.12) implicitly describesthisdependence. Itisinthesesituationsthatweresortto implicit dierentiation. Keepinmindthatwedonotneedtoexplicitlyknowhow y depends on x ,thatis,wedonotneedanexplicitexpressionforthefunction y ( x );weonlyneedtoknowthederivative dy/dx ofthatfunctionat x =2. ConsiderEquation(3.12),anddierentiatebothsideswithrespect tothevariable x toget d dx x3+ y3 = d dx (4 xy ) Nowrecallthat y = y ( x )isafunctionof x .So,whencomputing ( d/dx ) y3ontheleft-handside,weneedtousethechainrule.Onthe right-handside,weneedtousetheproductruleandthechainrule. Usingtheserules,weget 3 x2+3 y2dy dx =4 y +4 x dy dx .

PAGE 78

723.RULESOFDIFFERENTIATION Expressing dy/dx fromthisequation,weget dy dx = (4 y Š 3 x2) (3 y2Š 4 x ) Atthepoint(2 2),theright-handsideis Š 4 / 4= Š 1,sotheslopeof thetangentlineat(2 2)is Š 1. Notethatthefactthatthetangentlineat(2 2)hasslope Š 1makes (intuitively)perfectsense,sincethecurveinquestionis symmetric in x and y .Thatis,if( x,y )isonthecurve,then( y,x )isalsoonthe curve.17.2.DerivativesofInverseTrigonometricFunctions.Oneplacewhere implicitdierentiationisaverypowerfultoolisinthecomputationof thederivativesofinversetrigonometricfunctions.RecallthattanŠ 1x = y isthefunctionthatistheinverseoftherestrictionofthefunction tan x totheinterval( Š / 2 ,/ 2).Thatis,if tanŠ 1x = y, then (3.13) x =tan y, where y ( Š / 2 ,/ 2). Ourgoalistodetermine d dx tanŠ 1x = dy dx Tothatend,letustakethederivativeofbothsidesof(3.13)with respectto x .Recallingthat d dz tan z =sec2z and y = y ( x ) weget 1=sec2y dy dx Solvingfor dy/dx andrecallingtheidentitysec2z =1+tan2z ,we obtain dy dx = 1 sec2y = 1 1+tan2y = 1 1+ x2. Inotherwords,weprovedthesuprisinglysimpleformula (tanŠ 1x )= 1 1+ x2.

PAGE 79

17.IMPLICITDIFFERENTIATION73 Thisformulaisinterestingfortworeasons.First,itissurprisinglysimple.Second,itdoesnotevencontaintrigonometricfunctions.Imagine tryingtogetthisresult without implicitdierentiation,usingjustthe de“nitionofderivatives. Youwillbeaskedtocomputethederivativesoftheotherinverse trigonometricfunctionsintheexercises.17.3.Exercises.(1)Let C bethecirclegivenbytheequation x2+ y2=169.Use implicitdierentiationto“ndtheslopeofthetangentlineto C atthepoint(5 12). (2)Let E betheellipsegivenbytheequation x2+ xy + y2=108. Useimplicitdierentiationto“ndtheslopeofthetangentline to E atthepoint(6 6). (3)Let y beimplicitlyde“nedbytheequation x5+ y5=1.Compute dy/dx (4)Let y beimplicitlyde“nedbytheequation y =sin( xy ).Compute dy/dx (5)Let y beimplicitlyde“nedbytheequation x + y = y .Compute dy/dx (6)Let y beimplicitlyde“nedby x + y =4.Compute dy/dx (7)Letusassumethat h ( x )+ x cos h ( x )= x3forallrealnumbers x .Isthissucientinformationtodeterminethevalueof h(0)? (8)Letusassumethat h ( x )+ eh ( x )=5forallrealnumbers x Provethatthen h mustbea constant function. (9)Provethat(sinŠ 1x )=1 1 Š x2. (10)Provethat(cosŠ 1x )= Š1 1 Š x2. (11)Provethat(cotŠ 1x )= Š1 1+ x2. (12)Provethat(secŠ 1x )=1 x x2Š 1. (13)Provethat(cscŠ 1x )= Š1 x x2Š 1. (14)Compute(tanŠ 1 x ). (15)Compute(cosŠ 1(2 x +0 1)). (16)Compute(sinŠ 1( x2)). (17)Compute(sinŠ 1(1 /x )). (18)Compute(tanŠ 1(2 x )). (19)Compute(cscŠ 1( x/ 3)). (20)Compute(secŠ 1( x Š 0 01)).

PAGE 80

743.RULESOFDIFFERENTIATION 18.DerivativesofLogarithmicFunctions18.1.TheFormulafor(logax ).Asanotherpowerfulapplicationofimplicitdierentiation,wecomputethederivativeofthefunction f ( x )= ln x Theorem 3.13 Wehave (ln x )= 1 x Proof. Set y =ln x .Then ey= x .Dierentiatingbothsideswith respectto x ,weget ey dy dx =1 dy dx = 1 ey. However, ey= x byde“nition,so dy dx = 1 x asclaimed. Notethatthefunctionln x isde“nedfor positive valuesof x .If x< 0,thenthefunction y =ln( Š x )isde“ned.Thereaderisinvitedtopracticethemethodofimplicitedierentiationbyshowing that(ln( Š x ))=1 /x forall negative realnumbers x .Inotherwords, if y =ln( | x | ),then dy/dx =1 /x aslongas x isa nonzero realnumber. Itisnowabreezetodeterminethederivativeoflogarithmicfunctionsof any base. Corollary 3.1 Let a =1 bea“xedpositiverealnumber.Then (logax )= 1 x ln a Proof. Notethat x = eln alogax= e(ln a )(logax ). Soln x =(ln a )(logax )and f ( x )=logax = ln x ln a Asln a isaconstant,itfollowsthat f( x )= 1 ln a (ln x )= 1 x ln a asclaimed.

PAGE 81

18.DERIVATIVESOFLOGARITHMICFUNCTIONS75 18.2.TheChainRuleandlnx.Aninterestingconsequenceof Theorem3.13isthefollowing. Corollary 3.2 Let f ( x ) beadierentiablefunctionthattakes positivevaluesonly.Then d dx ln f ( x )= f( x ) f ( x ) Proof. Bythechainrule, d dx ln f = d df df dx = f( x ) f ( x ) Example 3.6 Let f ( x )=cos x .Then d dx ln(cos x )= Š sin x cos x = Š tan x.18.3.LogarithmicDifferentiation.Sometimesweneedtocomputethe derivativeofacomplicatedproduct.Thisissometimeseasierbytaking thelogarithmoftheproduct,whichwillbeasum,andusingimplicit dierentiation.Thisprocedure,whichiscalled logarithmicdierentiation ,hastheinherentadvantagethatitdealswithsumsinsteadof products,andsumsaremucheasiertodierentiatethanproducts. Example 3.7 Let y = x3 x +1 x Š 2 Compute dy/dx Solution: Takinglogarithms,weget ln y =3ln x + 1 2 ln( x +1) Š 1 2 ln( x Š 2) Nowtakingderivativeswithrespectto x andusingCorollary3.2,we have dy dx 1 y = 3 x + 1 2( x +1) Š 1 2( x Š 2) Finally,wecansolvethisequationfor dy/dx toget dy dx = y 3 x + 1 2( x +1) Š 1 2( x Š 2) = x3 x +1 x Š 2 3 x + 1 2( x +1) Š 1 2( x Š 2)

PAGE 82

763.RULESOFDIFFERENTIATION 18.4.PowerFunctionsRevisited.Recallthatinanearliersection,we provedthatif n isa“xedpositiveinteger,then( xn)= nxn Š 1.We statedthatthiswasthecaseforallnonzerorealnumbers n ,notjust positiveintegers,butwehavenotprovedthatclaim.Nowwehavethe tools,namelylogarithmicdierentiation,toproveit. Theorem 3.14 Let n beanynonzerorealnumber.Thenwehave d dx xn= nxn Š 1. Proof. Set y = xn.Letusassumeforthecaseofsimplicitythat x ispositive.Takinglogarithms,wehave ln y = n ln x. Dierentiatingbothsideswithrespectto x ,weget dy dx 1 y = n x Solvingfor dy/dx yields dy dx = ny x = nxn x = nxn Š 1asclaimed. 18.5.TheNumbereRevisited.Recallthatwehavede“nedthenumber e ,thebaseofthenaturallogarithm,asthenumberforwhich limh 0( ehŠ 1) /h =1.Ournewknowledgeletsusexpress e moredirectly,asalimit. Notethatif f ( x )=ln x ,then f( x )=1 /x ,so f(1)=1.Bythe de“nitionofderivatives,thismeansthat limh 0ln(1+ h ) Š ln1 h =1 Observingthatln1=0andusingthepowerruleoflogarithms,weget limh 0ln(1+ h )1 /h=1 or,applyingtheexponentialfunction eztobothsides,wehave limh 0(1+ h )1 /h= e. Equivalently,setting x =1 /h ,weget limx 1+ 1 x x= e. Eitherofthelasttwoformulascanhelptodeterminetheapproximate value2.712828of e .

PAGE 83

19.APPLICATIONSOFRATESOFCHANGE77 18.6.Exercises.(1)Let h ( x )=2x+3x.Compute h( x ). (2)Let h ( x )=4Š x.Compute h( x ). (3)Let h ( x )=2 x.Compute h( x ). (4)Let h ( x )= e2x.Compute h( x ). (5)Computed dxln( x +1). (6)Compute(ln | x | ). (7)Compute( xx). (8)Let h ( x )=3 x +1.Compute h( x ). (9)Let h ( x )=4 x2+2.Computer h( x ). (10)Let h ( x )=5 2x+3.Compute h( x ). (11)Compute f( x )if f ( x )= x43 x +4 x +1. (12)Computelimx 1+1 x2 x. (13)Computelimx 1+1 xŠ x. (14)Computelimx 1 Š1 xx. (15)Let h ( x )=ln sinŠ 1x .Compute h( x ). (16)Let h ( x )=sinŠ 1(ln x ).Compute h( x ). (17)Let h ( x )=ln 1 1 Š x .Compute h( x ). (18)Uselogarithmicdierentiationto“nd y( x )= dy/dx if y ( x )= xln x. (19)Uselogarithmicdierentiationto“nd y( x )= dy/dx if y ( x )= x3 x. (20)Uselogarithmicdierentiationto“nd y( x )= dy/dx if y ( x )= x2cos x. 19.ApplicationsofRatesofChange Inthissection,weconsiderafewapplicationsofderivativesinvariousdisciplines.19.1.Physics.Recallthatifanobjectmovesalongalineandthe distanceitcoversintime t isdescribedbythefunction s ( t ),then (3.14) v ( t )= ds dt = s( t )=limh 0s ( t + h ) Š s ( t ) h isthe instantaneousvelocity oftheobjectattime t Wecantakethisconceptonestepfurther.Iftheobjectmovesat achangingvelocity,thenthe rateofchangeofthevelocity itselfcan beimportantinformation.Forinstance,whenconsideringavehicles performance,wemaybeinterestedinhowfastitcanreachitstop speed,notonlywhatitstopspeedis.

PAGE 84

783.RULESOFDIFFERENTIATION Thecorrespondingnotioninphysicsiscalled acceleration ,andis denotedby a ( t ).Thatis,keepingthepreviousnotation,wehave (3.15) a ( t )= v( t )= dv dt = s( t ) Example 3.8 Thepositionofaparticleisdescribedbytheequation (3.16) s ( t )= 1 3 t3Š 3 t2+5 t. Here s ismeasuredinmetersand t inseconds. (I) Whatisthevelocityoftheparticleafter3seconds? (II) Findtheaccelerationoftheparticleafter10seconds. (III) Whendoestheparticlemovebackward? Additionalquestionsaboutthemovementofthisparticlewillbe givenintheexercises. Solution: (I)Thevelocityoftheparticleisdescribedbythefunction v ( t )= s( t )= t2Š 6 t +5.Thisyields v (3)=9 Š 18+5= Š 4.So thevelocityoftheparticleafter3secondsis Š 4m / s,meaning thattheparticleismovingbackwardataspeedof4meters persecondafter3seconds. (II)Theaccelerationoftheparticleisgivenbytheformula a ( t )= v( t )=2 t Š 6.So,after10seconds,theparticleisaccelerating at14m / s2. (III)Theparticleismovingbackwardwhenitsvelocity v ( t )isnegative.Thathappenswhen v ( t )= t2Š 6 t +5=( t Š 1)( t Š 5) < 0, thatis,when t (1 5).Inotherwords,theparticleismoving backwardbetweenthe“rstand“fthseconds. 19.2.Economics.Letussaythatacompanyestimatesthatitcosts C ( x )dollarstoproduce x unitsofanewproduct.Itisoftenthecase that C ( x ),whichiscalledthe costfunction ,canbedescribedbya polynomialfunction,suchas C ( x )= a + bx + cx2+ dx3. Thereasonforthisisasfollows.Therewillbesomecosts,suchas designingtheproductandobtainingpermits,thatwillbepresentregardlessofthenumberofunitsproduced.Thesewillberepresentedby theconstantterm a .Thentherewillbecosts,suchasrentingalocation andbuyingsupplies,thatwillbemoreorlessindirectproportionto thenumberofunitsproduced.Thesewillberepresentedbythelinear

PAGE 85

19.APPLICATIONSOFRATESOFCHANGE79 term bx .Thentherewillbeotherfactors,suchashiringworkers,marketingtheproduct,andorganizingproduction,thatwillbeindirect proportiontoahigherpowerof x asthedierencesinsizeturninto dierencesinkind.Taxesmayfactorinatanevenhigherrate. Becausethecostfunction C ( x )isnotalinearpolynomial,producing the1001stunitdoesnotcostofthesameasproducingthe“rstunit orthe5001stunit.Thecostofincreasingproductionfrom n unitsto n +1units,inotherwords,thecostofproducingthe( n +1)thunit, canbecomputedbytheformula M ( n )= C ( n +1) Š C ( n ) The marginalcostfunction C( x )describeshowthecostfunction changes.Inthat, C( x )and M ( n )aresimilar.Thereisoneimportant dierence.Asweknow,thederivative C( x )isgivenby (3.17)lim x 0C ( x + x ) Š C ( x ) x However,itcouldwellbethatthesmallestmeaningfulpositivevalue of x is1,incasetheproductsaresuchthatfractionalunitsdonot makesense(e.g.,automobiles).Inthatcase, x 0isimpossible initsprecisemathematicalmeaning;theclosestthat x cangetto 0iswhen x =1.Inthatcase,however,theexpressionafterthe limitsymbolin(3.17)simpli“esto C ( x +1) Š C ( x ),justifyingthe approximation (3.18) M ( x )= C ( x +1) Š C ( x ) C( x ) Example 3.9 Thecostfunctionofabottleofanewmedicationis givenby C ( x )=106+20 x +0 001 x2+0 000001 x3.Findtheapproximate costofproducingthe101standthe1001stbottles. Solution: Bytheprecedingdiscussion,weneedtocomputethefunction C( x ).Bytherulesofdierentiatingapolynomialfunction,we get C( x )=0 000003 x2+0 002 x +20.Sothe101stbottlecosts 0 0003 1002+0 002 100+20=20 23dollarstoproduce,whilethe 1001stbottlecosts0 000003 10002+0 002 1002+20=43dollarsto produce. Itisimportanttonotethattheresultofthepreviousexample,that is,thefactthatitcostsmoretoproducethe1001stbottlethanthe 101stbottledoes not meanthatthemorebottlesareproduced,the moreexpensiveitistoproducethe average bottle.Thisisbecause thecostofproducingthe “rst bottleisastronomical,since C (1) > 106. Comparedtothat,thecostofeachofthe“rstthousand,oreven,“rst

PAGE 86

803.RULESOFDIFFERENTIATION tenthousandbottlesisverysmall,sotheproductionofeachofthem willbringthecostofproducingthe average bottledown.(Thecost ofproducingtheaveragebottleif n bottlesareproducedisofcourse C ( n ) /n .) Intheexercises,youareaskedtocomparetheseresultstotheresults obtainedbyusingtheformula C ( n +1) Š C ( n ).19.3.Exercises.(1)ConsidertheparticleofExample3.8.After6seconds,howfar fromitsstartingpointisthatparticle?Inwhatdirection? (2)Considertheparticleofthepreviousexercise.Arethereany momentswhentheparticleisnotmoving? (3)Thelocationofanobjectmovingverticallyisdescribedbythe function s ( t )= t Št2 5for t [0 5],wheretimeismeasured insecondsanddistanceismeasuredinmeters.Whenwillthe objecthaveaninstantaneousvelocityof0.2m/s? (4)Considertheobjectofthepreviousexercise.Whendoesit havethegreatestspeedgoingup?Whendoesithavethe greatestspeedgoingdown? (5)ConsidertheobjectofExercise19.3.3.Willitsacceleration everbe1m/s2? (6)ConsidertheobjectofExercise19.3.3.Whenwillitsaccelerationbenegative? (7)Usetheformula M ( n )= C ( n +1) Š C ( n )to“ndthecostof producingthe101stand1001stunitsinExample3.9.Compareyourresultswiththeestimatesthatwefoundusingthe function C( x ). (8)Tworacecarsspeedupfromastandingstartto60m/sso thateachcarhasconstantacceleration.The“rstcarreaches one-thirdofitstopvelocityin4seconds,whilethesecondcar reachesone-fourthofitstopvelocityin3seconds.Whichcar willhavecoveredmoredistancebythetimeitreachesitstop velocity? (9)Aballisrollingdownaslopesothatitsdistancefromits startingpointisdescribedbythefunction s ( t )=2 t2+6 t where0 t 10,thetime t ismeasuredinseconds,andthe distance s ( t )ismeasuredinmeters.Whatwillbethevelocity oftheballafter3seconds? (10)Considertheballofthepreviousexercise.Whenwillitsvelocityreach40m / s?

PAGE 87

19.APPLICATIONSOFRATESOFCHANGE81 (11)Considertheballofthepreviousexerciseanddescribeitsaccelerationasafunctionofthetime t passedfromthestartof theballsmovement. (12)Thecostfunctionforacompanytoproduce x bicyclesis C ( x )=2400+3 x +0 6 x2+0 002 x3.Findthemarginalcost functionofthisproduct. (13)Considerthecostfunction C ( x )ofthepreviousexerciseand useittodeterminetheactualcostofproducingthe1001st bicycle. (14)Thecostfunctionforacompanytoproduce x laptopsis C ( x )=1500+2 x +0 4 x2+0 01 x3.Findthemarginalcost functionforthisproduct. (15)Considerthecostfunctionofthepreviousexercise.Explain themeaningof C(200). (16)Letussaythatthefunction f ( t )describesthegrowthofa certainbacteriapopulationovertime.Thatis, f (0)isthesize ofthepopulationatthebeginningoftheobservationperiod, while f ( t )isthesizeofthepopulation t hoursafterthat. Explainwhy f( t )describesthe growthrate ofthepopulation inthemomentoftimecorrespondingto t (17)Acertaininsectpopulationhasbeenexposedtoaninsecticide,whichresultsinthepopulationchangingaccordingto thefunction f ( t )=10 000 Š 1000 t Š 500 t2,where t ismeasuredinhours.Findthegrowthrateoftheinsectpopulation after1hourandafter5hours. (18)Acertainbacteriapopulationhasaninitialsizeof1000,and itdoublesineachhourforthenext10hours.Describethe growthofthisbacteriapopulationbyafunction,thenuse thatfunctiontodeterminethegrowthrateofthepopulation after3.5hours. (19)Acertainrumorspreadsaccordingtothefunction (3.19) P ( t )= 1 1+30 eŠ 2 t. Here P ( t )istheproportionoftheobservedpopulationthat heardtherumor t daysafteritstartedcirculating.Explain why P( t )shouldbede“nedastherateatwhichtherumoris spreading.Thencomputethatrateat t =4. (20)Considertherumordiscussedinthepreviousexercise.If(3.19) remainscorrectas t goestoin“nity,whatfractionoftheobservedpopulationwilleventuallyheartherumor?

PAGE 88

823.RULESOFDIFFERENTIATION 20.RelatedRates20.1.Preliminaries.Anintuitiveideaofthenotionof relatedrates comesfromasimplefactofeverydaylife:Iftherearetworelatedquantitiesthatarechangingwithtime,thentheirratesofchangeshould alsoberelated.Forexample,thevolume V ofwaterinapoolofarea 20m2isrelatedtothewaterlevel h (thepooldepthinmeters)as V =20 h .Supposethewaterlevelislowandneedstobeincreased.A hoseisputintothepoolthatcanpumpwateratarateof0 2m3/ h. Atwhatratedoesthewaterlevelincrease?Thevolumeandthewater levelarebothfunctionsoftime, V = V ( t )and h = h ( t ).Forevery instanceoftime t ,theirvaluesarerelatedas V ( t )=20 h ( t )andso mustbetheirderivativesorratesofchange: (3.20) V ( t )=20 h ( t )= V( t )=20 h( t ) Nowthequestioniseasytoanswer.Since V( t )=0 2m3/ h, h( t )= V( t ) / 20=0 01m / h=1cm / h.Thewaterlevelrisesby1cmevery hour.Asomewhatpracticalestimate!Youwouldknow exactly when tocomebackandturnothewaterifyouneededaninchorsoofthe waterlevelincrease.Apparently,thesameideaofrelatedrateswould workforloweringthewaterlevelafterrain.20.2.Units.Itisimportanttobringallthequantitiestothesame systemofunits.Forexample,intheaboveproblemthepoolarea isoftengiveninsquarefeet,forexample,200ft2,whilethepumpingrateisgiveningallonsperhour,forexample, V=60gal / h. Onegallonis3 785 10Š 3m3andtherefore V=60 3 785 10Š 3= 0 2271m3/ h.Onesquarefootis9 29 10Š 2m2,sothepoolareais 200 9 29 10Š 2=18 58m2.Hence, h=0 2271 / 18 58 1 2cm / h. In1999,NASAlosta$125millionMarsorbiterbecauseaLockheed MartinengineeringteamusedEnglishunitsofmeasurementwhilethe agencysteamusedthemoreconventionalmetricsystemforakey spacecraftoperation.20.3.FormalDenitionofRelatedRates.Definition 3.2(Relatedquantities) Twoquantities y and x are saidtoberelatedifthereisafunction f suchthat y = f ( x ) Inthepreviousexample, V = f ( h )=20 h .Supposenowthatthe quantities y and x arefunctionsofanothervariable t (e.g., t istime): x = x ( t )and y = y ( t ).Thentherateofchangeof x or y withrespect to t isnothingbutthederivative x( t )or y( t ).Theproblemofrelated

PAGE 89

20.RELATEDRATES83 ratesŽcannowbecastinthepropermathematicalterms:Whatisthe relationshipbetweenthederivatives x( t )and y( t )ifthevaluesof x ( t ) and y ( t )arerelatedby y = f ( x )?Thevaluesofthefunctions x ( t )and y ( t )arerelatedas y ( t )= f ( x ( t ))forany t .Takingthederivativeof bothsideswithrespectto t bymeansofthechainrule(Theorem3.11), weobtainageneralizationof(3.20): (3.21) y ( t )= f ( x ( t ))= y( t )= f( x ( t )) x( t ) Equation(3.21)establishesthesought-afterrelationbetweentherates yand x.However,itseemssomewhatdierentfrom(3.20):Therates arestillproportionaltooneanother,buttheproportionalitycoecient f( x )isnolongeraconstant,butafunction.Howdoweuseit?Take aparticularvalueof t = t0.Letthevaluesof x and y at t = t0be x0= x ( t0)and y0= y ( t0).Thenumber a = f( x0)canbecalculated.Then theequality y( t0)= ax( t0)determinestherelationbetweentherates yand xattheinstancewhen x hasthevalue x0(or y hasthevalue y0= f ( x0)). Example 3.10 Letalaserpointerbepositionedatadistance D =1 mfromawall.Thepointercanberotatedsothatthebright spotcreatedbythelaserbeamtravelshorizontallyonthewall. (I) Atwhatspeeddoesthebrightspottravelalongthewallifthe pointerrevolvesataconstantrate rad/s? (II) Atwhatdirectionofthelaserbeamdoesthebrightspottravel atthespeed v =4 m/s if = rad/s ? Solution: (i)Theanalysisofanyproblemonrelatedratesmustbeginwith de“ningthequantitieswhoseratesarebeingstudied.Inother words,onehastoanswerthequestion:Howarethesequantitiesmeasured?Theorientationofthelaserbeamcanbe describedbytheangle betweentheperpendiculartothe wallandthelaserbeam.Thepositionofthebrightspotmay besetbythedistance y traveledbyitfromthepointonthe wallwhen =0,thatis,whenthelaserbeamisperpendiculartothewall.Ifthepointerrotates,theanglebecomesa functionoftime, = ( t ),andsodoesthepositionofthe brightspot, y = y ( t ).Thus,thequestionisabouttherelation betweentherates y( t )= v (thespeedatwhichthebrightspot travels)and ( t )= (therateatwhichthepointerrotates). (ii)Thenextstepisto“ndafunctionthatdeterminestherelation betweenthequantitiesofinterest,thatis,betweenthedistance

PAGE 90

843.RULESOFDIFFERENTIATION D y v y Figure3.1. Alaserpointerispositionedatadistance D fromawallandrotatesclockwise.Itsbeammakesa brightspotthatmoveswiththespeed v totherightalong thewall.Thelaserbeamdirectionisdeterminedbythe angle andthepositionofthebrightspotisdetermined bythedistance y y andtheangle : y = f ( ).Itisclearthat D and y are relatedasthecathetioftherighttrianglewhosehypotenuse isthelaserbeam: y = D tan = f ( ). (iii)Oncetherelationbetweenthequantitiesofinteresthasbeen established,therelationbetweentheirratescanbefound. Since(tan )=1 / cos2 ,Equation(3.21)yields (3.22) y = D tan = y= D cos2 = v = D cos2 The“rstquestionisanswered. (iv)Notethattherate y= v isnotconstanteveniftherate = isconstant.Toanswerthesecondquestion,onehasto“nd thevalueof when v =4 m / s, D =1m,and = rad / s. ItfollowsfromEquation(3.22)that cos2 = D v = 1 4 = = 3 ;

PAGE 91

20.RELATEDRATES85 thatis,thebrightspotmovesatthespeed4 m / swhenthe laserbeammakes60withtheperpendiculartothewall. 20.4.CanAnythingTravelFasterThanLight?Thesolution(3.22)has aninterestingfeature.When approaches90,thatis,thelaserbeam isgettingclosertobeingparalleltothewall,thecosine,cos ,tendsto 0inEquation(3.22),andhencetherate y= v growsunboundedly.It seemslikejustwithmerelyalaserpointer,asuperluminalobjectcan becreatedinalecturehall!Letusinvestigatethis.Thespeedoflightis c 300 000km / s 186 000mi / sec.Thelightcanmakeatriparound theworldinmerely0 13seconds!Example3.10isnowsupplemented bytwoadditionalquestions: (III) Isitpossiblethat v canexceedthespeedoflight?Ifso,at whichdirectionofthelaserbeamdoesithappen? (IV) Atwhichpositionofthebrightspotdoesithappen? Theanswersread: (III)Setting D =1m=10Š 3km(watchtheunits:alldistances arenowinkilometers!)and v = c =3 105km / s,theangle atwhichthebrightspotexceedsthespeedoflightsatis“es theequationcos2 = D/c 1 05 10Š 8,andhence 89 99414.Sothebrightspotbecomessuperluminalif > 89 99414! (IV)Since y = D tan v>c if y> 9772m.Well,alecture hallappearstobeabitŽsmallforthisexperiment!Take aDremelminiaturegrinder(soldinLowesstores)forwhich 103rad / s(itcanbeusedtorotatethepointer),andset D =0 1m,then v>c if y> 98m;notyetexactlyalecture hallexperiment,butitcanbemanagedonthecampus! Einsteinstheoryofrelativitystatesthatnomaterialobjectcan travelfasterthanlight.HasacounterexampletoEinsteinstheoryjust beenfound?Theanswerisno.ŽInthemotionofthebrightspot, nomaterialobjectactuallymovesalongthewall.Brightspotsat y and y + y arecreatedby dierent portionsofthelaserbeamthat areemittedbythelaserattwodistinctmomentsoftime.Alumpof lightthatarrivedat y wasre”ectedbythewall(thatiswhywesee thebrightspot!),andhenceitcouldnotappearatthenextposition y + y (atthispositionarriveda dierent lumpoflightemittedbythe laseratalatertime).Sotherate y/ t cannotpossiblybeassociated withthemotionofanymaterialobjectalongthewall.

PAGE 92

863.RULESOFDIFFERENTIATION 20.5.RelatedProblem.ThenexttimeyouwatchaFloridasunset,look atyourshadow.DoesthereexistapositionoftheSunabovethe horizonatwhichyourshadowextendsfasterthanthespeedoflight?20.6.MoreThanTwoRelatedRates.Therearesituationswhenseveral quantitiesarerelatedamongthemselves.Ifthesequantitiesbecome functionsofavariable t ,thentheirratesare linearly related.Aproof ofthisstatementisgiveninCalculus3,wherefunctionsofseveralvariablesarestudied.However,thebasicideaof“ndingrelationsbetween therateshasnotchanged:Theyareobtainedbydierentiatingthe relationsbetweenthequantitiesinquestionwithrespectto t .The procedureisillustratedinthefollowingexample. Example 3.11 Considerarectanglewithsides x and y .Supposethat x and y changewithtime.Findtheirratesofchangewhen x =3 cm and y =1 cm if,atthatmoment,theareaoftherectangle decreasesatarateof 2 cm2/s whiletheperimeterdoesnotchange. Solution: (i)Therearefourquantitiesinvolved:therectangledimensions x and y ,thearea S ,andtheperimeter P (ii)Therearetworelationsbetweenthem: S = xy,P =2( x + y ) (iii)If x = x ( t )and y = y ( t ),then S ( t )= x ( t ) y ( t )and P ( t )= 2( x ( t )+ y ( t )).Usingthederivativeoftheproductandthe sumoftwofunctions,thelinearrelationsbetweentherates areobtained S= xy + xy,P=2( x+ y) (iv)Since P=0(theperimeterdoesnotchange), x= Š yand S=( x Š y ) y.Nowlet S= Š 2cm2/ sbecause S decreases ( Smustbenegative).With x =3cmand y =1cm,one has Š 2=(3 Š 1) yand y= Š 1cm / s.Itthenfollowsthat x= Š y=1cm / s. 20.7.Exercises.(1)Consideratrianglewithvertices A B C suchthat | AB | = | AC | =2cm.Let betheangleatthevertex A .Iftheangle decreasesattherate0 3rad/s,whatistherateofchangeof thelengthoftheside BC attheinstancewhen = / 3.

PAGE 93

20.RELATEDRATES87 (2)Consideratrianglewithvertices A B C suchthat | AB | = | AC | =3cm.Let betheangleatthevertex A .Iftheangle increasesattherate0 3rad/s,whatistherateofchangeof theareaofthetriangleattheinstancewhen = / 3. (3)Thesidesofarectanglechangewithtimesuchthatthearea oftherectangledoesnotchange.Iftherateofchangeofone sideis1m/s,“ndtherateofchangeoftheothersideatthe instancewhenbothsidesareofequallength. (4)Thesidesofarectanglechangewithtimeattherates1m/s and Š 3m/s.Findtherateofchangeofthediagonalofthe rectangleattheinstancewhenbothsidesareofequallength. (5)Atwhatratedoestheareaofadiskincreaseattheinstance whentheradiusis R =10cmiftheradiusincreasesata constantrateof2cm/s? (6)Atwhatratedotheareaandthelengthofthediagonalofa rectanglechangeattheinstancewhenonesideis x =20m andtheotheris y =15miftheformerisdecreasingatarate of1m/s,whilethelatterisincreasingatarateof2m/s? (7)Twoships, A and B ,leaveaharboratthesametime,one headingnorthandtheotherheadingeast.Atwhatrateisthe distancebetweentheshipsincreasingifthespeedofship A is 30km/handthespeedofship B is40km/h? (8)Thesurfaceareaofaballisincreasingataconstantrateof4 m2/min.Atwhatratedotheradiusandvolumeoftheball changeattheinstancewhentheballhasradius3m? (9)Aladder24ftlongleansagainstaverticalwall.Ifthelower endisbeingmovedawayfromthewallatarateof3ft/sec, howfastisthetopdescendingwhenthelowerendis8ftfrom fromthewall?Whenarethelowerandupperendsmovingat thesamerate? (10)Aman6fttallwalksawayfromanarclight15fthighata rateof3milesperhour.Howfastisthefartherendofhis shadowmoving?Howfastishisshadowlengthening? (11)Thevolumeofasphereisincreasingatarateof16cm3/s. Howfastistheradiusincreasingwhenitis6cm?Howfastis thesurfaceareaincreasingwhenitis36cm2? (12)Sandisbeingpouredonthegroundfromanelevatedpipeand formsapilethatalwayshastheshapeofacircularconewhose heightisequaltotheradiusofthebase.Ifthesandfallsata rateof0.5m3/min,howfastistheheightofthepileincreasing whenitis2m?

PAGE 94

883.RULESOFDIFFERENTIATION (13)Aparticlemovesalongthecurvede“nedbythealgebraicequation x2Š 2 y3=9sothatthecoordinate x increasessteadily atarateof3unitsoflengthpersecond.Findtherateof changeofthecoordinate y whentheparticleisatthepoint ( x,y )=(5 2). (14)Thevelocityofaparticlemovingalongastraightlinesatis“es thecondition v2= c +2 b/s ,where a and b areconstants and s isthedistancetraveledbytheparticle.Showthatthe acceleration(therateofchangeinvelocitywithrespectto time)is a = dv/dt = Š b/s2. (15)Considertwolines y + x =2 a and y Š x =0,where a isanumber.Supposethataparticlemovesalongthe“rstlinetoward thepointofintersectionofthelinesataconstantspeed v1, whileanotherparticlemovesalongthesecondlineinthedirectionawayfromthepointofintersectionataconstantspeed v2. Findtherateofchangeofthedistancebetweentheparticles whenthe“rstandsecondparticlesareatthedistances s1and s2fromthepointofintersection,respectively.Inparticular, whatisthevalueofthisrateif s1= s2and v1= v2? (16)Thebladesofapairofscissorshavewidth2 h .Findtherate atwhichthepointofintersectionoftheedgesoftheblades ismovingiftheanglebetweenthebladesdecreasesataconstantrate .Assumethatthebladesareattachedbyascrew throughthemidpointofeachblade(i.e.,throughapointthat isatdistance h fromtheedgesoftheblade).If h =4mmand = Š 2rad/s,howlongshouldthebladesbetoseethepoint ofintersectiongoingsuperluminal? (17)If y2=2 x and x isdecreasingsteadilyatarateof0.25units persecond,“ndhowfasttheslopeofthegraphischangingat thepoint( x,y )=(8 Š 4). (18)Apoolhasasphericalbottomofradius R andthemaximal depth h
PAGE 95

21.LINEARAPPROXIMATIONSANDDIFFERENTIALS89 increasesatarateof2m/s.Findtherateofchangeofthe angle andtheradius R atthemomentwhen =30and R =10m. 21.LinearApproximationsandDifferentials21.1.TangentLineApproximation.Thederivativeofafunction f ( x ) atapoint x = x0de“nestheslopeofthelinetangenttothegraph y = f ( x )atthepoint( x0,f ( x0))(seeEquation(2.5)).Theequation ofthetangentlineis y Š f ( x0) x Š x0= f( x0)or y = f ( x0)+ f( x0)( x Š x0) Definition 3.3 Suppose f ( x ) isdierentiableat x = x0.The linearfunction (3.23) L ( x )= f ( x0)+ f( x0)( x Š x0) iscalledthe linearization of f ( x ) inaneighborhoodof x0. Sincethevaluesof f and L coincideat x = x0,onemightexpect thatthedierence f ( x ) Š L ( x )issmall,provided x iscloseenoughto x0.Moreprecisely,considerthelimit limx x0f ( x ) Š L ( x ) x Š x0=limx x0f ( x ) Š f ( x0) x Š x0Š f( x0)= f( x0) Š f( x0)=0 wherethede“nitionofthederivativehasbeenused.Thisshowsthat theerroroftheapproximationof f ( x )by L ( x )decreasestozero faster than x Š x0as x approaches x0: f ( x )= L ( x )+( x Š x0) ( x Š x0) where ( x Š x0) 0as x x0Sothelinearfunction L ( x )maybeusedtoapproximatevaluesof f ( x ) inasmallneighborhoodof x0.Thisapproximationiscalledthe linearapproximation or tangentlineapproximation .Theconceptofthe tangentlineapproximationisillustratedinFigure3.2. Example 3.12 Usethelinearapproximationtoestimatethe value 3 92 Solution: (i)Consider f ( x )= x .Theclosestvalueof x to3 92atwhich thesquarerootcanbeevaluatedwithoutacalculatoris x0=4: f ( x0)=2.Notethetwoimportantstepshere:thechoiceof f ( x )suitablefortheproblemandthechoiceof x0nearwhich thelinearapproximationistobeused.

PAGE 96

903.RULESOFDIFFERENTIATION f ( x0)x0x0x0x yy L ( x ) y f ( x ) Figure3.2. Tangentlineapproximation.Inaneighborhoodof x0(aninterval[ x0Š ,x0+ ]),thetangent line y = L ( x )staysclosetothegraph y = f ( x ).Byreducingthewidthoftheinterval ,onecanmaketheerror ofthetangentlineapproximationassmallasdesired, i.e., | f ( x ) Š L ( x ) | forall x [ x0Š ,x0+ ]. (ii)Since f( x )=( x )=1 / (2 x )and f(4)=1 / 4,byEquation (3.23)thelinearizationof x near x =4is L ( x )=2+ 1 4 ( x Š 4) (iii)Thelinearapproximationmeansthatthevalue f (3 92)= 3 92isapproximatedbythevalue L (3 92): 3 92 L (3 92)=2+ 1 4 (3 92 Š 4)=1 98 Acalculatorgives 3 92 1 9799.Sotheapproximationerroris | 3 92 Š L (3 92) | < 1 02 10Š 4.Itiseasytoseethat L (4 08)=2 02 and | 4 08 Š L (4 08) | < 1 02 10Š 4.Innotationsgiveninthecaption ofFigure3.2,thisobservationcanbesummarizedbythefollowing inequality: | x Š L ( x ) | < 1 02 10Š 4= if | x Š 4 | 0 08= Inotherwords,thevaluesof x anditslinearizationdierbynomore than1 02 10Š 4forall3 92 x 4 08.Naturally,adecrease(increase) intheupperboundfortheerrorwouldleadtoadecrease(increase)in thesizeofaneighborhoodof x =4wherethelinearapproximationis accurate.

PAGE 97

21.LINEARAPPROXIMATIONSANDDIFFERENTIALS91 21.2.AccuracyoftheLinearApproximation.Thepreviousexampleleads toaproblemthatisextremelyimportantinapplications:Givenanupperboundfortheerror ofthelinearapproximationofafunction f ( x ) near x0,“nd suchthat | f ( x ) Š L ( x ) | if | x Š x0| or,alternatively,given ,thatis,theneighborhood x0Š x x0+ estimatetheerror ofthelinearapproximation.Thefollowingtheorem isusefultoanswerthesequestions. Theorem 3.15 Supposeafunction f ( x ) istwicedierentiablein ( a,b ) suchthat | f( x ) | M forall x ( a,b ) andsomenumber M Let L ( x ) bethelinearizationof f ( x ) at x0 ( a,b ) .Then | f ( x ) Š L ( x ) |1 2M ( x Š x0)2,x ( a,b ) ThistheoremisasimplerversionoftheTaylortheorem,which isprovedinadvancedcalculuscourses.Thefollowingexampleillustratestheuseofthistheoremtoassesstheaccuracyofthelinear approximation. Example 3.13 Considerthelinearizationof sin x at x =0 .Find aninterval | x | inwhichtheerrorofthelinearapproximationdoes notexceed =0 5 10Š 3. Solution: (i)Since f( x )=(sin x )=cos x f(0)=1,and f (0)=0,the linearizationis L ( x )= x (ii)InTheorem3.15,let a = Š and b = .Next,onehasto“nd M .Thesimplestwaytodothisistotakethemaximalvalue of | f( x ) | intheinterval | x | .Notethatthereshouldbe
PAGE 98

923.RULESOFDIFFERENTIATION followsfromEquation(3.24)thatalargervalueof M leadsto asmaller .Sothisoptionshouldnotbeabused.ŽAgood M isnottoolargeandyetissimpleenoughtosolveEquation (3.24).Thisrequiressomeskillstoachieve. (iv)Agoodcompromiseistousetheinequalitysin .So thechoice M = alsoful“llstheconditionsofTheorem3.15. Equation(3.24)becomes 3=10Š 3and =0 1,whichisto becomparedwith =0 0362when M =1and 0 100057 when M =sin Theconverseproblemissimpler:Findanupperboundforthe errorofthelinearapproximationofsin x at x =0intheinterval | x | 0 2.Bymonotonicityofsin x intheinterval( Š / 2 ,/ 2), | (sin x )| = | sin x | sin(0 2)= M for | x | 0 2and,hence, | sin x Š x | =1 2M2=0 5 sin(0 2) (0 2)2 3 9734 10Š 3.21.3.Differential.Forarealvariable x ,the dierential dx isde“ned asanincrementof x .Itcanbegiventhevalueofanyrealnumber independentlyofthevalueof x ;thatis, dx isconsideredasan independentvariable .So,witheveryrealvariable,onecanassociateanother realvariable,calledthe dierential .Iftworealvariablesarerelated, thefollowingrulepostulatestherelationbetweentheirdierentials. Definition 3.4 Lettwovariables y and x berelatedas y = f ( x ) where f isadierentiablefunction.Thedierential dy = df ( x ) is de“nedbythelineartransformationof dx : (3.25) dy = df ( x )= f( x ) dx. Notethatthevariables x and dx ontheright-handsideareindependent variables.Equation(3.25)statesthat,ifthevariables y and x are related,thenthedierential dy isnolongeranindependentvariable andisdeterminedby x and dx ;speci“cally, dy depends linearly on dx .21.4.GeometricalSignicanceoftheDifferential.Put dx = x ,where x isarealnumber.Fix x = x0andconsideranincrementofthe variable y = f ( x )between x0+ x and x0: y = f ( x0+ x ) Š f ( x0)= f ( x0) Thedierential df ( x0)= f( x0) x doesnotgenerallycoincidewiththe increment f ( x0).Forexample,put f ( x )= x2, x0=1, x =0 2,then f (1)=(1+0 2)2Š 1=0 44,whereas df (1)= f(1) x =2 0 2=0 4.

PAGE 99

21.LINEARAPPROXIMATIONSANDDIFFERENTIALS93 f ( x0 x ) y L ( x ) y f ( x ) y df f f ( x0) x0L ( x0 x ) x0 x x Figure3.3. Geometricalsigni“canceofthedierential. Thedierential df ( x0)= f( x0) dx istheincrementalong thetangentline: df ( x0)= L ( x0+ x ) Š L ( x0), dx = x .Thedierential df ( x0)doesnotcoincidewiththe incrementofthefunction f ( x0)= f ( x0+ x ) Š f ( x0) = df ( x0).Onlywhen x becomesin“nitesimallysmall, x 0,doesitcoincideuptotermsthatgoto0faster than x ,i.e.,[ f ( x0) Š df ( x0)] / x 0as x 0. Sincethederivative f( x0)determinestheslopeofthetangentline L ( x )= f ( x0)+ f( x0)( x Š x0)tothegraph y = f ( x ),thedierential df ( x0)istheincrementofthelinearization y = L ( x )ofthefunction at x = x0intheinterval[ x0,x0+ x ];thatis,foraparticularvalue x = x0andanarbitrarilychosenincrement dx = x df ( x0)= L ( x0+ x ) Š L ( x0)= f( x0) x. Thus, df ( x0) = f ( x0)becausethetangentlinedoesnotgenerally coincidewiththegraph.ThisobservationissummarizedinFigure3.3. Inparticular,thetangentlineapproximationcannowbestatedas f ( x0+ x ) L ( x0+ x )= f ( x0)+ df ( x0) ,dx = x. Anintuitiveunderstandingofthedierentialstemsfromitsgeometricalinterpretation.Let x tendto0.Theratio y Š dy x = f ( x ) Š f( x ) x x = f ( x ) x Š f( x ) 0 convergestozeroas x 0becausebytheexistenceof f( x ), f ( x ) / x f( x )as x 0.Thismeansthatthedierence y Š dy mustgoto0fasterthan x .Anincrement x issaidtobe in“nitesimallysmall if( x )n, n> 1,canalwaysbeneglected.Soonemight thinkofdierentialsasin“nitesimalvariationsofvariables.Fromthis

PAGE 100

943.RULESOFDIFFERENTIATION pointofview,thede“nition(3.25)looksrathernatural:In“nitesimal variationsoftworelatedvariablesmustberelatedlinearlyastheir higherpowerscanalwaysbeneglected.Theconceptofthedierential becomesratherpracticalwhenonehastoestablishrelationsbetween variationsofrelatedquantitiesinsituationswhenthesevariationsmay beviewedasin“nitesimal.21.5.InverseFunctionandtheDifferential.Theconceptofthedierentialoersasimplewayto“ndthederivativeofaninversefunction. Supposethatafunction f hastheinverse g = fŠ 1and g isdierentiable(conditionsunderwhich g existsandisdierentiablearestated laterintheinversefunctiontheoremofSection23).If y = f ( x ), thenthedierentialsarerelatedas dy = f( x ) dx .Ontheotherhand, x = fŠ 1( y )= g ( y )andhence dx = g( y ) dy .Since theratioofthe dierentialsisthederivative ,itfollowsthat dx dy = 1 dy dx g( f ( x ))= 1 f( x ) g( y )= 1 f( g ( y )) Forexample, f ( x )=tan x and g ( y )=tanŠ 1y .Then f( x )= 1 cos2x =1+( f ( x ))2= g( y )= 1 f( g ( y )) = 1 1+ y2, wheretherelation f ( g ( y ))= y hasbeenused.21.6.RelatedErrors.Everyphysicalquantityisknownonlywitha certaindegreeofaccuracy.Errorsareinherentintheveryprocessof takingmeasurements.Asapointoffact,avalueofaphysicalquantitygivenwithoutitsmeasurementerrordoesnotmakemuchsense; neithershouldonedrawanyconclusionfromdatawithoutaproper analysisoftheerrors.Oneoftheimportantpracticalapplicationsof thedierentialliesintheerroranalysisofrelatedquantities. Supposethereisarelationbetweentwoquantities y and x y = f ( x ).Let x bemeasuredwithanerror.Thismeansthefollowing.After n measurements,onegets n values x1,x2,...,xn.Theaverageis x = ( x1+ x2+ + xn) /n isviewedastheactualvalue.Themeasuredvalues deviatefromtheaveragebyamounts x1= x1Š x,..., xn= xnŠ x If x =max {| x1| ,..., | xn|} (i.e., x isthemaximalofthe absolute errors ),thenallmeasuredvalueslieintheinterval[ x Š x,x + x ].The quantity x isthe maximumuncertainty inthevalueof x (oran error bound ).Onewritesforthemeasuredvalue x x toindicatethe averagemeasuredvalueanditsmaximumuncertainty.Thenumber

PAGE 101

21.LINEARAPPROXIMATIONSANDDIFFERENTIALS95 x isusuallyknownanddeterminedbytheveryprocessoftaking measurements. Astandardquestioninerroranalysis:Whatistheaccuracyof thevalue y = f ( x )?Apparently, x and x areindependentvariables astheerrorbound x dependsonthewayinwhichthevariableis measured(therearemoreandlessaccuratemethodswhichwouldlead tosmallerandhighervaluesof x independentlyofthevalueof x ). Naturally,onemightassumethattheerrorsaresmall;thatis,they arein“nitesimalvariationsofmeasuredquantities.Thentheerrorsof therelatedquantitiesmustberelatedastheirdierentials.Thisisa standardassumptionoftheerroranalysis.Inotherwords,if y = f ( x ) where x isthemeasuredmeanvalue,thentheerrorinthevalueof y ineachmeasurementisassumedtobe yi= yiŠ y = f ( xi) Š f ( x )= f ( x + xi) Š f ( x ) = df ( x )= f( x ) xiwhichisnothingbut dy = df ( x )where dy = yiif dx = xi.The absolutevalueofthedierential | dy | representsan absolute errorof y = f ( x ).Thequantity | dy/y | 100%iscalleda relativeerror .The absoluteandrelativeerrorboundsare,respectively y = | f( x ) | x, y | y | 100%= | f( x ) | | f ( x ) | x 100% andonewritesforthemeasuredvalue y y toindicatethemaximum uncertaintyinthevalueof y Example 3.14 Whataretheabsoluteandrelativeerrorboundsof thevolumeofacubeifitssideis 10 0 1 cm ? Remark .Whenmeasuringthelengthbyarulerwithagrid,the measurementerrorshouldnotexceedtherulergridspacing(e.g.,a rulerwithamillimetergrid). Solution: Thevolume V andside x arerelatedas V = x3.So dV = 3 x2dx .Setting dx =0 1cmand x =10cm, dV =30cm3and V =1000 30cm3.Therelativeerrorboundis dV/V =0 03or3% (notethat dx/x =0 01,i.e.,only1%). Theerroranalysisforseveralrelatedquantitiesisstudiedinmultivariablecalculuscourses.Itisbasedontheconceptofthedierential offunctionsofseveralvariables.

PAGE 102

963.RULESOFDIFFERENTIATION 21.7.Exercises.(1)Findthelinearizationsofeachofthefollowingfunctionsatthe speci“edpoint: (i)cos x x = / 4 (ii)tan x x =0 (iii) eŠ x2, x =0 (iv)ln x x = e (v) 1+ x x =3 (2)Estimatetheerrorofthetangentlineapproximationofeach ofthefollowingfunctionsoveraninterval | x Š x0| forthe speci“edpoint x0andthewidth : (i) 1+ x x0=3, =0 1 (ii)ln x x =1, =0 2 (iii)tan x x =0, = / 4 (3)Findthedierentialsofeachofthefollowingfunctions: (i) x (1 Š x2)3(ii)( y Š 2) / ( y +1) (iii) 1+ x2/x (iv)sin2t +cos( t2) (v)ln( x +2)+ xex(vi) aŠ 1tanŠ 1( x/a ), a =0 (vii)ln | x + x2+ a | (viii)(2 a )Š 1ln | ( x Š a ) / ( x + a ) | (4)Usedierentials(thetangentlineapproximation)toestimate thefollowingnumbersandassesstheaccuracyoftheestimates: (i) 24 6 (ii) e0 08(iii)3 1 02 (iv)sin29(v)tanŠ 11 05 (vi)log1011 (5)Provetheapproximationformulan an+ x a + x nan Š 1,a> 0 anduseittocalculateapproximatelythenumbers (i)3 9 (ii)4 80 (iii)7 100 (iv)10 1000

PAGE 103

21.LINEARAPPROXIMATIONSANDDIFFERENTIALS97 (6)Calculate f (1)and df (1)forthefunction f ( x )= x3Š 2 x +1 andcomparetheminthefollowingthreecases: x =1, x = 0 1,and x =0 01. (7)Let u v ,and w bedierentiablefunctions.Find dy if (i) y = uvw (ii) y = u/v2(iii) y =( u2+ v2)Š 1 / 2(iv) y =ln u2+ v2(v) y =tanŠ 1( u/v ) (8)Find dy intermsof x y ,and dx if (i) x + y =4 (ii) y3+ x3=2 xy (iii)cos( x +3 y )=sin( xy ) (9)Findanapproximateformulafortheareaofacircularringof radius r andwidth dr .Whatistheexactformula? (10)Findanapproximateformulaforthevolumeofaspherical shellofradius r andthickness dr .Assesstheaccuracyofthe approximationbystatingtheconditionon r and dr sothat therelativeerrordoesnotexceed =0 01(i.e.,1%). (11)Whatisanadmissiblerelativeerrorinmeasurementsofthe radiusofaballinorderfortherelativeerrorofthevolumeto belessthan1%? (12)Asectorofadiskofradius R =100cmhasanangle =60. Howmuchistheareaofthesectorchangedif (i)theradius R isincreasedby1cm? (ii)theangle isdecreasedby30? Givetheexactandapproximatesolutions.Comparethem. (13)Theperiodofapendulumisdeterminedbytheequation T =2 l g where l isthelengthofthependulum(incm)and g =981 cm/s2isthefree-fallacceleration.Howmuchshouldthelength l =20cmbechangedinordertoincreasetheperiodby0 05 s? (14)Todeterminethefree-fallacceleration,theperiodofapendulumismeasuredsothatbytheaboveequation g =4 2l2/T2. Howdothemeasuredvaluesof g varyif (i) T ismeasuredwitharelativeerrorbound ? (ii) L ismeasuredwitharelativeerrorbound ?

PAGE 104

983.RULESOFDIFFERENTIATION (15)Findtheabsoluteerroroflog10x ( x> 0)iftherelativeerror of x is (16)Usedierentialsto“ndthederivativesoftheinversefunctions sinŠ 1x andcosŠ 1x (17)Provethatthelinearizationofadierentiablefunction f ( x ) inaneighborhoodof x0isuniqueinthesensethatif L ( x )= b + m ( x Š x0)and limx x0f ( x ) Š L ( x ) x Š x0=0 then b = f ( x0)and m = f( x0).Inotherwords, thelinearizationistheonlylinearapproximationwhoseerrordecreasesto zerofasterthan x Š x0as x approaches x0. (18)Findthetangentlineandthenormalline(thelineperpendiculartothetangentline)tothecurve y =( x +1)3 3 Š x at thepoints( Š 1 0),(2 3),and(3 0). (19)Findthepoint(s)oftheparabola y =2+ x Š x2atwhichthe tangentlineis(i)paralleltothe x axisand(ii)paralleltothe line y = x (20)Provetherelation a2+ x = a + x 2 a Š R, 0 0and x> 0.

PAGE 105

CHAPTER4 ApplicationsofDifferentiation 22.MinimumandMaximumValues Someofthemostimportantapplicationsofcalculusareoptimizationproblems.Anexampleofanancientoptimizationproblem:Aman canthrowastoneataspeedof v0.Atwhatangleshouldthestonebe throwninordertogetthemaximalrange?Anexampleofamodern optimizationproblem:Howcanoneoptimizetheinformation”owin theWorldWideWebtoavoidcrashesofservers?Manyoftheseproblemscanbereducedto“ndingthemaximalandminimalvaluesofa givenfunction. Definition 4.1(AbsoluteMaximumandMinimum) Afunction f hasanabsolutemaximumat c if f ( x ) f ( c ) forall x inthedomain D of f .Similarly,thevalue f ( c ) iscalledthe maximumvalue of f .A function f hasanabsoluteminimumat c if f ( x ) f ( c ) forall x in thedomain D of f .Thevalue f ( c ) iscalledthe minimumvalue of f Themaximumandminimumvaluesof f arecalledthe extremevalues of f Forexample,thefunction f ( x )=cos x attainsitsmaximumvalue 1at x =2 n ,where n =0 1 2 ,... ,anditsminimumvalue Š 1at x = +2 n .Afunctiondoesnotalwayshaveamaximumorminimum value.Forinstance,thefunction f ( x )=1 /x de“nedforallreal x =0 hasneithermaximumnorminimumvaluebecause,foranyreal M ,one canalways“nd x suchthat f ( x ) >M (0
PAGE 106

1004.APPLICATIONSOFDIFFERENTIATION relative)minimumat c if f ( x ) f ( c ) forall x insomeopeninterval containing c Example 4.1 Doesthefunction f ( x )= x3Š x = x ( x2Š 1) have anabsolutemaximum(minimum)valueandrelativemaxima(minima) ontherealaxis? 3 3 1 1 1 1y x Figure4.1. Graphofthefunction f ( x )= x3Š x = x ( x2Š 1).Itdoesnothaveanabsolutemaximumor minimumvalue.However,itdoeshavearelativemaximumat x = Š 1 / 3andarelativeminimumat x = 1 / 3. Solution:1 .Thefunctionhasneitheranabsolutemaximumnoran absoluteminimumbecauseitgrowsunboundedlywithincreasing x and itdecreasesunboundedlyas x attainslargernegativevalues. 2 .Thefunctionvanishesatthreepoints x =0 1.Itcanhave relativeminimaandmaximabetweenitszerosbecausethevaluesof f areboundedfromaboveandbelow: | f ( x ) || x |3+ | x | 2for | x | 1, thatis, Š 2 f ( x ) 2if Š 1 x 1. 3 .Considertheopeninterval x (0 1).Thefunctionisstrictly negativeinitandboundedfrombelow: M
PAGE 107

22.MINIMUMANDMAXIMUMVALUES101 can“ndapoint c ( Š 1 0)suchthat f ( x ) f ( c )forall x ( Š 1 0); thatis, f hasarelativemaximumin( Š 1 0). Remark .Theactualvalueis c = Š 1 / 3(seebelow). Oneofthelessonsthatcanbelearnedfromthisexampleisthatone canthinkofarelativeminimum(maximum)asanabsoluteminimum (maximum)when f isrestrictedtoasucientlysmallsubsetinits domain.Thisobservationisaccuratelystatedbythefollowingtheorem. Theorem 4.1(TheExtremeValueTheorem) If f isacontinuous functiononaclosedinterval [ a,b ] ,then f attainsitsabsolutemaximum andminimumvaluesin [ a,b ]; thatis,thereexist c1and c2in [ a,b ] such that f ( c1) f ( x ) f ( c2) forall x in [ a,b ] c1f ( x ) Figure4.2. Extremevaluetheorem.Anexampleofa continuousfunctionwithseverallocalminimaandmaxima.Theminimalvaluecoincideswithoneofthelocalminima,whilethemaximalvalueisreachedatthe endpointoftheinterval: f ( c1) f ( x ) f ( b )forall x [ a,b ].Thehypothesisoftheclosednessoftheintervaliscrucial.Ifthepoint b isexcluded,then f hasno maximalvalueon[ a,b ). Thecontinuityhypothesisis essential .Infact,thecontinuityof f ( x )= x3Š x was implicitly usedinExample4.1toestablishthe existenceofitsrelativemaximumandminimum!Thefollowingexampleillustratesthepoint.Considerthefunction f ( x )=2 x if x [0 1) and f ( x )=1if x [1 2].Sothefunctionisde“nedontheclosed

PAGE 108

1024.APPLICATIONSOFDIFFERENTIATION interval[0 2]andboundedfromabove f ( x ) 2).An attempttoestablishtheexistenceofamaximumvalueof f bylowering M fails!Indeed,thelowestupperboundis M =2,butthereisno c suchthat f ( c )=2.Thevaluesof f approach2as x approaches1from theleft,but f (1)=1!Foranypositive > 0, f (1 Š )
PAGE 109

22.MINIMUMANDMAXIMUMVALUES103 positive h> 0suchthat c
PAGE 110

1044.APPLICATIONSOFDIFFERENTIATION (IV)Afunctionde“nedonaclosedinterval[ a,b ]canhaveitsabsolutemaximumorminimumattheendpoints.When“nding theabsolutemaximumandminimumvalues,thevaluesof f atthecriticalpointsmustbecomparedwith f ( a )and f ( b ). Thelargest(smallest)ofthemistheabsolutemaximum(minimum)value. Example 4.2 Ifastoneisthrownataspeed v0m/s andanangle withthehorizontalline,thenitstrajectoryisaparabola: (4.4) y = x tan Š x2g 2 v2 0cos2 where y isthestoneheight(verticalposition), x isthehorizontalposition(allthepositionsareinmeters),and g =9 8 m/s2isaconstant universalforallobjectsnearthesurfaceoftheEarth(thefree-fallacceleration).ThisisaconsequenceoftheNewtonssecondlaw.Atwhat angleshouldonethrowastonetoreachthemaximalrangeatagiven speed v0? Solution:1 .Therangeasafunctionoftheangle hastobefound “rst.Thestonelandswhenitsheight y vanishes.Theequation y =0 hastwosolutions x =0(naturally,thisiswherethestonewasthrown) and x = L ( ),where L ( )= 2 v2 0 g tan cos2 = 2 v2 0 g sin cos = v2 0 g sin(2 ) 2 .Therange L ( )isadierentiablefunctionof sothevaluesof at which L attainsitsextremevaluesmaybefoundfromtheequation L( )=0= v2 0 g 2cos(2 )=0= cos(2 )=0 Thisequationhascountablymanysolutions2 = / 2+ n ,where n isanyinteger.Butintheintervalofthephysicalvaluesof [0 ,/ 2], ithasonlyonesolution = / 4.Sincesin(2 / 4)=1(theabsolute maximumofthesine), L attainsitsmaximumvalueat = / 4.So therangeismaximal, Lmax= v2 0/g ,whenastoneisthrownat45. Remark.Theconclusionintheprecedingexampleisindependentof thestonesmassanditsinitialspeed v0.Inreality,forlargervaluesof v0,likeaprojectileshotbyagun,trajectorywoulddeviatefromthe parabola(duetofrictionwiththeairthatincreaseswithincreasing thespeed).Sotheoptimalanglewoulddeviatefrom / 4.Thedeviationwouldalsodependonthemassandtheinitialspeed.Therange

PAGE 111

22.MINIMUMANDMAXIMUMVALUES105 optimizationproblembecomesmoreinvolvedandwouldrequirethe theoryof dierentialequations .Itshouldalsobenotedthattheangle atwhichthemaximalrangeisattaineddependsontheinitialheight atwhichthestoneisthrown.Sotheanglewouldbedierentfrom45when,forexample,thestoneisthrownfromacli.22.3.Exercises.(1)Examinethefollowingfunctionsformaximaandminima. Drawthegraphineachcase. (i) y =2+ x Š x2(ii) y =( x Š 1)3(iii) y =( x +1)4(iv) y = x2Š 5 x +3 (v) y =2 x3Š 3 x2+6 x Š 3 (vi) y = x2+16 /x (vii) y = x2Š 1 /x2(viii) y =4 x/ ( x2+1) (ix) y =sin x +cos x (x) y = xex(xi) y = xn(1 Š x )m,where n and m arepositiveintegers (xii) y = x1 / 3(1 Š x )2 / 3(2)Findallcriticalpointsofthefollowingfunctionsanddetermine whetherthereisalocalmaximum,alocalminimum,ornone oftheaboveateachcriticalpoint. (i) y = | x Š 3 | (ii) y = | x2Š 4 | +2 x (iii) y = | x3Š 1 | (iv) y = | sin(2 x ) | (v) y = | x3| ex(vi) y =( x Š 1)1 / 3(vii) y = x ( x +1)2 / 3(viii) y =(1 Š x2)3 / 2(3)Findtheextremevaluesofthefollowingfunctionsonthespeci“edintervalorshowthatsuchvaluesdonotexist. (i) y = x4Š 4 x2, Š 3
PAGE 112

1064.APPLICATIONSOFDIFFERENTIATION (5)Alineisdrawnthroughapoint( a,b )suchthatthepartinterceptedbetweentheaxeshasaminimumlength.Provethat theminimumlengthis( a2 / 3+ b2 / 3)3 / 2. (6)Findthemaximumareaofanisoscelestrianglewith“xed perimeter p (7)Letthesumoftwonumbersbe s .Findthenumbersineach ofthefollowingcases: (i)Thesumoftheirsquaresisaminimum. (ii)Thesumoftheircubesisaminimum. (iii)Theirproductisamaximum. (iv)Thedierencebetweenoneandthereciprocaloftheother isamaximum. (8)Canoneclaimthatifthefunction f ( x )hasamaximum x = x0, theninasucientlysmallneighborhoodtotheleftof x0the function f ( x )increasesandinasucientlysmallneighborhoodtotherightof x0itdecreases?Considertheexample: f ( x )=2 Š x2 2+sin 1 x if x =0and f (0)=2 (9)Doesthefunction f ( x )= | x | 2+cos 1 x if x =0and f (0)=0 havealocalextremevalueat x =0?Graphthefunction. 23.TheMeanValueTheorem Theorem 4.3(RollesTheorem) Let f beafunctionthatsatis“es thefollowingthreehypotheses: (I) f iscontinuousontheclosedinterval [ a,b ] (II) f isdierentiableontheopeninterval ( a,b ) (III) f ( a )= f ( b ) Thenthereisanumber c in ( a,b ) suchthat f( c )=0 Thistheoremprovidesausefulmethodtoprovetheexistenceofalocalmaximumorminimumofafunction f whenanalyticsolutionsofthe equation f( x )=0arehardto“nd.Infact,ithasalreadybeenusedin Example4.1:Thefunction f ( x )= x3Š x ontheintervals[ Š 1 0],[0 1], [ Š 1 1]satis“esthehypothesesofRollestheorembecause f ( 1)= f (0)=0.TheprooffollowscloselytheargumentsofExample4.1.

PAGE 113

23.THEMEANVALUETHEOREM107 y x a cb Figure4.3. Rollestheorem.Thecontinuityof f guaranteestheboundednessof f .Sothegraphof f liesbetweentwohorizontallines.Byloweringanupperbound orincreasingalowerbounduntiloneofthehorizontal lines(orboth)touchesthegraphandbecomesitstangentline,dierentiabilityof f ensurestheexistenceof thetangentlineateverypointin( a,b ).Theslopeofthe horizontaltangentlineis0andsoisthederivativeat thatpoint. ProofofTheorem4.3. 1 .If f ( x )= f ( a )= k isaconstantfunction,then f( x )=0everywhere. 2 .Let f ( x ) >f ( a )forsome x ( a,b )(cf.Example4.1for x [ Š 1 0]). Since f iscontinuous,theextremevaluetheoremapplies,andtherefore f hasamaximumin[ a,b ].Since f ( a )= f ( b ),themaximalvaluemust beattainedat c ( a,b ).ByFermatstheorem, f( c )=0because f is dierentiablein( a,b ). 3 .If f ( x ) 0.Bycontinuity, f hastotakeallintermediatevalues

PAGE 114

1084.APPLICATIONSOFDIFFERENTIATION between Š 4and2(theintermediatevaluetheorem).So f has atleast onerootin( Š 1 1). 2 .Supposeithastworoots a and b ,thatis, f ( a )= f ( b )=0.Then, byRollestheorem, f( x )hastovanishsomewherein( a,b ).Butthis isnotpossiblebecause f( x )=5 x4+3 x2+1 > 0forany x .Thus, f hastheonlyrealroot. Theorem 4.4(TheMeanValueTheorem) Let f beafunction thatsatis“esthefollowinghypotheses: (I) f iscontinuousontheclosedinterval [ a,b ] (II) f isdierentiableontheopeninterval ( a,b ) Thenthereisanumber c ( a,b ) suchthat (4.5) f( c )= f ( b ) Š f ( a ) b Š a or f ( b ) Š f ( a )= f( c )( b Š a ) Thegeometricalinterpretationofthetheoremissimple(see Figure4.4).Considerthelinethroughthepoints( a,f ( a ))and( b,f ( b )). Itsslopeis( f ( b ) Š f ( a )) / ( b Š a ).Thetheoremassertstheexistence ofapointwherethegraph y = f ( x )hasatangentlinewiththesame slope(cf.Equation(4.5))(as f( c )istheslopeofthetangentlineat x = c ).Letusturntoaformalproof. ProofofTheorem4.4. 1 .Considerthelinethroughthepoints( a,f ( a ))and( b,f ( b )).Its equationis y = L ( x )= f ( a )+ f ( b ) Š f ( a ) b Š a ( x Š a ) (4.6) L ( a )= f ( a ) ,L ( b )= f ( b ) Next,considerthefunction (4.7) h ( x )= f ( x ) Š L ( x )= f ( x ) Š f ( a ) Š f ( b ) Š f ( a ) b Š a ( x Š a ) Itsvaluesdeterminethedeviationofthegraph y = f ( x )fromthesecantline y = L ( x )ontheclosedinterval[ a,b ]. 2. Thefunction h ( x )satis“esthethreehypothesesofRollestheorem. First,itiscontinuouson[ a,b ]asthesumoftwocontinuousfunctions f ( x )and Š L ( x )(alinearfunctioniscontinuous).Second,itisdierentiableon( a,b )asthesumoftwodierentiablefunctions: (4.8) h( x )= f( x ) Š f ( b ) Š f ( a ) b Š a ,x ( a,b ) .

PAGE 115

23.THEMEANVALUETHEOREM109 x y c ( a, f ( a )) ( b, f ( b )) Figure4.4. Meanvaluetheorem.Thesecantlineof thegraphof f throughthepoints( a,f ( a ))and( b,f ( b )) hastheslopetan =[ f ( b ) Š f ( a )] / ( b Š a ),where istheanglebetweenthesecantlineandthehorizontal line.If f doesnotcoincidewiththesecantline,thennear x = a theslopeofthetangentlinedoesnotcoincidewith tan .Herethecasewhenthisslopeisgreaterthantan isshown.Thenthegraphof f liesabovethesecantline near x = a .Butthegraphhastoreturntothesecantline again.Nearthepointwherethegraphandthesecant linesmeetagain,thetangentlinehastohaveasmaller slopethantan .Soatsomepoint c thetangentlinehas tobeparalleltothesecantline,meaningthat f( c )= tan Finally,by(4.6)and(4.7), h ( a )= f ( a ) Š L ( a )=0and h ( b )= f ( b ) Š L ( b )=0,thatis, h ( a )= h ( b ). 3 .ByRollestheorem,thereisanumber c ( a,b )suchthat h( c )=0= f( c )= f ( b ) Š f ( a ) b Š a whereEquation(4.8)hasbeenused. Example 4.4 Aspeedingcarwaspulledoveronaninterstateroad andastatetroopergaveawarningtothedriver.Forty“veminutes laterandpassed65milesontheroad,thecarstoppedatarestarea.

PAGE 116

1104.APPLICATIONSOFDIFFERENTIATION Anotherstatetrooperapproachedthedriverandissuedaspeedingticket, claimingthatthedriverignoredthewarningandexceeded 86 milesper hour.Wasthetroopersclaimcorrect? Solution: Let s ( t )bethedistancetraveledbythecarafteritwas pulledoverthe“rsttime.Therateofchange s( t )= v ( t )isthespeed ofthecaratanymomentoftime.Thefunction s ( t )isde“nedbetween t =0and t =45min=0 75hrsothat s (0)=0and s (0 75)=65mi. Itisdierentiableas s( t )isthecarspeed!Bythemeanvaluetheorem, thereisatimemoment t = c (0 0 75)when s( c )= v ( c )= s (0 75) Š s (0) 0 75 Š 0 = 65 0 75 86 7mi / hr Thespeedingticketisjusti“ed. Foranytwomomentsoftime a and b ,theratio( s ( b ) Š s ( a )) / ( b Š a ) istheaveragespeedonthetimeinterval[ a,b ].Themeanvaluetheorem simplystatesthatamovingobjectalwaysattainsitsaveragespeedat leastatonemomentoftimebetween a and b .So,ifattimemoment b theobjectappearstobetravelingslowerthanitsaveragespeed,prior tothatitmusthavebeentravelingfasterthanitsaveragespeed. Example 4.5 Supposethederivative fexistsandisboundedon ( a,b ) ,thatis, m f( x ) M .If f ( a ) isgiven,howsmallandhow largecan f ( b ) possiblybe? Solution: Bythemeanvaluetheorem,thereisa c ( a,b )suchthat f ( b )= f ( a )+ f( c )( b Š a ).Since m f( c ) M f ( a )+ m ( b Š a ) f ( b ) f ( a )+ M ( b Š a ) Thisequationiseasytounderstandwiththehelpofamechanical analogy:Howfarcanacartravelintime b Š a ifitsspeedisnotlower than m ,butcannotexceed M ? 23.1.PropertiesoftheFirstDerivative.Thederivativeofaconstant functionvanishes.Howabouttheconverse?Thefollowingtheorem answersthisquestion. Theorem 4.5 If f( x )=0 forall x inaninterval ( a,b ) ,then f is constanton ( a,b ) Proof. Takeanytwonumbers x1and x2between a and b .Bythe meanvaluetheorem,thereisanumber c between x1and x2suchthat f ( x1) Š f ( x2)= f( c )( x1Š x2).Byhypothesis, f( c )=0forany c .

PAGE 117

23.THEMEANVALUETHEOREM111 Thus, f ( x1) Š f ( x2)=0or f ( x1)= f ( x2)forany x1and x2in( a,b ); thatis, f isconstant. Thehypothesisthat f( x )=0ina singleinterval iscrucial.For example,thesignfunction f ( x )=1if x> 0,and f ( x )= Š 1if x< 0, haszeroderivativeatanypointofitsdomain,butitisnotconstant. Thekeypointtonoteisthatthedomainisnotasingleinterval,but aunionoftwodisjointintervals( Š 0)and(0 ).Sothemean valuetheoremisnotapplicabletoanyintervalcontaining x =0.This exampleiseasilyextendedtothecasewhenthedomainisanycollection ofdisjointintervalsand f takesdierentconstantvaluesondierent intervals. Corollary 4.1 If f( x )= g( x ) forall x inaninterval ( a,b ) then f Š g isconstant,thatis, f ( x )= g ( x )+ k ,where k isaconstant. Proof. Let h ( x )= f ( x ) Š g ( x ).Since h= fŠ g=0in( a,b ), h isconstant,andtheconclusionfollows. Thesingofthe“rstderivativede“nesintervalsofgrowthanddecreaseofafunction. Theorem 4.6(Increasing-DecreasingTest) (I) If f> 0 onaninterval,then f isincreasingonthatinterval. (II) If f< 0 onaninterval,then f isdecreasingonthatinterval. Proof. Take any twonumbers x1and x2intheintervalsothat x1f ( x2).Since f isdierentiable,themeanvaluetheoremstates thatthereisanumber c between x1and x2suchthat (4.9) f ( x2) Š f ( x1)= f( c )( x2Š x1) If f> 0,thenitfollowsfrom(4.9)that f ( x2) Š f ( x1) > 0because,by assumption, x2>x1;thatis,thefunctionisincreasing.Similarly,for f< 0, f ( x2) Š f ( x1) < 0,andthefunctionisdecreasing. Thefunction f issaidtobe monotonic inanintervalifitisincreasingordecreasinginthisinterval.Bytheincreasing-decreasingtest,the functionismonotonicinanintervalifitsderivative fdoesnotvanish intheinterval. Theincreasing-decreasingtestisfurtherillustratedontheinteractivewebsiteathttp://www.math.u”.edu/ mathguy/ufcalcbook/inc dec.html.

PAGE 118

1124.APPLICATIONSOFDIFFERENTIATION 23.2.TheInverseFunctionTheorem.ABabyVersion.Givenafunction f itsinversefunctionexistsif f isone-to-oneasexplainedinSection5. AsimpleruletocalculatethederivativeoftheinversefunctionwaspresentedinSections17and21.However,theveryquestionofwhetherthe inversefunctionisactuallydierentiablehasnotbeenaddressed.Itappearsthatif f isdierentiable,thenthequestionsabouttheexistence oftheinversefunction fŠ 1anditsdierentiabilitycanbeansweredby lookingatthesignofthederivative f. Theorem 4.7 (ABabyVersionoftheInverseFunctionTheorem). Let f beafunctionon Š a 0 (or f( x ) < 0 )forall x ( a,b ) .Then f hastheinverse g = fŠ 1on ( c,d ) forsome Š c 0.Theothercaseissimilar.Bytheincreasingdecreasingtest, f ( x1) f ( x1)foranyinterval[ x1,x2] in( a,b ).Thisshowsthattherangeof f isasingleinterval( c,d )for some Š c 0forall Š
PAGE 119

23.THEMEANVALUETHEOREM113 y f ( x ) y f 1( x ) f f f 1 x Figure4.5. Inversefunctiontheorem.Anincreasing function f f> 0,isone-to-oneandhencehastheinverse g = fŠ 1.Thegraphsof f and g areobtainedfrom oneanotherbythere”ectionabouttheline y = x .Asecantlineofthegraphof f withtheslopetan = f/ x ismappedonthesecantlineofthegraphof g withthe slopetan = fŠ 1/ f bythisre”ection.Theangles and arerelatedas + = / 2andhence tan =1 / tan .Inthelimit x 0,whichalso implies f 0,thesecantlinesbecomethetangent linessothattan f( x )andtan g( y ),where y = f ( x ).Hence, g( f ( x ))=1 /f( x ). negativeon,forexample,(0 ).Sincetheimageof(0 )is( Š 1 1), theinversecosŠ 1x existsandisadierentiablefunctionon( Š 1 1).23.3.Exercises.(1)VerifyRollestheoremforthefunction f ( x )=( x Š 1)( x Š 2)( x Š 3) (2)Thefunction f ( x )=1 Š3 x2vanishesat x = a Š 1and x = b =1;nevertheless, f( x ) =0intheinterval( Š 1 1). DoesthisexamplecontradictRollestheorem?Explain.

PAGE 120

1144.APPLICATIONSOFDIFFERENTIATION (3)Isthefollowingassertiontrue?Ifso,proveit. Ifafunction f hasaderivativeateachpointofanopeninterval ( a,b ),if f iscontinuousat a andat b ,andif f ( a )= f ( b )=0, thenthereisapoint c ,with a 0,then f ( a )isarelativeminimumvalueof f Remark .Thederivativeat a isde“nedastherightlimit, f( a )=limh 0+f ( a + h ) Š f ( a ) h (6)Let f beafunctionwhosedomainofde“nitionistheclosed interval[ a,b ],whichiscontinuousat a and b ,whichisdierentiableintheinterval( a,b ),andforwhich f( a )=0ordoes notexist.Showthatifthereisapoint c suchthat f( x ) > 0 for a
PAGE 121

23.THEMEANVALUETHEOREM115 c ,with a 1 (iii) | tanŠ 1a Š tanŠ 1b || a Š b | (15)Supposethat f( x )= m =constfor Š
PAGE 122

1164.APPLICATIONSOFDIFFERENTIATION (iii) f ( x )=2 x3Š 9 x2+12 x +1 (iv) f ( x )=2 x/ (1+ x2) (v) f ( x )= x/ ( x +100), x> 0 (vi) f ( x )=cos2x +1 (vii) f ( x )=cos( /x ) (viii) f ( x )=sin x + x/ 2 (ix) f ( x )= ex+3 eŠ x+2 x (x) f ( x )= xneŠ x, n> 0, x 0 (xi) f ( x )= x2Š ln( x2) (xii) f ( x )= x22Š x(19)Showthatthefunction(1+ xŠ 1)xincreasesintheintervals ( Š Š 1)and(0 ). (20)Isthederivativeofamonotonicfunctionmonotonic?Consider theexample f ( x )= x +sin x (21)Showthatthefunction f ( x )= x + x2sin 2 x if x =0and f (0)=0 isincreasingat x =0( f(0) > 0)butisnotincreasinginany interval( Š a,a ),where a> 0canbearbitrarilysmall.Graph thefunction. (22)Supposethat f ( x )and g ( x )aredierentiablesothat f ( a )= g ( a )and f( x ) >g( x )for x>a .Showthat f ( x ) >g ( x )for x>a (23)Usetheresultofthepreviousexercisetoestablishtheinequalities: (i) ex> 1+ x x =0 (ii) x Š1 2x2< ln(1+ x ) 0 (iii) x Š1 6x3< sin x 0 (iv)tan x>x +1 3x3,0
PAGE 123

24.THEFIRSTANDSECONDDERIVATIVETESTS117 whetheritisalocalmaximum,localminimum,ornoneoftheabove? Itturnsoutthatthisquestioncanbeansweredbystudyingthederivatives fand f.Inaddition,manyqualitativefeaturesofthegraph y = f ( x )canbededucedfrompropertiesofthederivativesof f .24.1.TheFirstDerivativeTest.Bytheincreasing-decreasingtest, f is increasingonintervalifitsderivativeispositive,and f isdecreasing onanintervalifitsderivativeisnegative.Suppose fiscontinuous suchthat f( a )= m and f( b )= M .Then,ontheinterval[ a,b ], fmusttakeallintermediatevaluesbetween m and M .Suppose m< 0 and M> 0or m> 0and M< 0,thatis,thederivativechanges itssignontheinterval[ a,b ],then fmustvanishbetween a and b Thismeansthat f hasacriticalpoint af ( c + h )forsomesmallpositive h .Wecanthenconclude that f attainsitslocalmaximumat c .Similarly,ifthederivative fchangesfromnegativetopositiveat c ,then f changesfromdecreasing toincreasingat c f ( c Š h ) >f ( c )and f ( c ) f ( c ) >f ( c + h );thatis,ineithercasethefunction f hasneitheralocalminimumnoralocalmaximum.The“ndingsare summarizedinthefollowingtheorem. Theorem 4.8(TheFirstDerivativeTest) Supposethat c isacriticalpointofacontinuousfunction f (I) If fchangesfrompositivetonegativeat c ,then f hasalocal maximumat c (II) If fchangesfromnegativetopositiveat c ,then f hasalocal minimumat c (III) If fdoesnotchangesitssignat c ,then f hasneitheralocal maximumnoralocalminimumat c Itisimportanttonotethattheveryexistenceof fat c is not requiredinthe“rstderivativetest.Recallthede“nitionofacritical point( f( c )=0or f( c )doesnotexist).Infact,intheprecedingproof ofthe“rstderivativetest,thecondition f( c )=0canbedropped becauseallthatisneededtoapplytheincreasing-decreasingtestis thesignofthederivative f( x )for xc .Forexample,

PAGE 124

1184.APPLICATIONSOFDIFFERENTIATION f ( x )= | x | .Then f( x )= Š 1for x< 0(thefunctionisdecreasing) and f( x )=1for x> 0(thefunctionisincreasing).Hence, f ( x ) hasaminimumat x =0,eventhough fdoesnotexistat x =0.The continuityhypothesis isalsocrucial.Considerthefunction f ( x )=1 /x2for x =0and f (0)=0.Then f( x )= Š 2 /x3for x =0and f(0)does notexist.So x =0isacriticalpoint.Thefunctionisincreasingfor x< 0because f> 0,anditisdecreasingfor x> 0because f< 0. However, f hasnomaximumat x =0because f isdiscontinuousat x =0.Infact,itattainsitsabsoluteminimumat x =0.The“rst derivativetestissummarizedinthefollowingtable. Therstderivativetest Case sign f, xc Conclusion I + + Noextremum II + … Maximum III … + Minimum IV … … Noextremum Thereareplentyofmechanicalanalogiesofthe“rstderivativetest.Let H ( t )betheheight(relativetotheground)ofastonethrownupward asafunctionoftime t .Atthebeginning,thestonemovesupward so H> 0(theheightisincreasing).Whenthestonecomesbackto theground,itmovesdownwardso H< 0(theheightisdecreasing). Naturally,atsomemomentoftime,thestonehastoreachthemaximal height.Analyzethemotionofapendulum(orasee-saw)fromthis pointofview!Theheightwouldhavetwomaximaandoneminimum. Example 4.6(Example4.1Revisited) Findalllocalmaximaand minimaof f ( x )= x3Š x andtheintervalsonwhichthefunctionis increasingordecreasing(thefunctionisdepictedinFigure4.1). Solution:1 .Since f isdierentiable(itisapolynomial),allitscritical pointssatisfytheequation f( x )=3 x2Š 1=3 x Š 1 / 3 x +1 / 3 =0 Hence,thecriticalpointsare c1= Š 1 / 3and c2=1 / 3. 2 .For x 0( f isincreasingon( Š ,c1)). For c1
PAGE 125

24.THEFIRSTANDSECONDDERIVATIVETESTS119 ofanegativeandpositivenumber),andhence f< 0( f isdecreasing on( c1,c2)).For x>c2,theproduct( x Š c1)( x Š c2)ispositive(asthe productoftwopositivenumbers),andhence f> 0( f isincreasingon ( c2, )). 3 .Thederivativechangesfrompositivetonegativeat c1.Therefore, f hasalocalmaximumat c1.Thederivativechangesfromnegativeto positiveat c2.Therefore, f hasalocalminimumat c2. 24.2.PropertiesoftheSecondDerivative:InectionPoints.Definition 4.4(Concavity) Thegraphofafunction f iscalled concaveupward onaninterval I ifitliesaboveallofitstangentlines on I .Thegraphiscalled concavedownward on I ifitliesbelowallof itstangentlineson I Notethatthenotionofconcavityimpliesthat f isdierentiable (otherwise,thetangentlinesdonotexist).If f istwicedierentiable, thentheconcavityisdeterminedbythesignofthesecondderivative f.Supposethatthegraphof f isconcaveupwardon I .Considerthe tangentlinesattwopoints c and c + h in I : L1( x )= f ( c )+ f( c )( x Š c ) ,L2( x )= f ( c + h )+ f( c + h )( x Š c Š h ) Thegraphof f liesabovethelines L1and L2,thatis, f ( x ) Š L1( x ) > 0 and f ( x ) Š L2( x ) > 0forall x in I .Putting x = c inthelastinequality and x = c + h intheformerone,weobtain f ( c ) Š L2( c )= f ( c ) Š f ( c + h )+ f( c + h ) h> 0 f ( c + h ) Š L1( c + h )= f ( c + h ) Š f ( c ) Š f( c ) h> 0 Thesumoftheright-handsidesoftheseinequalitiesispositiveasthe sumoftwopositivenumbers: (4.10) h [ f( c + h ) Š f( c )] > 0= f( c + h ) Š f( c ) h > 0 wherethe“rstinequalityhasbeendividedbyapositivenumber h2. Inequality(4.10)istrueforany h .Therefore,bytakingthelimit h 0,wecanconcludethat f( c ) > 0ifthegraphisconcaveupward. Inequality(4.10)showsthat f( c + h ) >f( c )for h> 0and f( c ) > f( c + h )for h< 0.Inotherwords,thederivative f,ortheslopeofthe tangentlineofthegraphof f ,increasesfortheupwardconcavity,and hence( f)= fmustbepositivebytheincreasing-decreasingtest. Similarly,thedownwardconcavityimpliesthat fisnegative.Itturns outthattheconverseisalsotrue.

PAGE 126

1204.APPLICATIONSOFDIFFERENTIATION Theorem 4.9(TheConcavityTest) Let f betwicedierentiable onaninterval I (I) If f( x ) > 0 forall x in I ,thenthegraphof f isconcave upwardon I (II) If f( x ) < 0 forall x in I ,thenthegraphof f isconcave downwardon I x0x0x x f" ( x ) > 0 f" ( x ) < 0 y y Figure4.6. Concavitynearapoint x0atwhich f( x0)=0.Thegraphisconcavedownwardif f( x ) > 0 for x>x0and xx0and x 0forsomesmall h> 0.This meansthatthegraphisconcaveupwardtotheleftandrightof c .As anexample,consider f ( x )= x4.Second, f( c h ) < 0.Thisimplies thatthegraphisconcavedownwardtotheleftandrightof c .Asan example,take f ( x )= Š x4.ThesetwocasesaredepictedinFigure4.6. Third, f( c Š h ) > 0and f( c + h ) < 0,thatis,theconcavitychanges fromupwardtodownward(e.g., f ( x )= Š x3).Fourth, f( c Š h ) < 0 and f( c + h ) > 0,thatis,theconcavitychangesfromdownwardto upward(e.g., f ( x )= x3). Definition 4.5(In”ectionPoint) Apoint P onthegraph y = f ( x ) iscalledan in”ectionpoint if f iscontinuousthereandthegraph

PAGE 127

24.THEFIRSTANDSECONDDERIVATIVETESTS121 changesfromconcaveupwardtoconcavedownwardorfromconcave downwardtoconcaveupward. y x x0f" > 0 f" < 0 Figure4.7. Concavitynearapoint x0atwhich f( x0)=0(thecasenotdepictedinFigure4.6).Anin”ectionpoint.Thesecondderivativechangesitssignat thein”ectionpoint x = x0.Theconcavityofthegraphof f alsochangesatthein”ectionpoint.Suchalocalbehaviorcanbeillustratedby f ( x )= c + a ( x Š x0)+( x Š x0)3, where c and a areconstants,sothat f( x )=6( x Š x0) changesitssignat x0.Notethat f( x0)= a ;i.e., a de“nestheslopeofthegraphat x = x0. Let c beacriticalpointof f .Suppose fiscontinuousnear c Whatcan f( c )tellusaboutthenatureofthecriticalnumber(local minimumormaximum)?Therearethreepossibilities.First, f( c ) > 0. Thismeansthat f( x ) > 0forall x insomeneighborhoodof c (by thecontinuityof f).Hence, f isconcaveupwardnear c ;thatis, itsgraphlies above thetangentlineat c ,whichisahorizontalline because f( c )=0.So f musthavealocalminimum.Similarly,the condition f( c ) < 0impliesthattheconcavityisdownwardnear c and f hasalocalmaximum.If f( c )=0,thentheconcavitymay ormaynotchangeat c asdiscussedearlier.Thefunctionmayhave alocalmaximum,alocalminimum,oranin”ectionpoint;thatis,no conclusionaboutthenatureofthecriticalpointcanbereached.

PAGE 128

1224.APPLICATIONSOFDIFFERENTIATION Theorem 4.10(TheSecondDerivativeTest) Suppose fiscontinuousnear c (I) If f( c )=0 and f( c ) > 0 ,then f hasalocalminimumat c (II) If f( c )=0 and f( c ) < 0 ,then f hasalocalmaximumat c (III) If f( c )=0 and f( c )=0 ,then f mayhavealocalmaximum, alocalminimum,oranin”ectionpoint. yf" ( c ) < 0 f" ( c ) > 0 f" ( c ) = 0 y xx x c c c y(inflection) Figure4.8. Secondderivativetest.Thegraphof f has ahorizontaltangentlineatacriticalpoint, f( c )=0. If f( c ) < 0and f( x )iscontinuous,then f( x ) < 0 issomeopenintervalcontaining c .Hence,thegraphis concaveupwardnear x = c and f hasalocalmaximum at c (leftpanel).If f( c ) > 0,thegraphisconcave downwardnear x = c and f hasalocalminimumat c (middlepanel).If f( c )=0,thegraphmayhave anin”ectionpointwhen fchangesitssignat c (right panel),but fmaynotchangeitssignat c ,andhence thebehaviordepictedintheleftandmiddlepanelsisalso possibleinthecase f( c )=0.Thesecondderivativetest isinconclusive.Thefunctionmayhavealocalminimum, alocalmaximum,oranin”ection.Examplesaregiven inthecaptionsofFigures4.6and4.7(with a =0). InExample4.6,thefunction f ( x )= x3Š x isshowntohavetwo criticalpoints: x = 1 / 3asdepictedinFigure4.1.Since f( x )=6 x f( Š 1 / 3)= Š 2 3 < 0(alocalmaximum)and f(1 / 3)=2 3 > 0 (alocalminimum).Thefunctionalsohasanin”ectionpointat x =0: f( x )=6 x< 0if x< 0and f( x )=6 x> 0if x> 0.Notethat anin”ectionpointmay not beacriticalpoint!Inotherwords,the tangentlineatanin”ectionpointcanhave any slope.Intheexample discussed, f(0)= Š 1(seealsoFigure4.7).

PAGE 129

24.THEFIRSTANDSECONDDERIVATIVETESTS123 24.3.Exercises.(1)Findallcriticalpointsofthegivenfunctioninitsdomain.Use eitherthe“rstorsecondderivativetesttodeterminewhether thereisalocalmaximum,alocalminimum,oranin”ection ateachcriticalpoint. (i) f ( x )=( x Š 1)2(ii) f ( x )=( x Š 1)3(iii) f ( x )= a ( x Š 1)4(iv) f ( x )=( x Š a )( x Š b )( x Š c ), a
PAGE 130

1244.APPLICATIONSOFDIFFERENTIATION (iv) f ( x )= x +1 x,[0 01 100] (v) f ( x )= 5 Š 4 x ,[ Š 1 1] (3)Provetheinequalities: (i) | 3 x Š x3| 2for | x | 2 (ii) | a sin x + b cos x | a2+ b2(4)Findthegreatestlowerboundandthesmallestupperbound ofthefunctiononagiveninterval: (i) f ( x )= xeŠ 0 01 x,(0 ) (ii) f ( x )=(1+ x2) / (1+ x4),(0 ) (5)Provetheinequality 2 3 x2+1 x2+ x +1 2 Š
PAGE 131

25.TAYLORPOLYNOMIALSANDTHELOCALBEHAVIOR125 intervalsuchthat f ( b )= f ( a )+ f( a )( b Š a )+ f( c ) 2 ( b Š a )2. Hint :Considerthefunction h de“nedby h ( x )= f ( x ) Š f ( a ) Š f( a )( x Š a ) Š k ( x Š a )2, wherethenumber k ischosensuchthat h ( b )=0. 25.TaylorPolynomialsandtheLocalBehaviorofaFunction Thetangentlineapproximation L ( x )isthebestlinearapproximationof f ( x )near x = a because L ( x )and f ( x )havethesamerate ofchangeat a .Intheprevioussection,itwasshownthatthesecond derivativeat a providesimportantinformationaboutthebehaviorof f ( x )near a ,namelytheconcavity.Thetangentline L ( x )hasnoconcavityas L( x )=0.Thequestionariseswhetherthereisasystematic methodtoimprovetheaccuracyofthetangentlineapproximationto capturemoreessentialfeaturesofthebehaviorof f ( x )near a (i.e.,the localbehavior of f ).25.1.TaylorPolynomials.Thefunction L ( x )isapolynomialofthe“rst degree.Considerthesecond-degreepolynomial T2( x )= f ( a )+ f( a )( x Š a )+ c2( x Š a )2= L ( x )+ c2( x Š a )2, where c2isanarbitrarycoecient.Thispolynomialhasthesame featuresas L ( x ),thatis, T2( a )= L ( a )= f ( a )and T 2( a )= L( a )= f( a )because T 2( x )= f( a )+2 c2( x Š a ).Soitmightprovideabetter approximationof f ( x )than L ( x )near a ifthecoecient c2ischosen sothat T2( x )hasthesameconcavityas f ( x )near a .Bytheconcavity test,itisthenreasonabletoassumethat T 2( a )= f( a ),whichyields 2 c2= f( a )or c2= f( a ) / 2.Theideacanbeextendedtoapolynomial ofdegree n : Tn( x )= c0+ c1( x Š a )+ c2( x Š a )2+ + cn( x Š a )n, wherethecoecientsare“xedbytheconditions Tn( a )= f ( a ) ,T n( a )= f( a ) ,T n( a )= f( a ) ,...,T( n ) n( a )= cn. Theresultingpolynomialiscalled the n th-degreeTaylorpolynomial : Tn( x )= f ( a )+ f( a )( x Š a )+ f( a ) 2! ( x Š a )2+ + f( n )( a ) n ( x Š a )n.

PAGE 132

1264.APPLICATIONSOFDIFFERENTIATION 25.2.AccuracyofTaylorPolynomials.Theaccuracyofthetangentline approximationisassessedinTheorem3.15.Letuscompareitwith theaccuracyofhigher-degreeTaylorpolynomials.ConsiderTaylor polynomialsoftheexponentialfunction exnear x =0.Since( ex)= exand e0=1,theTaylorpolynomialsare f ( x )= ex: Tn( x )=1+ x + 1 2 x2+ 1 6 x3+ + 1 n xn. Letustakeafewvaluesof x near x =0andcomparethevaluesofthe Taylorpolynomialswiththevalueofthefunction: x =1 : f =2 718 T1=2 000 T2=2 500 T3=2 667 x = Š 0 5 : f =0 607 T1=0 500 T2=0 625 T3=0 604 x =0 25 : f =1 284 T1=1 250 T2=1 281 T3=1 284 Twoobservationscanbemadefromthistable.First,theaccuracy increaseswithincreasingthedegreeoftheTaylorpolynomial(reading therowsofthetablefromlefttoright).Second,lower-degreeTaylor polynomialsbecomemoreaccurateastheargumentgetsclosertothe pointatwhichtheTaylorpolynomialsareconstructed(readingthe columnsofthetablefromtoptobottom).Forexample,theapproximation ex T3( x )isaccurateuptofoursigni“cantdigitsif | x | 1 / 4. Sotheaccuracyoftheapproximation ex T2( x )isdeterminedby thedierence T2Š T3= Š x3/ 6,thatis,bythenextmonomialtobe addedto T2togetthenextTaylorpolynomial.Thisobservationisa characteristicfeatureofTaylorpolynomials: Theorem 4.11 Let f becontinuouslydierentiable n +1 timeson anopeninterval I containing a .Let f( n +1)beboundedon I | f( n +1)( x ) | M .Then (4.11) | f ( x ) Š Tn( x ) M ( n +1)! | x Š a |n +1, where TnistheTaylorpolynomialat a Theorem3.15isaparticularcaseofthistheoremfor n =1.Inequality(4.11)isaconsequenceoftheTaylortheoremwhoseproofis givenamoreadvancedcalculuscourse.Forexample,whatistheaccuracyoftheTaylorpolynomial T5( x )near a =0fortheexponential exintheinterval[ Š 1 1]?Togettheupperboundonerrors,oneshould takethemaximalvalueoftheright-handsideof(4.11)for n =5inthe interval,thatis,( ex)( n )= ex M = e ,and | x | 1,sotheabsolute errorcannotexceed e/ 6! 0 0038.

PAGE 133

25.TAYLORPOLYNOMIALSANDTHELOCALBEHAVIOR127 25.3.TaylorPolynomialsofBasicFunctions.Itisusefultomakealist ofafewTaylorpolynomialsoflowerdegreesforbasicfunctionsnear x =0.Thederivationofthefollowingrelationsisgivenasanexercise. ex T3( x )=1+ x +1 2x2+1 6x3sin x T3( x )= T4( x )= x Š1 6x3cos x T4( x )= T5( x )=1 Š1 2x2+1 24x4(1+ x )p T3( x )=1+ px +p ( p Š 1) 2x2+p ( p Š 1)( p Š 2) 6x3tan x T3( x )= T4( x )= x +1 3x3tanŠ 1x T3( x )= T4( x )= x Š1 3x3ln(1+ x ) T3( x )= x Š1 2x2+1 3x3Notethatsin x ,tan x ,andtanŠ 1x areoddfunctionsandtheirpolynomialapproximationscannotcontainevenpowersof x ,hence, T3( x )= T4( x ).Similarly,cos x isanevenfunctionanditspolynomialapproximationcannotcontainoddpowersof x .25.4.TaylorPolynomialsnearCriticalPoints.Let a beacriticalpoint of f .Provided f isenoughtimesdierentiable(seethehypothesesof Theorem4.11),Taylorpolynomialscanbeconstructednear a .The lineartermvanishesbecause f( a )=0.Thesecondderivativetestis easytounderstandbylookingat f ( x ) T2( x )= f ( a )+1 2f( a )( x Š a )2. If f( a ) > 0,thennear a thegraphof f lookslikeaparabolaconcave upward(seethemiddlepanelofFigure4.8)andhasalocalminimum. If f( a ) < 0,thennear a thegraphof f lookslikeaparabolaconcave downwardasdepictedintheleftpanelofFigure4.8(alocalmaximum). Forexample,cos x hasalocalmaximumat a =0,anditbehavesnear a =0ascos x T2( x )=1 Š x2/ 2. Thesecondderivativetestisinconclusiveif f( a )=0.Inthiscase, f ( x )behavesnear a as f ( x ) T3( x )= f ( a )+1 6f( a )( x Š a )3. So,if f( a ) =0, f hasanin”ectionpointat a asdepictedintheright panelofFigure4.8.If f( a )=0,oneshouldlookat f ( x ) T4( x )= f ( a )+1 24f(4)( a )( x Š a )4. Afunctionhasalocalmaximum(minimum)at a if f(4)( a ) < 0 ( f(4)( a ) > 0)astheconcavitydoesnotchangeat x = a .Thisistobe comparedwithexamplesgiveninthecaptionofFigure4.6.Itisnow

PAGE 134

1284.APPLICATIONSOFDIFFERENTIATION clearthatthelocalbehaviorof f nearitscriticalpointisdetermined byaTaylorpolynomialthathasthe“rstnonvanishingcorrectionto f ( a ),providedthefunctionisdierentiablesucientlymanytimes. Example 4.7 Investigate f ( x )= x Š tan x near x =0 Solution: FindaTaylorpolynomialfortan x withtwonontrivial terms.Inthiscase,itis T3:tan x T3( x )= x + x3/ 3(seeSection 25.3).Therefore, f ( x ) x Š T3( x )= Š x3/ 3.Sothereisanin”ection pointat x =0. 25.5.Asymptotes.Howcanthebehaviorofafunctionnear a beanalyzedifthefunctionisnotdierentiableat a ,ornotevende“nedat a ,orhowdoesitbehaveintheasymptoticregions x ? Definition 4.6(VerticalAsymptotes) Theline x = a isaverticalasymptoteofthegraph y = f ( x ) ifatleastoneofthelimits limx af ( x ) isin“nite( or Š ). Inotherwords,thefunction f ( x )increases(decreases)unboundedly as x approaches a fromeithertheleftortheright.Forexample,the function (4.12) f ( x )= x ( x2+3) x2Š 1 = x ( x2+3) ( x Š 1)( x +1) hastwoverticalasymptotesbecausethedenominatorvanishesat x =1 and x = Š 1.When x approaches Š 1fromtheleft, f ( x )tendsto Š whileittendsto if Š 1isapproachedfromtheright.Similarly, f ( x ) Š as x 1Šand f ( x ) as x 1+. Suppose f hasaverticalasymptoteat a .Howdoesitbehavenear a ?HowfastŽdoesitdivergewhen x getscloserto a ? Definition 4.7(AsymptoticBehavior) Thefunctions f ( x ) and g ( x ) onanopeninterval x>a (including x> Š )or x

PAGE 135

25.TAYLORPOLYNOMIALSANDTHELOCALBEHAVIOR129 withtheproperty(4.13).Inotherwords,onelooksforasimpleway toestimatethevaluesof f ( x )near a Example 4.8 Findtheasymptoticbehaviorofthefunction(4.12) at x = 1 Solution: Thefunctionhastobeinvestigatednear x = 1andalso when x 1 .Near x = Š 1,theunboundedgrowthof f ( x )isassociatedwiththe divergentfactor1 / ( x +1)sothat f ( x )= h ( x ) / (1+ x ),where h ( x )is “nitenear x = Š 1.Then f ( x ) h ( Š 1) / ( x +1)= g ( x ): f ( x )= 1 x +1 x ( x2+3) x Š 1 2 x +1 = g ( x ) Apparently,limx Š 1( f ( x ) Š g ( x ))=0.Thegraphsof f ( x )and g ( x )= 2 / ( x +1)areclosenear x = Š 1. 10 5510 20 10 10 20 Figure4.9. Graphof f ( x )givenin(4.12)(theblue solidcurve).Ithasaslantasymptote g ( x )= x as x (thedashedline).Intheseasymptoticregions, f ( x ) g ( x )= x .Thefunctionalsohastwovertical asymptotes x =1(theredverticalline)and x = Š 1(the blueverticalline).Theredsolidcurveisthegraphof g ( x )=2 / ( x Š 1),whichshowstheasymptoticbehaviorof f ( x )near x =1.Inaneighborhoodof x =1, f ( x ) g ( x )near x =1.Thefunction f exhibitsasimilarbehaviornear x = Š 1(notdepictedhere).

PAGE 136

1304.APPLICATIONSOFDIFFERENTIATION 2 .Similarly,near x =1 f ( x )= 1 x Š 1 x ( x2+3) x +1 2 x Š 1 = g ( x ) 3 .To“ndanasymptoticbehaviorwhen x islargeitisconvenientto factoroutthelargestpowerof x inthenumeratoranddenominator: f ( x )= x x2(1+3 x2) x2(1 Š1 x2) = x 1+3 x2 1 Š1 x2Since u =1 /x2issmallintheasymptoticregion,thefactor(1 Š u )Š 1 1+ u canbelinearized.Thereforetheasymptoticbehaviorof f ( x )is f ( x ) x 1+ 3 x2 1+ 1 x2 x 1+ 4 x2 = x + 4 x wheretheterms xŠ 4havebeenneglected.Thisshowsthatthegraph hastheslantasymptote y = x .Since f ( x ) Š x 4 /x ispositiveif x> 0 andisnegativeif x< 0,thegraphapproachestheslantasymptotefrom aboveas x andfrombelowas x Š 25.6.AsymptoticBehaviorandTaylorPolynomials.Taylorpolynomials alsoprovideapowerfultechniquetoinvestigateanasymptoticbehavior ofafunction.Thisisillustratedbythefollowingexample. Example 4.9 Investigate f ( x )= xŠ 8 / 3(1 Š cos x ) near x =0 Solution: Thefactor xŠ 8 / 3divergesas x 0,butcos x issmooth near x =0andcanbeapproximatedbytheTaylorpolynomial: cos x T4( x )= T4=1 Š1 2x2+1 24x4, f ( x ) xŠ 8 / 3(1 Š T4( x ))= xŠ 8 / 3(1 2x2Š1 24x4)=1 2xŠ 2 / 3Š1 24x4 / 3. Therefore,forasucientlysmall x f ( x ) 1 2xŠ 2 / 3= g ( x )because f ( x ) Š g ( x ) 1 24x4 / 3 0as x 0.Notethattheuseof T2inplaceof T4wouldnotbeenoughtoestablishtheasymptoticbehaviorof f 25.7.Exercises.(1)FindTaylorpolynomialsof f ( x )= x4Š x3+5 x2Š 2 x +1at x =0.WhatcanbesaidaboutTaylorpolynomialsofageneral polynomialfunction Pn( x )= anxn+ an Š 1xn Š 1+ + a1x + a0at x =0? (2)Decomposethepolynomial P ( x )=1+3 x +5 x2Š 2 x3into thesumofpowersofthemonomials(1+ x )n,where n isa nonnegativeinteger.

PAGE 137

25.TAYLORPOLYNOMIALSANDTHELOCALBEHAVIOR131 (3)Approximate f ( x )=(1+ x + x2) / (1 Š x + x2)by T2( x )about x =0. Hint: Approximate“rst(1+ u )Š 1by T2( u )about u =0 andthenuse f ( x ) (1+ x + x2) T2( u ) where u = Š x + x2, retainingonly xn, n 2,intheproduct. (4)Find T2( x )about x =0for f ( x )=m am+ x a> 0. (5)Find T2( x )about x =0for f ( x )= 1 Š 2 x + x2. Hint: Use T2( u )for 1+ u about u =0andset u = Š 2 x + x2. (6)Find T3( x )for f ( x )= e2 x Š x2about x =0. Hint: Useasuitable approximationof eubyaTaylorpolynomialwhere u =2 x Š x2. (7)Find T4( x )for f ( x )= 1+ x2Š 1 Š x2about x =0. Hint: UseasuitableTaylorpolynomialtoapproximate 1 u ,where u = x2. (8)Findthe n th-degreeTaylorpolynomialsofthegivenfunction ataspeci“edpoint: (i) f ( x )=sin x x =0 (ii) f ( x )=cos x x =0 (iii) f ( x )=ln x x =1 (iv) f ( x )=1 /x x =2 (v) f ( x )=(1+ x )p,p> 0, x =0 (vi) f ( x )= exŠ eŠ x, x =0 (9)Estimatetheabsoluteerroroftheapproximationforagiven interval: (i)sin x x Š1 6x3, | x |1 2(ii)tan x x +1 3x3, | x | 0 1 (iii) 1+ x 1+1 2x Š1 8x2,0 x 1 Hint: In(i)and(ii),compare T3( x )and T4( x )about x =0. (10)Findanintervalinwhichtheapproximationcos x 1 Š1 2x2isaccuratewithintheabsoluteerror0 0001. (11)If f istwicecontinuouslydierentiablenear x =0and f (0)= 0,“ndthelocalbehaviorofthefunction F ( x )= f ( | x |p)near x =0,where p> 0. (12)Let f and g betwicedierentiableat a and g ( a )=0.Find thesecond-degreeTaylorpolynomialforthefunction F ( x )= f ( g ( x ))near x = a Hint :Use f ( u ) T2( u )= f (0)+ f(0) u + f(0) u2/ 2,where u = g ( x )and g ( x )isalsoapproximatedby thecorresponding T2near a (13)Findthethird-degreeTaylorpolynomialforthefollowingfunctionsataspeci“edpointbyusingtheresultsfromtheprevious

PAGE 138

1324.APPLICATIONSOFDIFFERENTIATION exercises(i.e.,byusingTaylorpolynomialsofasuitablychosenargument): (i) f ( x )=sin( x3), x =0 (ii) f ( x )=sin(sin x ), x =0 (iii) f ( x )=tan(1 Š cos x ), x =0 (14)UseTaylorpolynomialstoinvestigatethelocalbehaviorofa givenfunctionnearaspeci“edcriticalpoint(whetherithasa localmaximum,alocalminimum,oranin”ection): (i) f ( x )=sin( x4), x =0 (ii) f ( x )=1 Š x2/ 2 Š cos x x =0 (iii) f ( x )=ln(1+ x ) Š x + x2, x =0 (15)UseTaylorpolynomialsofsuccessivedegreesfor f ( x )=ln(1+ x )near x =0toevaluateln2.Whatdegreeisrequiredto calculateln2correctwithintheabsoluteerror10Š 4? (16)UseTaylorpolynomialsto“ndthenumbercorrectwithinthe givenabsoluteerror : (i) e =10Š 4(ii)sin1, =10Š 4(iii)cos9, =10Š 5(iv) 5, =10Š 4(v)log1011, =10Š 5(17)Findverticalandslantasymptotes,ifany,ofagivenfunction. Investigatetheasymptoticbehaviorofthefunctionnearthe pointswhereithasverticalasymptotesandintheasymptotic regions x (i) f ( x )= x +4 /x (ii) f ( x )= x2/ ( x2Š 1) (iii) f ( x )=( x3Š 3 x2) / ( x2Š 2 x +1) (iv) f ( x )= x2 / 3( x2Š 1)Š 1 / 3(v) f ( x )=(cos x Š 1) /x2(vi) f ( x )=( x Š sin x ) /x4(18)Approximatethegivenfunctionbyapowerfunctionneara speci“edpointbyusingTaylorpolynomials: (i) f ( x )= xŠ 2 / 3ln(1+ x ), x =0 (ii) f ( x )= xŠ 4 / 3sin2(2 x2 / 3), x =0 (iii) f ( x )= xŠ 5 / 3( x Š tan( x1 / 3)), x =0 (iv) f ( x )=[sin( x Š 1) Š x +1] / ( x Š 1)3, x =1 (19)Supposethatthefunctions f and g aresuchthat f ( a )= g ( a )=0, f( k )( a )=0for k =1 2 ,...,n ,and g( k )( a )=0 for k =1 2 ,...,m ,while f( n +1)( a ) =0and g( m +1)( a ) =0.

PAGE 139

26.L’HOSPITAL’SRULE133 Investigatethelocalbehaviorofthefunction h ( x )= f ( x ) /g ( x ) near x = a if n = m ,if n>m ,andif m
PAGE 140

1344.APPLICATIONSOFDIFFERENTIATION Forthespecialcaseinwhich f ( a )= g ( a )=0,thederivatives fand garecontinuous,and g( a ) =0,itisnotdiculttoseewhy lHospitalsrule(4.16)holds: limx af ( x ) g ( x ) =limx af ( x ) Š f ( a ) g ( x ) Š g ( a ) =limx a f ( x ) Š f ( a ) x Š a g ( x ) Š g ( a ) x Š a= limx a f ( x ) Š f ( a ) x Š a limx a g ( x ) Š g ( a ) x Š a= f( a ) g( a ) =limx af( x ) g( x ) The“rstequalityfollowsfrom f ( a )= g ( a )=0,thesecondandthird equalitiesaretheconsequenceofthelimitlawsandtheassumption that g( a ) =0,andthelastequalityfollowsfromthecontinuityofthe derivatives.Thissimpli“edversionoflHospitalsrulecanbeunderstoodgeometrically.Thefunctions f and g canbeapproximatedby theirtangentlinesat a f ( x ) f( a )( x Š a )and g ( x ) g( a )( x Š a ), sothat f ( x ) /g ( x ) f( a ) /g( a )near a ItisnotsoeasytoprovethegeneralversionoflHospitalsrule (theproofisomittedhere).LHospitalsruleisalsovalidforone-sided limits x aandforthelimitsat .ThehypothesesoflHospitals rulemustbeveri“edforthecorrespondinglimits. Whathappensif f( a )= g( a )=0?Apparently,theconditions oflHospitalsrulearesatis“edforthederivatives f( x )and g( x )if f and g aretwicedierentiable.SolHospitalsrulemaybeapplied againtotheratio f( x ) /g( x ).Forfunctionsdierentiablemanytimes, lHospitalsruleiseasytounderstandviatheTaylorpolynomials.Supposethatfunctions f and g arecontinuouslydierentiablesuciently manytimesnear a .ThenbyTheorem4.11thefollowingapproximation holds f ( x ) g ( x ) f ( a )+ f( a )( x Š a )+1 2f( a )( x Š a )2+ g ( a )+ g( a )( x Š a )+1 2g( a )( x Š a )2+ If f ( a )= g ( a )=0,thenthelimitoftheratioisdeterminedby f( a ) /g( a ).If f ( a )= g ( a )=0and f( a )= g( a )=0,thenthe limitisdeterminedby f( a ) /g( a )andsoon. Example 4.10 Investigatetheindeterminateforms(4.14)and (4.15). Solution:1 .Let f ( x )= exŠ 1and g ( x )= x .Then f (0)= g (0)=0 (theconditionsoflHospitalsruleareful“lled).Hence, limx 0exŠ 1 x =limx 0( exŠ 1) ( x )=limx 0ex 1 =1 .

PAGE 141

26.L’HOSPITAL’SRULE135 2 .Let f ( x )=1 Š cos x and g ( x )= x2sothat f (0)= g (0)=0. Then f( x )=sin x and g( x )=2 x .Since f(0)=0and g(0)=0, lHospitalsrulecanbeappliedagain: limx 01 Š cos x x2=limx 0sin x 2 x =limx 0(sin x ) (2 x )=limx 0cos x 2 = 1 2 3 .Let f ( x )=tan x Š x and g ( x )= x3sothat f (0)= g (0)=0. Thederivatives f( x )=sec2x Š 1and g( x )=3 x2vanishat x =0. LHospitalsrulecanbeusedagaintoresolvetheindeterminateform. Forcomplicatedfunctions,takinghigher-orderderivativesmightbe quiteanalgebraicexercise.Sometimes,simplealgebraictransformationsofanindeterminateformincombinationwithbasiclimitlaws mayleadtotheanswerfasterthanasuccessiveuseoflHospitalsrule: limx 0tan x Š x x3=limx 0sec2x Š 1 3 x2=limx 01 Š cos2x 3 x2cos2x =limx 0sin2x 3 x2= 1 3 limx 0sin x x 2= 1 3 Thethirdequalityfollowsfromcos x 1as x 0,andtherefore cos2x inthedenominatorcanbereplacedby1inaccordwiththe basiclimitlaws. 4 .Let f ( x )=ln x and g ( x )= xŠ 1sothat f ( x ) Š and g ( x ) as x 0+.SotheconditionsoflHospitalsruleareful“lled.Therefore, limx 0+ln x xŠ 1=limx 0+(ln x ) ( xŠ 1)=limx 0+xŠ 1 Š xŠ 2= Š limx 0+x =0 26.2.IndeterminateProducts0.Supposethat f ( x ) and g ( x ) 0as x a .Howcantheindeterminateproduct f ( x ) g ( x ) beinvestigatedwhen x a ?Itturnsouttheindeterminateproductcanbetransformedintooneoftheindeterminateformstowhich lHospitalsruleisapplicable: (4.17) fg = f 1 /g 0 or fg = g 1 /f 0 0 0 Thefunction x ln x isanindeterminateproductofthetype0 as x 0+.Itcanbetransformedintoanindeterminateformofthe type asin(4.15),whichisthenresolvedbylHospitalsrule(see Example4.10).Notethat,althougheitherofthetransformationsin (4.17)maybeappliedwiththesubsequentuseoflHospitalsrule,the

PAGE 142

1364.APPLICATIONSOFDIFFERENTIATION technicalitiesinvolvedmightdiersubstantially.Forinstance,ifthe secondoptionin(4.17)isappliedto x ln x = x/ (1 / ln x ),then limx 0+x ln x =limx 0+x 1 ln x=limx 0+1 Š1 ln2x 1 x= Š limx 0+x ln2x. Althoughourgoalhasnotbeenachieved,oureorthasnotbeenin vain.Sincetheleft-handsidevanishesbyExample4.10,itfollowsthat x ln2x 0as x 0+.Byrepeatingthisprocedurerecursively,one caninferthat limx 0+x (ln x )n=0 ,n =1 2 ,....26.3.IndeterminatePowers00,0,and1.Severalindeterminate formsarisefromthelimitsof[ f ( x )]g ( x )as x a : 00( f ( x ) 0 ,g ( x ) 0); 0( f ( x ) ,g ( x ) 0); 1( f ( x ) 1 ,g ( x ) 0) Note c0=1if c =0and c = .Similarly, c=0if0 c< 1and c= if c> 1.Theindeterminatepowerscanbetransformedinto anindeterminateproductwiththehelpoftheidentity y = eln y: limx a[ f ( x )]g ( x )=limx aeln([ f ( x )]g ( x ))=limx aeg ( x )ln( f ( x ))= elimx ag ( x )ln( f ( x )). Thelimitof g ( x )ln( f ( x ))isoftype0 andcanbetreatedbythe rule(4.17).Theprocedureisillustratedwithanexampleofthetype 0indeterminatepower: limx x1 /x=limx eln( x1 /x)=limx eln( x ) /x= elimx ln( x ) /x= e0=1 .26.4.IndeterminateDifferencesŠ.Suppose f ( x ) and g ( x ) as x a .Thelimitof f ( x ) Š g ( x )as x a iscalled an indeterminatedierence .Thefollowingtransformationsmightbe helpfultoinvestigateit: f Š g = f 1 Š g f = 1 Š g/f 1 /f or f Š g = g f g Š 1 = f/g Š 1 1 /g If f ( x ) /g ( x ) 1,thentheindeterminatedierenceisequivalenttoan indeterminateformoftype0 / 0andcanbeinvestigatedbylHospitals rule.Thelimitof f/g isanindeterminateformoftype / andcan alsobeinvestigatedbylHospitalsrule.Supposethat f ( x ) /g ( x ) k as x a ,where k canbeeitheranonnegativenumberor k = If k< 1,then f Š g = g ( f/g Š 1) ( k Š 1)= Š ;that is, g increasesfasterthan f as x a .If k> 1or k = ,then

PAGE 143

26.L’HOSPITAL’SRULE137 f Š g = g ( f/g Š 1) ( k Š 1)= ;thatis, f increasesfaster than g as x a .Forexample, limx 0+ ln x + 1 x =limx 0+1 x 1+ x ln x =limx 0+1 x (1+0)= If k =1,thenitisalsopossiblethat f Š g c ,where c isanumber.In thiscase, f and g increaseasymptoticallyatthesamerate: fŠ g 0. If c =0,thefunctions f and g havethesameasymptoticbehavior.For example, limx 0 1 sin x Š cot x =limx 01 sin x 1 Š cos x =limx 0sin x cos x =0 wherelHospitalsrulehasbeenusedinthesecondequality. Analternativesolutionisobtainedifthelocalbehaviorofthefunctionsnear x =0isapproximatedbytheTaylorpolynomials.Use T2toapproximatecos x and T3forsin x : 1 sin x Š cot x = 1 Š cos x sin x x2/ 2 x Š x3/ 6 = x/ 2 1 Š x2/ 6 x 2 where x2/ 6issmallascomparedto1when x iscloseenoughto0and canthereforebeneglectedinthedenominator.Thismethodisoften technicallyeasierthantheuseoflHospitalsrule.26.5.Exercises.(1)Findthelimits: (i)limx 0ln(1+ x ) x (ii)limx 0x Š sin x x3(iii)limx / 4sin x Š cos x cos(2 x ) (iv)limx 0sin( ax ) sin( bx ) (v)limx 01 Š cos( x2) x2sin( x2) (vi)limx 0sinŠ 1( ax ) sin( bx )

PAGE 144

1384.APPLICATIONSOFDIFFERENTIATION (vii)limx 0sinŠ 1(2 x ) Š 2sinŠ 1x x3(viii)limx 0+sin x ln x (ix)limx 0+[1 Š cos x ]ln x (x)limx 0sin x Š x exŠ 1 Š x2/ 2 (xi)limx 0tan x Š x ex3Š 1 (xii)limx 0xŠ 2 x +1 Š 1 x Š 1 2 (xiii)limx 1+ 2 x x(xiv)limx (1 Š eŠ x)ex(xv)limx x Š (ln x )n ,n> 0 (xvi)limx x Š ln x 1 /x(xvii)limx 0+(sin( ax ))sin( bx ),a> 0 ,b> 0 (xviii)limx 0+lnsin( ax ) lnsin( bx ) ,a> 0 ,b> 0 (xix)limx 0lncos( ax ) lncos( bx ) (xx)limx 0x cot x Š 1 x2(xxi)limx aaxŠ xa x Š a ,a> 0 (2)UseapproximationsofbasicfunctionsbyTaylorpolynomials to“ndthelimit: (i)limx 0cos x Š eŠ x2/ 4 x4(ii)limx 0exsin x Š x (1+ x ) x3(iii)limx x3 / 2( x +1+ x Š 1 Š 2 x )

PAGE 145

26.L’HOSPITAL’SRULE139 (iv)limx 0ax+ aŠ xŠ 2 x2,a> 0 (v)limx x Š x2ln 1+ 1 x (vi)limx 0 1 x Š 1 sin x (vii)limx 01 x 1 x Š cot x (3)Supposethat f ( x )hasthesecondderivative f( x ).Showthat f( x )=limh 0f ( x + h )+ f ( x Š h ) Š 2 f ( x ) h2. (4)Let y 0as x 0.Findtheasymptoticbehaviorof y ,that is,theleadingterm y Cxn, C =0,if (i) y =tan x Š x (ii) y =tan(sin x ) Š sin(tan x ) (iii) y =(1+ x )xŠ 1 (iv) y =1 Š eŠ 1(1+ x )1 /x(5)If y = x Š ( a + b cos x )sin x ,“nd a and b suchthat y Cx5as x 0, C =0. (6)Find a and b suchthattheapproximation cot x 1+ ax2 x + bx2iscorrectwhenthetermsoforder x5andhighercanbeneglectedas x 0. (7)Considerthefunction f ( x )= eŠ 1 /x2if x =0and f (0)=0. Show“rstthat limx 0xneŠ 1 /x2=0 ,n> 0 Usethisfactandthede“nitionofderivativestoshowthat f( k )(0)=0forall k .CanTaylorpolynomialsbeusedto investigatethelocalbehaviorofthefunctionnear x =0and establishthenatureofthecriticalpoint x =0? (8)Findallcriticalpointsofthefunction f ( x )= eŠ 1 /x,x> 0 0 ,x =0 Š e1 /x,x< 0 andinvestigatethebehaviorofthefunctionnearthem.

PAGE 146

1404.APPLICATIONSOFDIFFERENTIATION 27.AnalyzingtheShapeofaGraph Toanalyzetheshapeofagraph y = f ( x ),itisusefultohaveaclear ideaofhowthebasicfunctionsbehave.Forexample,sin x andcos x areregulareverywhere,bounded(e.g., | sin x | 1),andperiodicwitha periodof2 .Inaddition,sin x haszerosat x = n n =0 1 2 ,... whilecos x vanishesat / 2+ n .Thefunctionsin x isodd,while cos x iseven.Theirratiotan x =sin x/ cos x isnotde“nedatrootsof cos x .Howdoestan x behave,say,near x = / 2?Sincebothsin x and cos x aresmoothnear x = / 2,thebehavioroftan x near / 2canbe understoodwiththehelpofTaylorpolynomials.Put x = x Š / 2 (thedeviationof x from / 2).Letusapproximate sin x T1( x )=1+( x Š 2)=1+ x cos x T3( x )= Š ( x Š 2)+1 6( x Š 2)3= Š x +1 6( x )3. Then tan x 1+ x Š x +( x )3/ 6 = Š 1 x 1+ x 1 Š ( x )2/ 6 Š 1 x = Š 1 x Š / 2 wherethesecondratiointheproducthasbeenapproximatedby1 because x issmall.Sincetan( x + )=tan x ,thisbehaviorrepeats itselfatneareveryrootofcos x .27.1.GrowthofthePower,Exponential,andLogarithmicFunctions.Let uscomparethegrowthofthepowerfunction xn,theexponentialfunction ex,andthelogarithmicfunctionln x as x Theexponential functiongrowsfasterthanthepowerfunction .Let f ( x )= exand g ( x )= xn.Letusanalyzetheratio f/g as x .Theconditions oflHospitalsrulearesatis“ed: ex and xn as x LHospitalsrulecansuccessivelybeapplieduntiltheindeterminate formisresolved: limx ex xn=limx ex nxn Š 1=limx ex n ( n Š 1) xn Š 2= =limx ex n = Theconclusionistrueforanyreal n .Foranyreal n ,thereexists apositiveinteger N suchthat n 1.But exgrowsfasterthan xN.Similarly,itisstraightforwardtoshowthatthe logarithmicfunctiongrowsslowerthananypowerfunction : limx ln x xn=limx (ln x ) ( xn)=limx 1 x nxn Š 1=limx 1 nxn=0 forany n> 0( n maybeanypositiverealnumberhere).

PAGE 147

27.ANALYZINGTHESHAPEOFAGRAPH141 27.2.Asymptotesatx .Theasymptoticbehaviorofrational functionsiseasilydeterminedbythehighestpowersofthenumerator anddenominator,asinExample4.8.Ingeneral,iflimx f ( x )is in“nite,thenthelimitof f/g canbestudiedfortrial g swithdierent growth, g = mx (forslantasymptotes), g = xn, g =ln x ,andsoon. Suppose g ( x )isfoundsuchthat f ( x ) /g ( x ) 1as x .Doesthis meanthat g and f havethesameasymptoticbehavior?Theansweris no.ŽIftheindeterminateform f ( x ) Š g ( x )oftype Š converges to0as x ,thentheindeterminateform f ( x ) /g ( x )oftype convergesto1.Indeed,itfollowsfrom1 /g ( x ) 0and f ( x ) Š g ( x ) 0 that(1 /g ( x ))( f ( x ) Š g ( x ))= f ( x ) /g ( x ) Š 1 0. Theconverseisnot true .Considerthefollowingsimpleexample: f ( x )= x +sin x and g ( x )= x .Evidently, f ( x ) /g ( x )=1+sin x/x 1as x .But thelimitlimx ( f ( x ) Š g ( x ))=limx sin x doesnotexist.So,even if g isfoundtohavetheproperty f ( x ) /g ( x ) 1as x ,the indeterminateform f Š g oftype Š muststillbeinvestigatedin ordertodeterminewhetherornot g hasthesameasymptoticbehavior as f .27.3.GuidelinesforAnalyzingtheShapeofaGraph.Thefollowing guidelinesareusefulforsketchingthegraphofafunction.Itshould benotedthatnotallthestepscanalwaysbecarriedout.Thisdepends verymuchonthecomplexityofthefunctioninquestion.Sotheseare reallyguidelines,notamust-doŽalgorithm.Givenafunction f ,“nd: (I) Domain. Thedomainconsistsofallvaluesof x atwhich f ( x )isde“ned.Typically,itisacollectionofintervals.If f isde“ned for x>a or x

PAGE 148

1424.APPLICATIONSOFDIFFERENTIATION Thegraph y = f ( x )repeatsitselfonintervalsoflength p ,for example[ a,a + p ],[ a + p,a +2 p ],andsoonforany a .Examples aresin x p =2 ;tan x p = ;cos(4 x ), p =2 / 4= / 2. (IV) Asymptotesandasymptoticbehaviorof f If f isaratio f = h/g ,thenverticalasymptotesare x = c where c solves g ( c )=0and h ( c ) =0.If h ( c )=0,“ndthe limitslimx cf ( x ).Ifoneofthelimitsorbothisin“nite, investigatethelocalbehaviorof f near c (e.g.,withthehelp ofTaylorpolynomialsifpossible).Theasymptoticbehaviorof f ( x )near c andforlargepositiveandnegative x determines theshapeof y = f ( x )neartheverticalasymptotesandthe asymptoticshapeofthegraphwhen x (V) Criticalpointsof f Criticalpointsaresolutionsof f( x )=0orthevaluesof x where f( x )doesnotexist.If,forexample, f( x )tendsto (or Š )as x approaches c ,thenthelinetangenttothe graph y = f ( x )at x = c isvertical.Forexample, f ( x )= x1 / 3and f( x )=1 / (3 x2 / 3).So f( x )divergesas x 0.Thegraph y = x1 / 3hasaverticaltangentlineat x =0. (VI) Intervalsofpositiveandnegativevaluesof f Thesearetheintervalswherethegraph y = f ( x )liesaboveor belowthe x axis.Rootsof f generallyseparatetheintervalsof positiveandnegativevaluesof f .However,thisisnotalways thecase.Let c bearootof f .If f( c ) =0,thenthefunction f isincreasingordecreasingat c andhencemustchangeits sign.If f( c )=0or fdoesnotexistat c ,thatis,aroot of f coincideswithitscriticalpoint,then f isnegativenear c if f hasalocalmaximumat c and f ispositivenear c if ithasalocalminimumat c .Sothesignofthederivative fmustbeinvestigatednear c (the“rstderivativetest).Vertical asymptotescanalsoseparateintervalsofpositiveandnegative valuesof f .Forexample,thefunction(4.12)hasoneroot x =0 andtwoverticalasymptotesat x = Š 1and x =1.So f is negativeon( Š Š 1),positiveon( Š 1 0),negativeon(0 1), andpositiveon(1 ).ThegraphisshowninFigure4.9. (VII) Intervalsofincrease( f> 0)anddecrease( f< 0 ). If f> 0( f< 0)onaninterval,then f increases(decreases) onit(theincreasing-decreasingtest).Theseintervalsaregenerallyseparatedbycriticalpointsandverticalasymptotes.As aconsequenceofthisstudy,thenatureofeachcriticalpoint isestablishedbythe“rstderivativetest.

PAGE 149

27.ANALYZINGTHESHAPEOFAGRAPH143 (VIII) Intervalsofupwardanddownwardconcavity. Theseintervalsareseparatedbyin”ectionpointsandvertical asymptotes.Thesignof f( x )mustbestudied.Yet,thesecond derivativetestandTaylorpolynomialscanbeusedtoestablish thenatureofacriticalpointof f (IX) Valuesof f atcriticalpointsandinectionpoints. Thesevaluessetrelativescalesofthegraph(e.g.,theyshow howmuchthefunctionincreasesbetweentwocriticalpoints). Example 4.11 Sketchthegraphof f ( x )= x1 / 3( x Š 6)2 / 3. Solution: Followingtheprecedingguidelines: (I)Thedomainisthewholerealline. (II)Therootsof f are x =0and x =6(theinterceptswiththe x axis).Theinterceptwiththe y axisis f (0)=0. (III)Thefunctionisnotperiodic,anditisneitheroddnoreven. (IV)Thereisnoverticalasymptote.Tostudytheasymptoticbehavioras x ,itisconvenienttofactoroutthelargest powerof x : f ( x )= x (1 Š 6 /x )2 / 3andapproximatethesecond factorusingTaylorpolynomialsin u = Š 6 /x .Onehas (1+ u )2 / 3 T2( u )=1+2 3u Š1 9u2=1 Š 4 x Š 4 x2f ( x ) xT2( u )= x Š 4 Š 4 x Thisshowsthatthegraphhasaslantasymptoteoftheform y = x Š 4.Italsofollowsthat f ( x ) Š ( x Š 4) Š 4 /x< 0 if x islargeandpositive.Hence,thegraphapproachesthe slantasymptotefrombelow.Similarly, f ( x ) Š ( x Š 4) > 0if x islargeandnegative.Hence,thegraphapproachestheslant asymptotefromabove. (V)Thederivativereads f( x )= x Š 2 x2 / 3( x Š 6)1 / 3. Itvanishesat x =2anddoesnotexistat x =0and x =6. Thecriticalpointsare0,2,and6.Inparticular, f( x ) as x 0andittendsto as x 6,respectively.Therefore, thegraphhasverticaltangentlinesat x =0and x =6.Near x =0,thegraphlookslike y = f ( x ) 62 / 3x1 / 3,whilenear x =6,ithasadownwardcusp y = f ( x ) 61 / 3( x Š 6)2 / 3. (VI)Thegraphliesbelowthe x axison( Š 0)as f ( x ) < 0and aboveiton(0 )as f ( x ) 0.Thefunctiondoesnotchange

PAGE 150

1444.APPLICATIONSOFDIFFERENTIATION Figure4.10. Graphof f ( x )= x1 / 3( x Š 6)2 / 3.Theroots of f are x =0and x =6,andtheyde“netheintercepts withthe x axis.Ithastheslantasymptote f ( x ) x Š 4as x .Thederivativevanishesat x =2(a localmaximum).Itdivergesat x =0and x =6;the graphhasverticaltangentlinesatthesepoints.The secondderivativeisnegativeif x< 0sothatthegraph isdownwardconcave.Itispositiveon(0 6)and(6 ). Thegraphisconcavedownward.Thepoint x =0isan in”ectionpointastheconcavitychangesatit. itssignattheroot x =6( f musthavealocalminimumat6, whichisalsoveri“edbythe“rstderivativetestbelow). (VII)Thederivativeisaproductofthreefactors x Š 2, xŠ 2 / 3,and ( x Š 6)Š 1 / 3.Byinvestigatingthesignsofthesefactorsonthe intervalsseparatedbythecriticalpoints,wecanconcludethat f> 0( f isincreasing)on( Š 0), f> 0( f isincreasing) on(0 2), f< 0( f isdecreasing)on(2 6),and f> 0( f is increasing)on(6 ).Also, f hasalocalmaximumat x =2 andalocalminimumat x =6bythe“rstderivativetest.

PAGE 151

27.ANALYZINGTHESHAPEOFAGRAPH145 (VIII)Thesecondderivativereads f( x )= Š 8 x5 / 3( x Š 6)4 / 3. Thefactor( x Š 6)4 / 3cannotbenegative.Thesignof fis determinedonlybythatof x5 / 3.Thus, f> 0on( Š 0)(the graphisupwardconcave)and f< 0on(0 6)and(6 )(the graphisdownwardconcave).So x =0isthein”ectionpoint. Also,near x =2,thegraphlookslikethedownwardparabola y = T2( x )= f (2)+ f(2)( x Š 2)2/ 2=(4 Š1 4( x Š 2)2) /3 2. Intheageofgraphingcalculators,theprecedingguidelinesmight lookratherobsoletebecause“ndingtheshapeofagraphcanbedone justbyhittingtheappropriatecalculatorbuttons.Butwhatacalculatorcannotdoistoprovidedetailsofthelocalbehaviorofafunctionnearpointsofinterest(e.g.,criticalpoints,asymptotes,etc.).In scienceandengineering,thisisoftenmuchmoreimportantthanthe overallshapeofagraph.Inthepreviousexample,acalculatorwould showthatthereisaslantasymptote,acuspat x =6,andalocal maximumat x =2,butitwouldnotbeabletodeterminethelocal behaviorofthefunctionnearthecusp,oratthelocalmaximum,or intheasymptoticregion.Hereagoodworkingknowledgeofcalculus becomesindispensable,whileagraphingcalculatorisjustausefultool thatgreatlyfacilitatesthestudyofafunction.27.4.Exercises.(1)Sketchthegraphofeachofthefollowingfunctions: (i) f ( x )=3 x Š x3(ii) f ( x )=1+ x2Šx4 2(iii) f ( x )= x2( x Š 1)(2 Š x ) (iv) f ( x )=( x +1)( x Š 2)2(v) f ( x )= x2/ (4+ x2) (vi) f ( x )= x4/ (1+ x2)2(vii) f ( x )= x/ [(1+ x )(1 Š x2)] (viii) f ( x )= x3/ ( x2Š 3 x +2) (ix) f ( x )=( x5Š x2) / ( x +1) (x) f ( x )=[(1+ x ) / (1 Š x )]4(xi) f ( x )= ( x Š a )2+ b2(xii) f ( x )=( x Š 2) / x2+1 (xiii) f ( x )=3 x3Š x2Š x +1 (xiv) f ( x )=3 x2Š3 x2+1

PAGE 152

1464.APPLICATIONSOFDIFFERENTIATION (xv) f ( x )= | 1+ x |2 / 3/ x (xvi) f ( x )= xŠ 1 / 3( x Š 6)Š 2 / 3(xvii) f ( x )= | x2Š 1 |Š x (xviii) f ( x )=[( x Š 1) / ( x +1)]1 / 3(xix) f ( x )=sin x +cos2x (xx) f ( x )=cos x Š1 2cos(2 x ) (xxi) f ( x )=sin x/ (2+cos x ) (xxii) f ( x )=cos x/ cos(2 x ) (xxiii) f ( x )= x sin x (xxiv) f ( x )=2 x Š tan x (xxv) f ( x )=sin( nx ) / sin x n =2 3 4 (xxvi) f ( x )= x + eŠ x(xxvii) f ( x )= xex(xxviii) f ( x )= x2 / 3eŠ x(xxix) f ( x )=( ex+ eŠ x)cos x (xxx) f ( x )=sin2x/x2(xxxi) f ( x )=ln x/ x (xxxii) f ( x )=ln( x + x2+1) (2)Sketchthegraphofthepolynomialwith k realroots: f ( x )= A ( x Š x1)n1( x Š x2)n2 ( x Š xk)nk, where A> 0and n1,n2,...,nkarepositiveintegers.Investigate“rstthecasewhen n1= n2= = nk=1,thenthe casewhenoneofthepowers n1,n2,...,nkisgreaterthan1 (howdoesthegraphlookwhenthispowerisoddorwhenit iseven?).Thenproceedtothegeneralcase. (3)Let f and g besecond-degreepolynomialssuchthat f> 0 and g> 0.Sketchallpossibleshapesofthegraph y = f ( x ) /g ( x ). 28.OptimizationProblems Supposethataquantity Q dependsonsomevariables.Theproblemofoptimizing Q implies“ndingthevaluesofthevariablesatwhich thequantity Q attainsitmaximalorminimalvalue.Thesimplestoptimizationproblemariseswhen Q dependsonasinglevariable x such that Q isafunction f ( x ).Thentheoptimizationproblemisreduced totheproblemof“ndingextremevaluesof f ( x ).Thelatterproblem hasbeenanalyzedinSection22.Todetermineextremevaluesof f ,one hasto: (I)Findallcriticalpointsof f .

PAGE 153

28.OPTIMIZATIONPROBLEMS147 (II)Investigatethenatureofthecriticalpoints(localminimaand localmaxima).The“rstorsecondderivativetestscanbeused forthispurpose. (III)Calculatethevaluesof f attheendpointoftheinterval[ a,b ] (ifextremevaluesaresoughtonlyin[ a,b ])andcomparethem withvaluesof f atitslocalmaximaandminimatodetermine absoluteextremevaluesof f Thefollowingtestcanalsobeusedto“ndabsoluteextremevaluesof afunction. Theorem 4.13(FirstDerivativeTestforAbsoluteExtreme Values) Suppose c isacriticalpointofacontinuousfunction f de“nedonaninterval. (I) If f( x ) > 0 forall xc ,then f ( c ) istheabsolutemaximumvalueof f (II) If f( x ) < 0 forall x 0 forall x>c ,then f ( c ) istheabsoluteminimumvalueof f Theconclusionofthetheoremiseasytounderstand.Consider case(I).Since f( x ) > 0forall xc ,thefunctiondecreasesforall x>c .Bycontinuityof f ,thenumber f ( c )mustbethelargestvalue of f .Case(II)isprovedsimilarly. RecallExample4.2.Thisisatypicaloptimizationproblem.Its solutionisratherstraightforward,providedEquation(4.4)isgiven. Withoutit,theproblemof“ndinganoptimalangleforaprojectile becomesfarmoredicult.Itsmajorpartnowinvolvesaderivation ofEquation(4.4)!Thisisquitetypicalforoptimizationproblems.As arule,theyariseinvariousdisciplines,andtheirformulationasthe mathematical problemofextremevaluesrequiresaspeci“cknowledge outsidemathematics,forexample,thelawsofphysicsasinExample 4.2,chemistry,biology,economics,andsoon.Atypicaloptimization problemmaybesplitintothreebasicsteps: (I)Identifyavariablewithrespecttowhichaquantity Q istobe optimized. (II)Usethelawsofaspeci“cdisciplinetoexpress Q asafunction f ofthatvariable, Q = f ( x ). (III)Solvethemathematicalproblemofextremevaluesof f Example 4.12 Analuminumcanhastheshapeofacylinderof radius r andheight h .Designanaluminumcanofvolume V =300 cm3tominimizethecost(ortheamount)ofmaterialneededtomakethe can.

PAGE 154

1484.APPLICATIONSOFDIFFERENTIATION Solution: Followingtheprecedingguidelines: (I)Apparently,theleastamountofmaterialisusedwhenthe surfaceareaofthecanisminimal.Soonehastominimizethe surfacearea S ,whichdependson r and h .Butthevariables r and h arenotindependentbecausethevolumeis“xed. (II)Thesurfaceareaisthesumoftheareasoftheside,top,and bottomofthecan: S =2 rh + r2+ r2=2 rh +2 r2.The volumeis V = r2h .Sincethevolumeis“xed,thevariables r and h arerelatedas h = V/ ( r2).Hence, S canbewrittenas afunctionoftheradius r only: S ( r )=2 r V r2+2 r2= 2 V r +2 r2. Onehasto“ndthevalueof r> 0atwhich S ( r )attainsits absoluteminimum.Thecorrespondingvalueof h isthenfound fromtherelation h = V/ ( r2). (III)Thefunction S ( r )isdierentiableforall r> 0.Therefore,all itscriticalpointsarerootsofthederivative: S( r )= Š 2 V r2+4 r = 4 r2 r3Š V 2 =0 Sothecriticalpointis rc= V 2 1 / 3. Since S( r ) < 0forall0 0forall r>rc,thefunction S ( r )attainsitsabsoluteminimumat rcbythe“rstderivativetestforabsoluteextremevalues.The dimensionsofthecanwithminimalcostsofmaterialfora givenvolume V are r = V 2 1 / 3 3 6cm ,h = V r2 c= 4 V 1 / 3=2 rc 7 2cm Theanalysishasshownthattheheightanddiameterofacan ofagivenvolumemustbeequalinordertominimizethecost ofmaterial(orthesurfaceareaofthecan).Checkoutalocal supermarkettoseeifmanufacturersusethisfact! Thisexampleisfurtherillustratedontheinteractivewebsite athttp://www.math.u”.edu/ mathguy/ufcalcbook/optimize cylinder .html.

PAGE 155

28.OPTIMIZATIONPROBLEMS149 Remark.Inthepreviousexample, S hasbeenexpressedasafunction of r .Thesameconclusioncouldbereachedif S isexpressedasafunctionoftheheight h only,thatis,whentherelation r = V/ ( h )is substitutedintotheexpressionforthesurfaceareatoobtain S ( h ).The criticalpointof S ( h )canbeshowntobe hc=2 rc.Verifythis!ACuriousFact.Theprecedingproblemisessentialtoreducewaste fromplastic,glass,andaluminumcontainers.Itcanbestatedmore generally.Whatistheshapeofacontainerthathasthesmallestsurface areaatagivenvolume?Itcanbeprovedbythecalculusofvariations thatsuchacontainermustbeasphere.Evenintheexampleofan aluminumcan,theoptimaldimensionsappeartobeasclosetothose ofasphereasthecylindricalgeometrywouldallow:Theheightand diameterarethesame.Shouldonlysphericalcontainersbeusedtogo greenŽ?Toanswerthisquestion,afarmorecomplicatedoptimization problemmustbestudied.Forexample,spheresarenotoptimalfor storageandhencefortransportation;rectangularcontainersarefar better.Storagemaintenanceandtransportationrequireenergy(hence carbonemissions).Theproductionwasteforcontainersofdierent shapesisdierent.Finally,whataboutconsumersreactiontospherical Cokecansinavendingmachineorsphericalaluminumcansinthe supermarket?28.1.ApplicationstoEconomics.InSection19,weintroducedthecost function C ( x ),whichisthecostofproducing x unitsofacertainproduct.Thederivative C( x )isthe marginalcost .Itdeterminesthecostof increasingproductionfrom x unitsto x +1units.Let p ( x )betheprice perunitthatacompanycanchargeifitsells x units.Thefunction p ( x ) isalsocalledthe pricefunction .Naturally,itisgenerallyexpectedto beadecreasingfunctionbecausethepriceperunitusuallygoesdown whenalargernumberofunitsissold.Thetotalrevenue R ( x )= xp ( x ) iscalledthe revenuefunction .Thederivative R( x )iscalledthe marginalrevenuefunction .Itdeterminesthechangeintherevenuewhen thenumberofunitssoldincreasesfrom x to x +1.Finally,the pro“t function P ( x )= R ( x ) Š C ( x )= xp ( x ) Š C ( x ) determinesthetotalpro“tif x unitsaresold.Itsderivative P( x ) determinesthechangeinthetotalpro“twhenthenumberofunits soldincreasesfrom x to x +1.Thestandardoptimizationproblem hereistominimizecostsandmaximizerevenuesandpro“t.

PAGE 156

1504.APPLICATIONSOFDIFFERENTIATION Example 4.13 Asmallstoresellsjeansatapriceof $ 80per pair.Everyweek60unitsaresold.Thecosttothestorefor60unitsis $ 2500,includingthecostoftransportation.Amarketsurveyindicates that,foreach $ 10rebateoeredtobuyers,thenumberofunitssoldwill increaseby20aweek.Also,thepurchaseandtransportationcostswill godownby $ 2pereachweeklyorderincreaseof5units.Howlargea rebateshouldthestoreoertomaximizeitspro“t? Solution:1 .Whatisknownaboutthepricefunction p ( x )?First,its valueataparticularnumberofsoldunits x = x0=60is p0= p (60)= 80.Also,if x increasesbyanamountof x =20,thepricefunction decreasesby p =10(therebate).Thus,theratio m = Š p/ x = Š 1 / 2istherateofchangeof p ( x )(theminussignindicatesthedecrease in p ( x )).Sothepricefunctionis p ( x )= p0+ m ( x Š x0)=80 Š1 2( x Š 60)=110 Š1 2x. 2 .Whatisknownaboutthecostfunction C ( x )?First,itsvalueat aparticularnumberofsuppliedunits x = x0=60is C0= C (60)= 2500.Also,thecostfunctiondecreasesby C =20if x increasesby x =5.Sotheratio M = Š C/ x = Š 4istherateofchangeof C orthemarginalcost.Therefore, C ( x )= C0+ M ( x Š x0)=2500 Š 4( x Š 60)=2740 Š 4 x. 3 .Onehastomaximizethepro“tfunction: P ( x )= xp ( x ) Š C ( x )=114 x Š1 2x2Š 2740 Since P( x )=114 Š x, thefunctionhasonecriticalpoint x =114at which P ( x )attainsitsabsolutemaximalvaluebythe“rstderivative testforabsoluteextremevalues. 4 .If x =114unitscanbesold,thepriceperunitis p (114)=110 Š 57= 53;thatis,therebateshouldbe p (60) Š p (114)=80 Š 53=27.Thus, thestoreshouldoerarebateof$27tomaximizeitspro“t.Notealso theincreaseintheweeklypro“t: P (60)=$2300whereas P (114)= $3758. Remark.Infact,thelinear(tangentline)approximationhasbeenused togettheunknownpriceandcostfunctionsinthepreviousexample. Thisisabene“tofmarketsurveys:Theyestimatethederivatives(or trends)ofthepricefunctions.Naturally,anincreaseinsalesleadstoa decreaseinthedemandforthatparticularitem.So,afterasuccessful rebatecampaign,thestorewouldneedanewmarketsurveytoestimate p(114)andgetthelinearapproximationat x =114.Thepricemaygo

PAGE 157

28.OPTIMIZATIONPROBLEMS151 upthen.Similarly,thecostfunctionisgenerallyhighlynonlinear.Its linearizationnearaparticular x = x0cannotbevalidforall x> x0.Indeed,inthepreviousexample,itvanishesat x =685andbecomes negativeafterthat,whichcannotpossiblybetrue.28.2.Exercises.(1)Amongallrectanglesofagivenarea S ,“ndonewhoseperimeterisminimal. (2)Twoshipsmovealongstraightlineswithconstantspeeds u and v ,andtheanglebetweenthelinesis .Findtheminimal distancebetweentheshipsifatsomemomentoftimetheir distancesfromthepointofintersectionofthelineswereequal to a and b ,respectively. (3)Findallanglesofaright-angledtriangleofthemaximalarea ifthesumofitscathetusandhypotenuseisconstant. (4)Apieceofwire1mlongiscutintotwopieces.Oneisbent intoasquareandtheotherintoacircle.Whereshouldcuts bemadeifthesumoftheareasofthesquareandcircleisto beanextreme?Whichoftheseextremesarerelativemaxima andwhicharerelativeminima? (5)Showthatofalltrianglesinscribedinacircletheequilateral trianglehasthegreatestarea. (6)Atankhastheformofacylinderwithhemisphericalends. Ifthevolumeistobe V m3,whatarethedimensionsfora minimumamountofmaterial? (7)Thedemandforacertainarticlevariesinverselyasthecubeof thesellingprice.Ifthearticlecosts20centstomanufacture, “ndthesellingpricethatyieldsthemaximumpro“t. (8)Amanisinaboat1milefromthenearestpoint, A ,ofa straightshore.Hewishestoarriveassoonaspossibleata point, B ,3milesalongtheshorefrom A .Hecanrow2miles perhourandwalk4milesperhour.Whereshouldheland? (9)Thestinessofarectangularbeamvariesastheproductof thebreadthandthecubeofthedepth.Findthedimensionsof thestiestbeamthatcanbecutfromacylindricallogwhose radiusis R (10)Afactory A islocatedata(shortest)distance a milesfroma railroadthatgoesfromnorthtosouththroughatown B .If C isthepointontherailroadatadistance a from A ,then B is b milestothenorthof C .Thecostofshippingbyrailroad is p dollarsfor1tonpermileandtheshipmentbytruckcosts

PAGE 158

1524.APPLICATIONSOFDIFFERENTIATION q>p dollarsfor1tonpermile.Atwhatangletotherailroad shouldatruckroadfrom A beconstructedtominimizethe costsofshippingpertonfrom A to B ? (11)Ifthecostperhourforfuelrequiredtooperateagivensteamer variesasthecubeofitsspeedandis$40perhourforaspeed of10milesperhour,andifotherexpensesamountto$200per hour,“ndthemosteconomicalratetooperatethesteamera distanceof500miles. (12)Achannelofwidth b metersjoinsariverofwidth a metersat arightangle.Whatisthemaximallengthofashipthatcan enterthechannelfromtheriver? (13)Arailroadcompanyagreedtorunaspecialtrainfor50passengersatauniformfareof$10each.Inordertosecuremore passengers,thecompanyagreedtodeduct10centsfromthis uniformfareforeachpassengerinexcessofthe50(i.e.,if therewere60passengers,thefarewouldbe$9each).What numberofpassengerswouldgivethecompanythemaximum grossreceipt? (14)Asheetofpaperforaposteristocontain16squarefeet.The marginsatthetopandthebottomaretobe6inches,and thoseonthesides4inches.Whatarethedimensionsifthe printedareaistobemaximal? (15)Ataxicompanycharges15centsamileandlogs600passengermilesaday.Twenty-“vefewerpassenger-milesadaywouldbe loggedforeachcentincreaseintheratepermile.Whatrate yieldsthegreatestgrossincome? (16)Tworoadsintersectatrightangles,andaspringislocatedin anadjoining“eld10mfromoneroadand5mfromtheother. Howshouldastraightpathjustpassingthespringbelaidout fromoneroadtotheothersoastocutotheleastamountof land?Howmuchlandiscuto? (17)Illuminanceisameasureofhowmuchtheincidentlightilluminatesthesurface.Ifasourceoflightofluminosity k is positionedaboveaplane,theilluminanceatapointonthe planeis I = k cos /r2,where r isthedistancefromthesource tothepointand istheanglebetweenthelightrayfromthe sourcetothepointandthenormaltotheplane.Atwhat distanceabovethecenterofarounddiningtableofradius R shouldalightbulbbepositionedinorderforthetableborder tohaveamaximalilluminance?

PAGE 159

29.NEWTON’SMETHOD153 (18)Twolightsourcesofluminosity k1and k2arepositionedat points A and B ,respectively.Findthepointonthestraight linesegment AB thathastheleastilluminanceifthedistance betweenthesourcesis a (seethede“nitionofilluminancein thepreviousexercise). (19)Apointlikesourceoflightispositionedbetweentwononintersectingspheresonthelineconnectingthecentersofthe spheres.Iftheradiiofthespheresare R and r ( R>r ),“nd thepositionofthelightsourcesuchthatthesumofilluminatedareasofthespheresismaximal. (20)Arectangularboxwithasquarebaseandanopentopistobe made.Findthevolumeofthelargestboxthatcanbemade from A cm2ofmaterial. (21)Arectangular“eldcontaining S m2istobefencedoalong thebankofastraightriver.Ifnofenceisneededalongthe river,whatmustbethedimensionsrequiringtheleastamount offencing? (22)Ifastoneisthrownfromacliofheight h ataspeed v0m/s andanangle withthehorizontalline,thenitstrajectoryis aparabola: y = h + x tan Š x2g 2 v2 0cos2 where y isthestoneheight(verticalposition), x isthehorizontalposition(allthepositionsareinmeters),and g =9 8 m/s2isaconstantuniversalforallobjectsnearthesurface oftheEarth(thefree-fallacceleration).ComparewithEquation(4.4).Atwhatangleisastonetobethrowntoreachthe maximalrangeatagivenspeed v0? 29.Newton’sMethod Findingrootsofafunction f ( x )isanimportantprobleminvarious applications.Unfortunately,ananalyticsolutionoftheequation f ( x )= 0isimpossibleinmanypracticalcases.Forexample,consider f ( x )= x Š eŠ x.Theequation f ( x )=0isequivalentto x = eŠ x.Thegraphs y = x and y = eŠ xintersectatsome x between0and1.So f ( x )hasa root.Buthowcanitbecalculated?Herewepresentoneofthesimplest methods,knownasNewtonsmethod.Itprovidesarecurrencerelation thatallowsustocomputearootofadierentiablefunctionwithany desiredaccuracy.

PAGE 160

1544.APPLICATIONSOFDIFFERENTIATION 29.1.Newton’sRecurrenceRelationforFindingaRoot.Suppose f ( x ) hasarootnear x0.Considerthetangentlineapproximationof f near x0: L ( x )= f ( x0)+ f( x0)( x Š x0).Itiseasyto“ndtherootof L ( x ), whichisdenotedby x1: L ( x )=0= x = x1= x0Š f ( x0) f( x0) Notethattherootof L ( x )existsif f( x0) =0(otherwise,thetangent lineishorizontalandcannothaveanyroot). x2x1x0 Figure4.11. DiagramforNewtonsmethod.Pick x0neartherootof f .Findthetangentlineofthegraph of f at x0.Determinetheintersectionpoint x1ofthe tangentlinewiththe x axis.Findthetangentlineto thegraphof f at x1anditsintersection x2withthe x axis.Byrepeatingthisprocedureasequenceofnumbers x0, x1,x2,... isobtainedthatconvergestotherootof f provided x0waschosencloseenoughtotheroot. Since L ( x )isonlyanapproximationto f ( x ),thenumber x1is closertotherootof f than x0,butdoesnotcoincidewithit.Inother words,thevalue f ( x1)iscloserto0than f ( x0):0 < | f ( x1) | < | f ( x0) | (theabsolutevalueisnecessaryifthefunctiontakesnegativevalues). Therefore,thetangentlineconstructedat x = x1, L ( x )= f ( x1) Š f( x1)( x Š x1),canbeexpectedtoapproximate f ( x )evenbetternear itsrootbecause x1isclosertotherootthan x0.Therootofthenew tangentlineisgivenbythesameexpressionasbeforewhere x0should bereplacedby x1: x2= x1Š f ( x1) /f( x1).Theproceduremaybe

PAGE 161

29.NEWTON’SMETHOD155 recursivelyrepeatedtogenerateasequenceofvalues xn: (4.18) xn +1= xnŠ f ( xn) f( xn) ,n =0 1 2 ,..., provided f( xn) =0. Theorem 4.14 If f hasasingleroot r inanopenintervaland f( x ) =0 ontheinterval,thenthereexists x0sucientlycloseto r suchthatthesequence(4.18)convergestotheroot limn xn= r. Theconvergenceof xnto r meansthatforalllargeenough n the numbers xnlieinanysmallinterval r Š 0.Inpractice,theroot r needtobefoundonlywith someaccuracy.Sothesequenceelementsneedtobecalculatedwith aparticularnumberofsigni“cantdigits;thatis =10Š mwhere m is somepositiveinteger.ItissucienttoapplyNewtonsrecurrenceuntil xn +1and xnagreetoalltherelevantdecimalplaces.Then r = xn +1is correcttotherelevantdecimalplaces. Example 4.14 Findtherootof f ( x )= x Š eŠ xthatiscorrectto sixdecimalplaces. Solution:1 .Determinethepositionoftheroot“rst.Thegraphs y = x and y = eŠ xintersectonlyonceatapointbetween0and1.So, inanyopenintervalcontainingtheinterval(0 1), f hasonlyoneroot. 2 .Verifythecondition f( x ) =0: f( x )=1+ eŠ x> 0forall x 3 .PickaninitialvalueofNewtonssequenceasclosetotherootas possible,e.g. x0=0.ThenNewtonssequenceforsixdecimalplaces is: x0=0 ,x1=0 5 ,x2=0 566311 ,x3=0 567143 ,x4=0 567143 Sotheroot r =0 567143iscorrecttosixdecimalplaces(infact, f (0 567143)= Š 4 5 10Š 7). 29.2.PitfallsinNewton’sMethod.Unfortunately,thereisnounique recipeforchoosinganinitialpointinNewtonssequence.Thechoice dependsverymuchonthefunctioninquestion.Inpractice,itisdeterminedbytryingdierentvalues.Afewpossiblebadbehaviorsof Newtonssequenceareusefultokeepinmind.

PAGE 162

1564.APPLICATIONSOFDIFFERENTIATION ChoiceoftheInitialPointinNewton’sMethod.Apoorchoiceofthe initialpoint x0canproducethevalueof x1thatisaworseapproximationtotherootthan x0.Consider,forexample,thefunction f ( x )= x3Š 3 x2+2intheinterval[0 2]and f ( x )=2when x< 0and f ( x )= Š 2when x> 2.ThisisdepictedinFigure4.12.Thefunction iscontinuouslydierentiablebecause f( x )=3 x2Š 6 x approaches0as x 0+and x 2Š.Thefunctionhastheroot x =1and f( x ) < 0in theopeninterval(0 2).If0
PAGE 163

29.NEWTON’SMETHOD157 xn xn 2 xn 1 xn Figure4.13. AcycleinaNewtonssequencefor f ( x )= | x | .Thefunctionhastheroot r =0, f (0)=0,and italsohastheverticaltangentlineattherootbecause thederivativediverges f( x ) as x r =0.A Newtonssequenceexhibitsacycle;itoscillatesaround theroot r neverconvergingtoit.Ingeneral,suchacycle canoccurinaNewtonssequenceforafunctionwhose behaviorneararoot r is f ( x ) a ( x Š r )2 ,where a is aconstantand =1 / 4.Furthermore,if0 << 1 / 4, thenaNewtonssequencedoesnotconverge. Newtonssequencemayoscillatearound r ,neverconvergingtoit,or itmaydivergeforanyinitialpoint.Tounderstandthisphenomenon, suppose f ( x )behavesnearitsroot r as f ( x ) a ( x Š r )2 > 0,and a isconstant.Thederivative f( x ) 2 a ( x Š r )2 Š 1divergesas x r when < 1 / 2.Since f ( x ) /f( x )=1 2 ( x Š r ),Newtonssequence(4.18) xn +1= xnŠ1 2 ( xnŠ r )= xn(1 Š1 2 )+1 2 r canalsobewrittenas xn +1Š r = xn(1 Š1 2 )+1 2 r Š r = q ( xnŠ r ) where q =1 Š1 2 .Apparently,thecondition xn r isequivalentto yn= xnŠ r 0.Butthesequence yn +1= qyn= q2yn Š 1= = qn +1y0convergestozeroonlyif | q | = | 1 Š1 2 | < 1or > 1 / 4unless y0=0 (i.e.,iftherootisaccuratelyguessed!).Recalltheasymptoticbehavior oftheexponentialfunction ax as x if a> 0and ax 0 if0
PAGE 164

1584.APPLICATIONSOFDIFFERENTIATION initialpoint x0 =0because | qn| = | ( Š 2)n| =2n.For f ( x )= | x |1 / 2( =1 / 4),Newtonssequenceoscillates xn +1=(1 Š 2) xn= Š xn(see Figure4.13).29.3.UnderstandingMoneyLoans.Supposethatonetakesaloanof P dollars(the principal )for n monthswithanannualinterestrateof I %.Whatisthemonthlypayment?Itiscalculatedasfollows.The interestratepermonthis x = I/ 12.Forexample,anannualinterest rateof6%meansthat I =0 06and x =0 06 / 12=0 005.Each paymentincludesthepaymenttowardtheprincipalandtheinterest. Let Fkbetheamountyettobepaidafter k monthlypayments.It iscalledthe futurevalue oftheloan.Thesequence Fksatis“esthe conditions: F0= P and Fn=0(theloanandinterestarepaidoafter n payments).Let A bethemonthlypayment.Then F1= P + Px Š A,F2= F1+ F1x Š A,...,Fk= Fk Š 1+ Fk Š 1x Š A. Here F1isthefuturevalueoftheloanafteronepayment,whichisthe loan P plusthemonthlyinterest Px minusthepayment A .Afterone payment,theloanvalueis F1.So,afteronemorepayment,itsvalueis thevalue F1plusinterest F1x minusthepayment A ,andsoon.After n payments, Fn= Fn Š 1(1+ x ) Š A = Fn Š 2(1+ x )2Š A [(1+ x )+1]= = F0(1+ x )nŠ A [(1+ x )n Š 1+(1+ x )n Š 2+ +(1+ x )+1] = P (1+ x )nŠ A (1+ x )nŠ 1 x where,inthelastequality,thegeometricsumformula sn=1+ q + q2+ + qn Š 1= qnŠ 1 q Š 1 hasbeenused.Thelatterisprovedbynotingthatthesums qsn= q + q2+ + qn Š 1+ qnand snhavethesametermsexcept1in snand qnin qsnsothat qsnŠ sn= qnŠ 1or sn=( qnŠ 1) / ( q Š 1).Since Fn=0,themonthlypaymentis (4.19) A = Px 1 Š (1+ x )Š n. Forexample,aloanof$200,000for10yearsata“xedannualinterestrateof6%implies120monthlypaymentsof$2220 41.Indeed,in Equation(4.19),substitute x =0 06 / 12=0 005, n =120,and P =

PAGE 165

29.NEWTON’SMETHOD159 200 000,then A 2220 41004.Thetotalamountpaidafter10yearsis 120 A =$266,449.20.Theinterestpaidis nA Š P =$66,449.20. Whensellingacar,adealermightoeramonthlypaymentfora fewyearsifacustomercannotaordtopaythepriceinfull.Inthis case,theloanamount P isthepriceofthecar;themonthlypayment A anditsnumber n areknown.Toassessthedeal,onehasto“gureout theinterestratebeforesigningup.Itmightbethecasethattheloan forahigher-qualitycar,meaningahigherpriceandhighermonthly payments,mighthavealowerinterestrate,thantheloanforacheaper car(smallermonthlypayments).Knowingtheinterestrate,onecan alsoshopforaloanatalowerrateelsewhere(e.g.,banks)tobuya car.If A P ,and n aregiven,then x canbefoundbysolvingEquation (4.19),whichcanbewritteninamoreconvenientformas (4.20) f ( x )= Px (1+ x )nŠ A (1+ x )n+ A =0 Inotherwords,thisistheroot-“ndingproblem!Itcanbesolvedby Newtonsmethod.Thenumber x shouldbefoundupto“vedecimal places,whichissucientourpurposes. Example 4.15 Adealeroersacaratapriceof $ 10,000.Itcan alsobesoldforpaymentsof $ 217 42 permonthfor5years.Thereis anothercarbeingoeredatapriceof $ 15,000,whichcanalsobesold forpaymentsof $ 311 38 permonthfor5years.Whichloanhasalower interestrate? Solution:1 .Forthe“rstcar,onehasto“ndtherootofEquation (4.20)if A =217 42, P =10 000,and n =5 12=60.Itisconvenient toinitiateNewtonssequenceat x1=0 01,whichcorrespondstoan annualinterestrateof12%(i.e., I =0 12and x =0 12 / 12=0 01).Up to“vedecimalplaces,Newtonsmethodyields x =0 00917,which correspondsto I =12 x =0 11004,oranannualinterestrateof11%. 2 .Forthesecondcar,onehasto“ndtherootofEquation(4.20)if A =311 38, P =15 000,and n =5 12=60.Newtonsmethod, initiatedagainat x1=0 01,yieldstheroot x =0 00750(upto“ve decimalplaces).Thiscorrespondstoanannualinterestrateof9%.So thesecondloanhasalowerinterestrate. Itisinterestingtonotethatthecarpricesdierby50%.Similarly, themonthlypaymentsappearinasimilarproportion311 38 / 217 42 1 43.Theoersmightlooklikeasnearlythesamedeal.Infact,they arenot!

PAGE 166

1604.APPLICATIONSOFDIFFERENTIATION 29.4.Exercises.(1)Findthenumberofrealrootsoftheequationandestimate theintervalsthatcontainasingleroot: (i) x3Š 6 x2+9 x Š 10=0 (ii) x3Š 3 x2Š 9 x + h =0 (iii)3 x4Š 4 x3Š 6 x2+12 x Š 20=0 (iv) x5Š 5 x = a (v)ln x = kx (vi) ex= ax2(2)Underwhatconditiondoestheequation x3+ px + q =0have (i)asinglerealrootand(ii)threerealroots?Depictthe correspondingregionsof( p,q )inaplane. (3)Findtherootsofthegivenequationupto“vedecimalplaces: (i)cos x =2 x (ii) exŠ eŠ x=1 Š x2(iii)tanŠ 1x = x3(iv)tanŠ 1x =ln x (v)ln(1+ x2)=4 Š x (vi) x5+ x Š 4=0 (vii)(4 Š x2)2Š x +4=0 (4)Newtonsmethodisbasedonthelinearapproximationofthe functionatasamplepoint xntogeneratethenextpoint xn +1ofNewtonssequence.Thisapproximationdoesnottakeinto accounttheconcavityofthefunctionat xn.GeneralizeNewtonsmethodbyusingtheTaylorpolynomial T2( x )at xnto generate xn +1asarootof T2.TakeanyoftheaboveexercisesandcomparetheconvergenceofNewtonsmethodwith itsgeneralization(i.e.,thenumbersofstepsneededtoobtain therootcorrecttothesamenumberofdecimalplaces,e.g.,6, 7,or10,startingwiththesameinitialpoint x0). (5)Consideraloanof$250,000atanannuallowinterestrate of4%for15years.Findthemonthlypayments.Theinterest ratewasnot“xedandissubjecttochangesothatthemonthly paymentsmayincreaseupto20%.Howmuchmaytheannual interestrateincrease(percentagewise)? (6)Acardealeroersacaratapriceof$15,000for36monthly paymentsof$477.Whatistheinterestrate? (7)The“xedannualinterestrateonamortgageis7%.Forhow longshouldonetakealoanifonewantstopayintotalinterestnomorethanhalfoftheprincipal?Doesthemaximum loanperiodincreaseordecreasewithincreasingordecreasing

PAGE 167

30.ANTIDERIVATIVES161 interestrate?Doestheanswerchangeifthepaymentswillbe madeevery2weeks(i.e.,30paymentsperyearinsteadof12)? (8)FindtherootoftheequationtanŠ 1x =1 Š x correctupto fourdecimalplacesbyinitiatingNewtonssequenceat x0=1. Investigatethedependenceofthenumberofneedediterations toachievethisaccuracyontheinitialpointbytaking x0= n where n =1 2 ,..., 10. 30.Antiderivatives Inmanypracticalproblems,afunctionistoberecoveredfromits derivative.Forexample,ifthevelocityisgivenasafunctionoftime, v = v ( t ),onemightwantto“ndthepositionasafunctionoftime, s = s ( t ),where s( t )= v ( t ).Whatis s ( t )? Definition 4.8 Afunction F iscalledanantiderivativeof f on aninterval I if F( x )= f ( x ) forall x in I Formanybasicfunctions,itisnotdicultto“ndthecorresponding antiderivative.Forexample,fromtherule( xn +1)=( n +1) xn,itfollows thatif f ( x )= xn, n = Š 1,theantiderivativeis F ( x )= xn +1/ ( n +1).It hasalsobeenprovedthat(ln | x | )=1 /x .Sothefunction F ( x )=ln | x | istheantiderivativeof f ( x )=1 /x forall x =0.30.1.UniquenessoftheAntiderivative.Suppose F( x )= f ( x )forall x inaninterval( a,b ).Issuchan F ( x )unique?Thisquestionisanswered byCorollary4.1.Indeed,let F ( x )and G ( x )beantiderivativesof f ( x ), thatis, F( x )= G( x )= f ( x )on( a,b ).ByCorollary4.1, F and G mayonlydierbyaconstant: G ( x )= F ( x )+ C .RecallthatCorollary 4.1doesnotholdfortheunionof disjoint intervals.Thus,anytwo antiderivativesofthesamefunctionmaydieratmostbyaconstant onaninterval. Theorem 4.15 If F isanantiderivativeof f onaninterval I thenthemostgeneralantiderivativeof f on I is F ( x )+ C, where C isanarbitraryconstant. Forexample,thegeneralantiderivativeofthepowerfunction xn, where n isapositiveinteger,is F( x )= f ( x )= xn= F ( x )= xn +1 n +1 + C

PAGE 168

1624.APPLICATIONSOFDIFFERENTIATION because xnisde“nedonthesingleinterval( Š ).Thefunction f ( x )=1 /x isde“nedontheunionofdisjointintervals( Š 0)and (0 ).Sothegeneralantiderivativeis F( x )= f ( x )= 1 x = F ( x )= ln x + C1,x> 0 ln( Š x )+ C2,x< 0 where C1and C2arearbitraryconstants.Inwhatfollows, thedomain isalwaysassumedtobeasingleinterval,unlessstatedotherwise Thenonuniquenessoftheantiderivativeisnotadrawbackofthe conceptbutratheragreatadvantage.Thisisexplainedbythefollowing example.Thevelocityofapieceofchalkthrownverticallyupwardwith avelocityof v0is v ( t )= v0Š gt ,where g =9 8m / s2istheacceleration ofafreefall.At t =0,thechalkhasavelocityof v (0)= v0.Thenit beginstoslowdown( v ( t )decreasesbecauseofgravity).Eventually,at t = v0/g ,thechalkstopsandbeginstofallback.If h ( t )istheheightof thechalkrelativetothe”oor,then h( t )= v ( t );thatis,theheightis anantiderivativeof v ( t ).Itiseasyto“ndaparticularantiderivativeof v ( t )usingtheantiderivativeofthepowerfunction: h ( t )= v0t Š gt2/ 2 (indeed, h( t )= v0Š gt ).Whatisthephysicalsigni“canceofthegeneral antiderivative h ( t )= C + v0t Š gt2/ 2?Itappearsasifthepositionof thechalkrelativetothe”oorisnotuniquelydetermined.Inparticular, h (0)= C istheheightattheverymomentwhenthechalkwasthrown upward.Butthechalkcouldbethrownupwardat1mabovethe”oor or2maboveitwiththeverysameinitialvelocity.So,inbothcases, v ( t )isthesame,whilethe h ( t )arenot.Inthe“rstcase, h (0)=1, whereasinthesecondcase, h (0)=2. Thus,theconstant C canbe“xed byspecifyingthevalueoftheantiderivativeataparticularpoint Thisfeatureofthegeneralantiderivativecanalsobevisualizedby plottingthegraphs y = F ( x )+ C fordierentvaluesof C .Allsuch graphsareobtainedfromthegraph y = F ( x )byrigidtranslations alongthe y axis.Ifonedemandsthatthegraph y = F ( x )+ C should passthroughaparticularpoint( x0,y0),then C is“xed: y0= F ( x0)+ C or C = y0Š F ( x0).Forexample,“nd f ( x )if f( x )=3 x2and f (2)=1. Thegeneralantiderivativeof3 x2is f ( x )= x3+ C .From f (2)=1,it followsthat f (2)=8+ C =1or C = Š 7.Therefore, f ( x )= x3Š 7.30.2.LinearityoftheAntiderivative.Let F and G beantiderivativesof f and g ,respectively.Thenanantiderivativeof f + g is F + G .An antiderivativeof kf ,where k isanarbitraryconstant,is kF .These propertiesareeasilyveri“ed.Indeed,( F + G )= F+ G= f + g and( kF )= kF= kf ,wherethelinearityofthederivativehasbeen

PAGE 169

30.ANTIDERIVATIVES163 used.Inotherwords,antidierentiationisalinearoperationjustlike dierentiationitself.30.3.AntiderivativesofBasicFunctions.Anantiderivativeofthepower functionhasbeenfoundbystudyingthederivativeofthepowerand logarithmicfunctions.Theideaisusefulforotherbasisfunctions.Their antiderivativescanbefoundbyreadingthetableofderivativesofbasic functionsbackward,thatis,fromtherighttoleft. Tableofantiderivativesofbasicfunctions F ( x ) f ( x )= F( x ) F ( x ) f ( x )= F( x ) sin x + C cos x Š cos x + C sin x tan x + C (sec x )2 Š cot x + C (csc x )2 sinŠ 1x + C 1 1 Š x2 Š cosŠ 1x + C 1 1 Š x2 tanŠ 1x + C 1 1+ x2 Š cotŠ 1x + C 1 1+ x2 xn +1 n +1+ C xn,n = Š 1 ln | x | + C 1 x,x =0 ex+ C ex ax ln a+ C ax,a> 0 ,a =1 1 2ln 1+ x 1 Š x + C 1 1 Š x2 ln | x + x2 1 | + C 1 x2 1 Asnotedearlier,here F ( x )isgivenonasingleintervalinthedomain of f ( x ).Thetableofantiderivativesofbasicfunctionscombinedwith thelinearityofantidierentiationisagoodsourceofantiderivativesof morecomplicatedfunctions.Thechainrule( f ( g ( x ))= f( g ( x )) g( x ) canalsobeusedtoobtainantiderivatives. Example 4.16 Findthegeneralantiderivativeof x/ ( x2+1) Solution: Bythechainrule (ln( x2+1))= 1 x2+1 ( x2+1)= 2 x x2+1 Therefore F ( x )=1 2ln( x2+1)+ C Example 4.17 Findthegeneralantiderivativeof f ( x )= eŠ 2 x+ cos(4 x )+ x2/ (1+ x2) Solution:1 .Bythelinearityoftheantiderivative,itissucientto “ndantiderivativesof eŠ 2 x,cos(4 x ),and x2/ (1+ x2).Thegeneralantiderivativeisobtainedbyaddingageneralconstanttothesumofthe

PAGE 170

1644.APPLICATIONSOFDIFFERENTIATION particularantiderivativesofthepreviousthreefunctions. 2 .From( eŠ 2 x)= Š 2 eŠ 2 x,itfollowsthat( Š eŠ 2 x/ 2)= eŠ 2 x.Hence, anantiderivativeof eŠ 2 xis Š eŠ 2 x/ 2. 3 .Similarly,from(sin(4 x ))=4cos(4 x ),itfollowsthatanantiderivativeofcos(4 x )issin(4 x ) / 4. 4 .Thetableofderivativesdoesnotappearhelpfulinthecaseof x2/ (1+ x2).However,asimplealgebraicmanipulationleadstothe goal: x2 1+ x2= 1+ x2Š 1 1+ x2=1 Š 1 1+ x2. Soitsantiderivativeis x Š tanŠ 1x .Thus,thegeneralantiderivative reads: F ( x )= Š 1 2 eŠ 2 x+ 1 4 sin(4 x )+ x Š tanŠ 1x + C. because f ( x )isde“nedonthesingleinterval( Š ). 30.4.AntiderivativesofHigherOrder.Whatis F ( x )if F( x )= f ( x ) foragiven f ( x )?Or,moregenerally,whatis F ( x )if F( n )( x )= f ( x )? Afunction F thatsatis“esthelatterconditioniscalledanantiderivativeof f ofthe n thorder.To“ndit,onehastoantidierentiate f n times.Forexample, F( x )=6 x .Takingthe“rstantiderivativeof f ( x )=6 x ,onegets F( x )=3 x2.Takingtheantiderivativeonemore timeyields F ( x )= x3.Whatabouttheuniquenessofhigher-order antiderivatives?To“ndthegeneralantiderivativeofahigherorder, eachtimeantidierentiationiscarriedout,thecorrespondinggeneral antiderivativemustbeused.Intheprecedingexample,thegeneral antiderivativeof f ( x )=6 x is3 x2+ C1,where C1isanarbitrary constant.Hence, F( x )=3 x2+ C1.Itsgeneralantiderivativereads F ( x )= x3+ C1x + C2,where C2isanotherarbitraryconstant.Thus, thegeneralsecondantiderivativecanbeobtainedfromaparticularone byaddingageneralfunctionwhosesecondderivativeis0,whichisa generallinearfunction:( C1x + C2)=0.Similarly,if F ( x )isaparticularfunctionthatsatis“esthecondition F( n )( x )= f ( x ),thenthe generalantiderivativeofthe n thorderis F ( x )+ C1xn Š 1+ C2xn Š 2+ + Cn Š 1x + Cn, where C1,...,Cnarearbitraryconstants.Indeed,the n thderivativeof apolynomialofdegree n Š 1is0.Notethatthisanalysisisjusti“edonly when f wasde“nedinaninterval.Thereaderisinstructedtoanalyze thesituationwhenthedomainofthefunction f consistsofdisjoint

PAGE 171

30.ANTIDERIVATIVES165 intervalsand,inparticular,toconsiderhigherorderantiderivativesof theinversepowerfunction xŠ n,where n isapositiveinteger. Thefollowingexampleillustratesthesigni“canceofarbitraryconstantsingeneralhigher-orderantiderivatives. Example 4.18 Anyfree-fallingobjectnearthesurfaceoftheEarth hasthefree-fallaccelerationof 9 8 m/s2.Apieceofchalkisthrown verticallyupwardataspeedof 7 m/s andat 1 5 m abovethe”oor.When doesthechalkhitthe”oor? Solution:1 .Let h ( t )betheheightofthechalkrelativetothe ”oor.Thenitsvelocityis v ( t )= h( t ),anditsaccelerationis a ( t )= v( t )= h( t ).Sinceallfree-fallingobjectshaveanaccelerationof 9 8m / s2,onehas h( t )= Š 9 8.Theminussignindicatesthatthe accelerationisdirecteddownward. 2 .Thegeneralsecondantiderivativeoftheconstantfunction Š 9 8is h( t )= Š 9 8= h ( t )= Š 9 8 t2 2 + C1t + C2, where C1and C2arearbitraryconstants. 3 .To“x C1and C2,theinitialconditionsofthemotionmustbe used.Theinitialvelocityis v (0)=7.Since v ( t )= h( t )= Š 9 8 t + C1, oneinfersthat v (0)= C1=7.Theinitialheightis h (0)=1 5.Hence, h (0)= C1=1 5. 4 .Theheightis h ( t )= Š 9 8 t2/ 2+7 t +1 5.Thechalkhitsthe”oorwhen itsheightvanishes,thatis,atthetimemoment t> 0when h ( t )=0.A positiverootofthequadraticequation Š 9 8 t2/ 2+7 t +1 5=0is t 1 62s.Themaximumheightreachedbythechalkis4m.Why? 30.5.Exercises.(1)Usethetableofantiderivativesofbasicfunctionsto“ndan antiderivativeofeachofthefollowingfunctions: (i) f ( x )=(3 Š x2)2(ii) f ( x )= x2(1 Š x )2(iii) f ( x )=(1 Š x )(1 Š 2 x )(1 Š 3 x ) (iv) f ( x )=sin(4 x )+ x (v) f ( x )=1 / ( x2+4) (vi) f ( x )= e3 x+ eŠ 3 x+ x (vii) f ( x )=cos2x ( Hint: 2cos2x =1+cos(2 x )) (viii) f ( x )=sin( ax )cos( bx ) (ix) f ( x )=cos( ax )cos( bx ) (x) f ( x )=sin( ax )sin( bx )

PAGE 172

1664.APPLICATIONSOFDIFFERENTIATION Hint: Expresstheproductsoftrigonometricfunctionsviathe sumoftrigonometricfunctions. (xi) f ( x )=( x +1) / x (xii) f ( x )=3 x2(1+ x ) (xiii) f ( x )= x/ ( x2+1) (xiv) f ( x )=1 / ( x2+ a2) (xv) f ( x )=( x2+ a2) / ( x2+ b2) (xvi) f ( x )=( x4+1) / ( x2+1) Hint: x4= x4+2 x2Š 2 x2(xvii) f ( x )= | x | (xviii) f ( x )= xeŠ x2(xix) f ( x )= | x Š a | + | x Š b | ,where aa> 0 (xxiii) f ( x )=2x/ 3x(xxiv) f ( x )=(2x+3Š x)2(2)Usethetableofantiderivativesto“ndthegeneralsecondantiderivativeofeachofthefollowingfunctions: (i) f ( x )= x ( x Š 1) (ii) f ( x )= x1 / 3(iii) f ( x )=sin(2 x ) (iv) f ( x )= | x | (v) f ( x )= f0=constif x [ a,b ]and f ( x )=0otherwise (vi) f ( x )= | x | +1 (vii) f ( x )=sin x cos(2 x ) (viii) f ( x )= e3 x+ eŠ 3 x(ix) f ( x )= x/ ( x2+1)2(x) f ( x )=2xŠ 3Š x(3)Findthegeneralantiderivativeofthefunctionwhosedomain isnotasingleinterval: (i) f ( x )=1 / ( x Š 1) (ii) f ( x )=1 / ( x2Š 1) (iii) f ( x )= xŠ 4 / 3(iv) f ( x )=(1 Š x )3/x4 / 3(v) f ( x )=sec2x (vi) f ( x )=csc2x (4)Acarthatwasatrestacceleratesatarateof1m/s2for1 minute.Thenitdeceleratesatarateof0.5m/s2untilitstops. Findthedistancetraveledbythecar.

PAGE 173

CHAPTER5 Integration 31.AreasandDistances31.1.LimitofaNumericalSequence.Anumericalsequence a1,a2,a3,... isafunctionwhosedomainisthesetofallpositiveintegers.Inother words,itisarulethatassignsauniquenumber antoeachinteger n Forexample, an= n n +1 ,n =1 2 3 ,.... If f ( x )= x/ ( x +1),thenthesequence anisformedbythevaluesof f ( x )at x = n an= f ( n ).ItfollowsfromSection10that limx f ( x )=limx x x +1 =limx 1 1+1 x=1 Thismeansthatforanygiven > 0,thereexistsarealnumber N such thatvaluesof f ( x )liesintheinterval1 Š N Since isarbitrarily,onecansaythatthevaluesof f arearbitrarily closeto1.Inparticular,for n largeenough,thevalues an= f ( n )are arbitrarilycloseto1,andonecanconcludethat limn an=limn n n +1 =1 Sincenumericalsequencesarejustaspecialcaseofafunction,aformal de“nitionofthelimitofasequencecanbeadoptedfromSection10. Definition 5.1(Limitofanumericalsequence) Let an, n = 1 2 ,... ,beanumericalsequence.Thesequenceissaidtoconvergetoa number a ifforany > 0 thereisaninteger N suchthat | anŠ a | < forall n>N ,andinthiscaseonewrites limn 0an= a Ifthesequencehasnolimit,itiscalled divergent Thisde“nitionalsomeansthataconvergentsequencehasallbuta “nitenumberofitstermsinanarbitrarilysmallinterval( a Š ,a + ). Forexample,take an=1 /npwhere p> 0.Thenevidentlylimn an= 0.Indeed,givenanysmallnumber > 0,itfollowsthattheinequality | anŠ 0 | = an= 1 np< = n> 1 1 /p167

PAGE 174

1685.INTEGRATION holdsforall n>N where N isanintegersuchthat N> 1 /1 /p. Therefore,onlya“nitenumberoftermsofthesequencelieoutsidethe interval( Š )and,hence,thesequenceconvergesto0.Thissequence isformedbythevaluesofthefunction f ( x )=1 /xp, an= f ( n ).In Section10,ithasbeenshownthat f ( x ) 0as x ,whichalso impliesthat f ( n ) 0as n .Therelationbetweenthelimitsof sequencesandthelimitsoffunctionsatin“nity limx f ( x )= a = limn f ( n )=limn an= a isausefultooltocalculatethelimitsofsequences.Notethatthe converseoftheaboveassertionisnottrue.Why? A“nalremarkisthatthebasiclimitlawsandthesqueezeprinciple holdforthelimitsofsequences.31.2.AreaUnderaGraph.Considerthelinearfunction f ( x )= x Whatistheareabelowthegraph y = f ( x )andabovetheinterval 0 x 1?Thisquestioniseasytoanswerbecausetheareain questionistheareaoftherighttrianglewithcathetiofunitlength: A =1 / 2.Let f ( x )= x2.Whatistheareanow?Tocalculateit, considera partition oftheinterval[0 1]by n segmentsonlength1 /n Thepartitionisde“nedbythesetofpoints x0=0 ,x1=1 /n,x2= 2 /n,...,xn Š 1=( n Š 1) /n ,and xn= n/n =1,thatis, xk= k/n ,where k =0 1 2 ,...,n .Theareaundertheparabola y = x2overtheinterval [0 1]isthesumoftheareas Skundertheparabolaoverthepartition interval[ xk Š 1,xk]where k =1 2 ,...,n A = S1+ S2+ + Sn. Intheinterval[ xk Š 1,xx],thefunction f ( x )= x2attainsitsmaximum valueat x = xkanditsminimumvalueat x = xk Š 1.Thereforethe area Skcannotexceedtheareaofarectanglewithbase1 /n andheight f ( xk)=( k/n )2.Letusdenotethisupperboundby SU k= k2/n3.The area Skisgreaterthantheareaofarectanglewithbase1 /n andheight f ( xk Š 1)=( k Š 1)2/n2.Thelowerboundisdenotedby SL k=( k Š 1)2/n3. Thus, SL k= ( k Š 1)2 n3
PAGE 175

31.AREASANDDISTANCES169 1.0 0.8 0.6 0.4 0.2 01 3 4 1 2 1 4 1.0 0.8 0.6 0.4 0.2 01 3 4 1 2 1 4 Figure5.1. Theupperandlowerboundsforthearea underthegraph y = f ( x )= x2for n =4partitionintervalsin[0 1].Theupperboundisobtainedbytaking themaximumvalueof f oneachpartitioninterval(left panel).Thelowerboundisobtainedbytakingtheminimumvalueof f oneachpartitionsegment(rightpanel). When n increases,theupperbounddecreases,whilethe lowerbounddecreases,bothapproachingtheareaunder thegraphas n Letuscalculatethedierence 0
PAGE 176

1705.INTEGRATION forthesumofsquaresofthe“rst n positiveintegers: (5.1)12+22+ + n2= 1 6 n ( n +1)(2 n +1)= n 6 (2 n2+3 n +1) Indeed,bymakinguseofthisformula,onecaninferthat limn AU n=limn 1 n3(12+22+ + n2)=limn 2 n2+3 n +1 6 n2=limn 1 3 + 1 2 n + 1 6 n2 = 1 3 +limn 1 2 n +limn 1 6 n2= 1 3 Sotheareais A =1 3. Let x kbeanumberintheinterval[ xk Š 1,xk].Thenthearea Skcanalsobeapproximatedbythearea S kofarectanglewithbase 1 /n andheight f ( x k)=( x k)2,thatis, S k= f ( x k) /n .Thenthetotal areaunderthegraphisapproximatedbythesum A nofall Sk.Since SL k S k SU k(owingtothemonotonicityofthefunction x2ineach interval[ xk Š 1,xk]),thefollowinginequalityholdsforany n : AL n S 1+ S 2+ + S n= A n AU n. Takingthelimit n inthisinequalityandusingthesqueezeprinciple,aremarkableresultisobtained limn A n= A ; thatis,thelimitof A ndoesnotdependonthechoiceof samplepoints x k.Theareacouldhavebeenapproximatedby,forexample, A nwith thesamplepointsasthemidpoints x k=( xk+ xk Š 1) / 2,oranyother convenientchoice.Thisanalysiscanbeextendedtoanycontinuous function. Thecalculationoftheareaunderthegraphisfurtherillustratedin thevideowebsiteathttp://www.math.u”.edu/ mathguy/ufcalcbook/ riemann.html.31.3.TheAreaUndertheGraphofaContinuousFunction.Let f ( x )be continuouson[ a,b ].Considerapartitionof[ a,b ]by n segmentsof length x =( b Š a ) /n .Theendpointsofthepartitionssegmentsare xk= a + k x with k =0 1 2 ,...,n ,suchthat x0= a and xn= b .Let x kbeasamplepointintheinterval[ xk Š 1,xk].

PAGE 177

31.AREASANDDISTANCES171 Definition 5.2 Thearea A oftheregionthatliesunderthegraph ofacontinuousfunction f ( x ) 0 onaninterval [ a,b ] is (5.2) A =limn A n=limn f ( x 1) x + f ( x 2) x + + f ( x n) x foranychoiceofsamplepoints x k. Letusassessthisde“nition.Anycontinuousfunctionattainsits maximumandminimumvaluesonaclosedinterval.Let Mkand mkbe,respectively,themaximumandminimumvaluesof f ( x )onthe interval[ xk Š 1,xk].If Skistheareaunderthegraph y = f ( x )onthe interval[ xk Š 1,xk],then SL k= mk x Sk SU k= Mk x .Thearea S k= f ( x k) x oftherectanglewithbase x andheight f ( x k)isa continuousfunctionof x kontheinterval[ xk Š 1,xk].Therefore, S kmust takeallthevaluesbetweenitsminimumandmaximumvalues, SL kand SU k.Inparticular, S k= Skforsome x k [ xk Š 1,xk].Thus, forany “xed n ,thereisachoiceofsamplepointssuchthat A n= A Continuingtheanalogywiththeexampleof f ( x )= x2,letusshow thatthelimit(5.2)isindependentofthechoiceofsamplepoints, providedthe lower sums AL n= SL 1+ + SL nandthe upper sums AU n= SU 1+ + SU nconvergetothe same numberas n .Indeed, foranychoiceofsamplepoints SL k S k SU kand,hence,bytakingthesumoverthepartitioninthelatterinequality,oneinfersthat AL n A n AU n.Thereforeboththenumbers A nand A liebetween AL nand AU n: AL n A n AU nAL n A AU n= | A nŠ A | AU nŠ AL n(5.3) Thus,if AU nŠ AL n 0as n ,then A n A foranychoiceof partition.Thefollowingtheoremholds. Theorem 5.1 Let f beacontinuousfunctionon [ a,b ] .Suppose thatforanypartition x0= a n +1.Thentheupperandlowersumsconvergetothesamelimit as n Thistheoremjusti“esthede“nition(5.2).Notealsothatthepartitionisnotgenerallyrequiredtobeequispaced.Theabovetheoremonly requiresthatthelengthnofthelargestpartitionintervaldecreases withincreasingthenumberofpartitionintervals(n=( b Š a ) /n = x foranequispacedpartition).

PAGE 178

1725.INTEGRATION 31.4.ApproximatingtheAreaUnderaGraph.Inpractice,Equation(5.2) canbeusedto“ndtheareaunderthegraphthatiscorrecttoanydesirednumberofdecimalplaces.Takeapartitionoftheinterval[ a,b ], e.g.,“xsome n sothat x =( b Š a ) /n .Choosesamplepoints xkŠ 1 x k xk.Convenientchoicesmightbetheleftpoints x k= xk Š 1,the rightpoints x k= xk,orthemidpoints x k=( xk Š 1+ xk) / 2.Calculate thesum A n,keepingthedesirednumberofdecimalplaces.Re“nethe partitionby,forexample,doublingthenumberofsegments,andcalculate A 2 n.If A nand A 2 ncoincideinthedesirednumberofdecimal places,then A = A 2 niscorrecttothatnumberofdecimalplaces.If not,re“nethepartitionfurtherandcompute A 4 nandcompareitwith A 2 nandsoon,untiltheneededaccuracyisreached.Forany n ,the absoluteerroroftheapproximationmayestimatedbytheinequality ontherightin(5.3).31.5.SigmaNotationforSums.Toavoidwritinglengthyexpressions forsumsofanarbitrarynumberofterms,itisconvenienttoadoptthe followingnotation: A n= S 1+ S 2+ + S n=nk =1S k, wheretheindex k iscalledthe summationindex .Thesymbol means addingall S k,startingwith k =1upto k = n .Forexample,the geometricsumformulacannowbewrittenas (5.4)1+ q + q2+ + qn=nk =0qk= qn +1Š 1 q Š 1 .31.6.TheDistanceProblem.Ifanobjectmoveswithaconstantvelocity v duringatimeinterval a t b ,thenthedistancetraveledby theobjectis D = v ( b Š a ).Howdoesonecalculatethedistanceifthe speedisanonconstantcontinuousfunctionoftime v = v ( t ) 0? Let D ( t )bethedistanceasafunctionoftime a t b .It satis“esthecondition D ( a )=0.Since v ( t ) 0,theobjecttravelsin thesamedirectionallthetime,and D ( t )increasesbecause D( t )= v ( t ) 0.Thus, D = D ( b ).Tocalculate D ( b ),considerapartition of[ a,b ]byinterval[ tk Š 1,tk]where tk= a + tk t =( b Š a ) /n k =0 1 ,...,n .Thedistance Dk= D ( tk) Š D ( tk Š 1)traveledbythe objectinthetimeinterval[ tk Š 1,tk]canbefoundbythemeanvalue theorem: D ( tk) Š D ( tk Š 1)= v ( t k) t forsome t kin[ tk Š 1,tk].Recall that v ( t k)istheaveragevelocityoverthetimeinterval[ tk Š 1,tk].The totaldistanceis D = D1+ + Dn.Ontheotherhand,points

PAGE 179

31.AREASANDDISTANCES173 t krepresentaparticularchoiceofsamplepointsinthede“nition(5.2) appliedtoacontinuousfunction v ( t ).Therefore, D istheareaunder thegraphof v ( t )and,hence,canbecalculatedwith any choiceof samplepoints t k,notnecessarilywiththoseatwhich v coincideswith theaveragevelocityineachpartitioninterval: D =limn nk =1v ( t k) t, Furthermore,bythecondition D( t )= v ( t )thefunction D ( t )isthe antiderivativeof v ( t )satisfyingtheinitialcondition D ( a )=0.If F ( t ) isaparticularantiderivativeof v ( t ),then D ( t )and F ( t )candieronly byaconstant, D ( t )= F ( t )+ C for a
PAGE 180

1745.INTEGRATION Thus,thedistancetraveledis D =(1 Š eŠ 2) / 2. Alternativesolution :Usingthetableofantiderivatives F( t )= v ( t )= eŠ 2 t= F ( t )= Š 1 2 eŠ 2 t. ByEquation(5.5), D = F (1) Š F (0)=(1 Š eŠ 2) / 2.Notethatany particularantiderivativecanbeused. Whencomparedtotheprevioussolution,thisonelookslikecheating!Moretothepoint,take v ( t )= t2(theexamplediscussedatthebeginningofthissection).Itsparticularantiderivativeis F ( t )= t3/ 3.So thedistancetraveled,ortheareaunderthegraphof t2,is F (1) Š F (0)= 1 / 3.Itturnsoutthattherelation(5.5)betweenanantiderivativeofa functionandtheareaunderthegraphofthefunctionisnotspeci“cfor thedistanceproblem.Itsgeneralizationwillbeestablishedwiththe helpoftheconceptofthe de“niteintegral .31.7.Exercises.(1)Findexplicitformulasfortheupperandlowersums, AU nand AL n,for f ( x )=2 x +1on[0 2]usinganequispacedpartition.Findthelimitsof AU nand AL nas n .Whatisthe geometricalsigni“canceofthislimit? (2)Findexplicitformulasfortheupperandlowersums, AU nand AL n,for f ( x )= x3on[0 1]usinganequispacedpartition.Show that AU,L n 1 / 4as n .Whatisthegeometricalsigni“canceofthislimit? Hint :nk =1k3= 1 4 n2( n Š 1)2. (3)Findtheareaunderthegraphof f ( x )= eŠ x2on[ Š 1 1]correct upto“vedecimalplaces. (4)Findtheareaunderthegraph f ( x )= 1 Š x2,where Š 1 x 1,correctuptothreedecimalplaces.Usethegeometrical interpretationofthisareato“nditsexactvalue. (5)Findtheareaunderthegraphofeachofthefollowingfunctions onthegivenintervalusingtherelation(5.2): (i) f ( x )=3 Š 3 x ,0 x 1 (ii) f ( x )=1+ x + x2,0 x 2 (iii) f ( x )= e3 x, Š 1 x 1

PAGE 181

31.AREASANDDISTANCES175 (6)Usetherelationn Š 1k =0sin(2 kx )= sin( nx )sin[( n Š 1) x ] sin x to“ndthefollowing: (i)Theupperandlowersumsfor f ( x )=sin x ontheinterval [0 ].Calculate AU nŠ AL nandinvestigateitsbehavioras n increases.Whatisthesigni“canceofthisnumberfora“xed n ? (ii)Theareaunderthegraphof f ( x )=sin x ,0 x using(5.2). (iii)Theareaunderthegraphof f ( x )=cos x ,0 x / 2, using(5.2). (7)Anobjecttravelswithvelocity v ( t )=cos2t .Findthedistance passedbytheobjectoverthetimeinterval0 t 2 (8)Usethetableofantiderivativesto“ndtheareaunderthe graphofeachofthefollowingfunctions.Sketchthegraphof thefunctiononagivenintervalandexplainwhythismethod canbeusedto“ndthearea. (i) f ( x )=1 / ( x2+1), Š 1 x 1 (ii) f ( x )=sin( ax ),0 x /a (iii) f ( x )= x ,0 x 4 (iv) f ( x )=1 /x ,1 x 2 (v) f ( x )=tan x / 4 x / 3 (vi) f ( x )=1 / x2+1, Š 1 x 1 (vii) f ( x )=2x, Š 2 x 2 (viii) f ( x )= x/ ( x2+1),0 x 1 (ix) f ( x )= x4 / 3, Š 27 x 27 (x) f ( x )=1 / (1 Š x2), Š 1 / 2 x 1 / 3 (9)Acarstartsacceleratingat t =0sothatitsspeedis v = at a> 0.Atatime t = b ,itbeginstoslowdownsothatitsspeed becomes v ( t )= ab + c ( b Š t ), c> 0,untilitsspeedvanishes andthecarstops.Findthedistancetraveledbythecar. Hint: Sketchthegraphof v ( t )inthetimeintervalinwhichthecar wasmoving. (10)Anobjecttravelswithspeed v ( t )= a2Š t2,startingat t = Š a andstoppingat t = a .Sketchthegraphof v ( t )and“nd thedistancetraveledbytheobject. (11)Let f ( x )=( x5Š 1) / ( x Š 1)if0 x< 1and f (1)=5.Show that f ( x )iscontinuouson[0 1].Useantiderivativesto“nd theareaunderthegraphof f Hint :SeeEquation(5.4).

PAGE 182

1765.INTEGRATION (12)Findtheareaofaplanarregionboundedbythecurves y = 2 Š x2and y =1.Sketchtheregion. (13)Findtheareaofaplanarregionboundedbythecurves y = x2and y = x .Sketchtheregion. 32.TheDeniteIntegral Ageneralizationoftheconceptoftheareaunderagraphleadsto oneofthemostfundamentalconceptsincalculus,thede“niteintegral.32.1.SupremumandInmum.Theareaunderagraphisalsowellde“nedifthefunctionhassomenumberofboundedjumpdiscontinuities. Thedierencewiththecaseofacontinuousfunction f isthatnow f mayormaynotattainitsmaximumorminimumvaluesoneachpartitioninterval.Whatshouldbechangedinthede“nitionofthearea toaccommodatepossiblejumpdiscontinuitiesofthegraph?Suppose afunction f isboundedonaninterval[ a,b ];thatis,therearenumbers m and M suchthat m f ( x ) M forall x [ a,b ].If m isalower bound,thenanynumber m1m mayormaynotbealowerbound.Soonecan“ndthegreatest lowerboundthatisuniquefor f on[ a,b ].Similarly,onecan“ndthe leastupperboundof f on[ a,b ].Theseboundshavespecialnames. Definition 5.3 (In“mumandSupremum). Thenumber m is calledthe in“mum ofaboundedfunction f onaninterval I =[ a,b ] if m isalowerboundof f but m + isnotalowerboundforany > 0 .Thisnumberisdenotedas m =infIf .Thenumber M iscalled the supremum of f on [ a,b ] if M isanupperboundof f but M Š isnot anupperboundforany > 0 .Thisnumberisdenotedas M =supIf .Remark.The completenessaxiom forasetofrealnumberssaysthatif S isanonemptysetofrealnumbersthathasunupperbound M ,then S hastheleastupperboundinf S .If S hasalowerbound m ,thenit alsohasthegreatestlowerboundinf S .Thecompletenessaxiomisan expressionofthefactthatthereisnogaporholeintherealnumber line.Thenumberssup S andinf S are unique .Indeed,assumethat M1 = M2aretheleastupperboundsof S .Sincetheyarenotequal oneofthemshouldbelessthantheother,e.g., M1 0.Then M2Š = M1but M1is alsoanupperbound,hence,acontradiction.Theuniquenessofinf S isestablishedbysimilarlinesofreasoning.

PAGE 183

32.THEDEFINITEINTEGRAL177 Naturally,ifthefunctioniscontinuous,thensup f isnothingbut themaximumvalueof f andinf f isitsminimumvalue.However,if afunctionhasjumpdiscontinuities,thensup f andinf f alwaysexist,whilethemaximumandminimumvaluesmaynotexist.Thisis illustratedinFigure5.2.32.2.DenitionoftheDeniteIntegral.Let f beaboundedfunction onaninterval[ a,b ].Considerapartitionof[ a,b ]by n intervals Ik= [ xk Š 1,xk], k =1 2 ,...,n ,where a = x0 n +1;thatis,thelengthof thelargestpartitionintervaldecreaseswithincreasingthenumberof partitionintervals. Definition 5.4(TheDe“niteIntegral) Aboundedfunction f is saidtobe integrableonaninterval[ a,b ] ifthesequencesofitslower anduppersumsconvergetothesamenumber.Thisnumberiscalled the de“niteintegral of f from a to b andisdenotedby b af ( x ) dx =limn AL n=limn AU n; thenumbers a and b arecalledthe lowerandupperintegrationlimits respectively,andthefunction f iscalledthe integrand Apparently,fora continuousandnonnegative f on[ a,b ],thede“nite integralcoincideswiththeareaunderthegraphof f .Similarlyto theareaunderthegraphofacontinuousnonnegativefunction,an integrablefunctionhastheproperty AL n b afdx AU nforany n (seeExercise32.9.4).32.3.RiemannSums.ThereisananalogofEquation(5.2)forthe de“niteintegral.

PAGE 184

1785.INTEGRATION 1 acb 1/2 1 acb 1/2 1 acb 1/2 1 acb 1/2 sup f 1> f ( x ) f ( c ) 1/2 sup f max f 1 inf f mix f 1/2 inf f 1/2< f ( x ) Figure5.2. Relationsbetweenthesupremumandin“mumof f andthemaximumandminimumvaluesof f Upperleftpanel :Thevaluesofthefunctionapproach 1as x approaches c fromtheleft,but f ( c )=1 / 2 < 1. Themaximumvalueof f doesnotexist,buttheleast upperbounddoesexist,sup f =1. Lowerleftpanel :Thevaluesof f approach1 / 2as x approaches c fromtheright,but f ( c )=1.Thefunction hasnominimumvalue,butthegreatestlowerboundis inf f =1 / 2. Upperrightpanel :Thevaluesof f approach1as x approaches c fromtheleftand f ( c )=1.Inthiscase, themaximalvalue f ( c )=1coincideswithsup f =1. Lowerrightpanel :Thevaluesof f approach1 / 2as x approaches c fromtherightand f ( c )=1 / 2.Theminimumvalue f ( c )=1 / 2coincideswiththegreatestlower boundinf f =1 / 2.

PAGE 185

32.THEDEFINITEINTEGRAL179 Definition 5.5 Let Ikbepartitionintervalsof [ a,b ] xkbethe lengthof Ik,and x k Ik.Thesum Rn( f )=nk =1f ( x k) xkiscalleda Riemannsum ofafunction f on [ a,b ] Thesum Rn( f )isnamedaftertheGermanmathematicianBernhardRiemann(1826…1866).Evidently,thevalueoftheRiemannsum generallydependsonthechoiceofpartitionintervalsandsamplepoints x k.However,forintegrablefunctionsRiemannsumshavearemarkable property.LetasequenceofRiemannsums Rnbede“nedsimilarlyto thesequencesoftheupperandlowersums,i.e.,thelargestpartition segmentdecreaseswithincreasing n Theorem 5.2 If f isintegrableon [ a,b ] ,then,foranynumber > 0 ,thereexistsaninteger N suchthat b af ( x ) dx Š Rn( f ) < foreveryinteger n>N andforeverychoiceof x kin Ik. Aproofofthistheoremisgivenasanexercise(seeExercise32.9.5; seealsoExercises32.9.12and32.9.13).Thetheoremassertsthata Riemannsumforasucientlylarge n canapproximatethede“nite integralwithanydesiredaccuracy;thatis,forany(small)designated absoluteerror Rn( f )diersfrom b afdx nomorethan forasucientlylarge n .Inotherwords, (5.6)limn Rn( f )= b af ( x ) dx, foranychoiceofsamplepoints x k.Equation(5.6)istheanalogofEquation(5.2).Itcanbeunderstoodfromtheinequality AL n Rn( f ) AU n, whichfollowsfrom mk f ( x k) Mkforany x k(seeFigure5.3). Foranintegrablefunction, AL nand AU nconvergetothesamenumber,whichisthevalueofthede“niteintegral,and,bythesqueeze principle,soshould Rn( f )independentlyofthechoiceofsamplepoints.32.4.ContinuityandIntegrability.Therelation(5.6)canbeusedto calculatethede“niteintegral, provided thefunction f is integrable Thequestionofintegrabilityrequiresinvestigatingtheconvergenceof thesequencesoftheupperandlowersums,whichmightbeatedious taskevenforsuchsimplefunctionsas,forexample, f ( x )= x2,as

PAGE 186

1805.INTEGRATION ab x1 *x2 *x3 *x4 *x5 Figure5.3. Riemannsumfor n =5partitionintervals.Itsvaluealwaysliesbetweenthelowerandupper sums, AL 5 R5 AU 5,foranychoiceofsamplepoints x kbecause mk f ( x k) Mk. discussedintheprevioussection.Thefollowingtheoremishelpful whenstudyingthequestionofintegrability. Theorem 5.3 If f iscontinuouson [ a,b ] ,orif f hasonlya“nite numberofboundedjumpdiscontinuities,then f isintegrableon [ a,b ] ; thatis,thede“niteintegral b af ( x ) dx exists.AnExampleofaNonintegrableFunction.Aboundedfunction f with in“nitelymanyjumpdiscontinuitiesmayormaynotbeintegrable.So, ingeneral,theareaunderthegraphofsuchafunctioncannotbeunambiguouslyde“ned.Asanexample,consideraboundednonnegative function f on[0 1]suchthat f ( x )= 1if x isarationalnumber 0if x isairrationalnumber Thefunctionisnotcontinuousanywherein[0 1]andhasin“nitely manyjumpdiscontinuities.Forexample, f (1 / 2)=1,butwhen x approaches1 / 2,thevalue f ( x )keepsjumpingfrom0to1andback, nomatterhowclose x isto1 / 2because,forany > 0,theinterval

PAGE 187

32.THEDEFINITEINTEGRAL181 (1 2Š ,1 2+ ) always containsbothrationalandirrationalnumbers. Thisfunctionisnotintegrable.Indeed,takeapartition xk= k/n k =0 1 ,...,n .Anypartitioninterval[( k Š 1) /n,k/n ]containsboth rationalandirrationalnumbers.Therefore, mk=0and Mk=1. Hence,thelowersumvanishesforanypartition, AL n=0,whereas theuppersumis AU n= n k =1 x =1,thatis,limn AL n=0while limn AU n=1.Thefunctionisnotintegrable.Theintegraldoes notexist.NotethattheRiemannsumcanstillbede“ned,butits limitwould depend onthechoiceofsamplepoints(e.g.,take x ktobe rationalnumbersortake x ktobeirrationalnumbers;bothoptions arepossiblesinceanypartitionintervalalwayscontainsrationaland irrationalnumbers).Infact,withasuitablechoiceofsamplepoints, theRiemannsumscanconvergetoanyvaluebetween0and1(e.g.,in allpartitionsegmentstoleftof0
PAGE 188

1825.INTEGRATION and,inparticular, (5.10) a af ( x ) dx =0 Itcanbeprovedthat (5.11) b af ( x ) dx = c af ( x ) dx + b cf ( x ) dx for f integrableon[ a,b ]andany a c b .Theproofisrather technicalandisomitted.If f iscontinuousandpositiveon[ a,b ],then theproperty(5.11)istrivial:Theareaunderthegraphof f on[ a,b ] isthesumoftheareasunderthegraphof f on[ a,c ]and[ c,b ].32.6.GeometricalSignicanceoftheDeniteIntegral.Asalreadynoted, thede“niteintegralof f from a to b coincideswiththeareaunderthe graphof f foracontinuousandpositive f .Suppose f iscontinuousand negative on[ a,b ].Considerthefunction g ( x )= Š f ( x ).Theintegralof g isthearea A underthegraphof g and,hence, A alsocoincideswith thearea above thegraphof f andbelowthe x axis.Bythelinearity oftheintegral, b af ( x ) dx = Š b ag ( x ) dx = Š A .So,foranegative f theintegralof f coincideswiththe negative areaoftheregionbounded belowbythegraphof f andabovebythe x axis.Nowlet f becontinuouson[ a,b ].Letitbepositiveon[ a,c ]andnegativeon[ c,b ],that is, f ( c )=0.Thenitfollowsfromtheproperty(5.11)that b af ( x ) dx = c af ( x ) dx + b cf ( x ) dx = A1Š A2, where A1isthearea under thegraphof f on[ a,c ]and A2isthearea above thegraphof f on[ c,b ].ThispropertyisillustratedinFigure5.4.32.7.ComparisonPropertiesoftheIntegral.Thefollowingadditional propertiesofthede“niteintegralcanbeestablished: b af ( x ) dx 0 if f ( x ) 0in[ a,b ] (5.12) b af ( x ) dx b ag ( x ) dx, if f ( x ) g ( x )in[ a,b ] (5.13) m ( b Š a ) b af ( x ) dx M ( b Š a ) if m f ( x ) M in[ a,b ] (5.14) Theproperty(5.12)followsdirectlyfromthede“nition.Since0 mk Mkforanypartitionif f ( x ) 0,theupperandlowersums

PAGE 189

32.THEDEFINITEINTEGRAL183 a b A1A2I A2 A1 Figure5.4. Geometricalinterpretationofthede“nite integral.Ifanintegrablefunction f isnonnegativeonan interval[ c,b ],thenitsintegralover[ c,b ]isthearea A2underthegraphof f abovetheinterval[ c,b ].If f isnonpositiveonaninterval[ a,c ],thenitsintegralover[ a,c ]is Š A1,where A1istheareaabovethegraphof f andbelow theinterval[ a,c ]onthe x axis.Bytheadditivityofthe integral,theintegralof f overtheinterval[ a,b ]beingthe unionofintervals[ a,c ]and[ c,b ]isthedierence A2Š A1. Thisshowsthatthevalueontheintegralcanbeanyreal number. arenonnegativeandsomustbetheintegral.If f iscontinuous,the property(5.12)statestheobviousthattheareaunderthegraphof f is nonnegative.Theproperty(5.13)followsfrom(5.12)forthefunction f ( x ) Š g ( x ) 0andthelinearityoftheintegral(5.8).Theproperty (5.14)isalsoaconsequenceofthede“nition.Indeed,foranypartition, m mk Mk M .Hence, m ( b Š a ) AL n AU n M ( b Š a )for any n .Inthelimit n ,thisinequalityturnsinto(5.14). Theorem 5.4(Integrabilityoftheabsolutevalue) Ifabounded function f ( x ) isintegrableoveraninterval [ a,b ] ,thenitsabsolutevalue | f ( x ) | isalsointegrableover [ a,b ] and b af ( x ) dx b a| f ( x ) | dx Aproofofthistheoremunderasimplifyinghypothesisthat f is continuousisgivenasanexercise(Exercise32.9.11).Theconverseof Theorem5.4isnottrue.Theintegrabilityoftheabsolutevalue | f ( x ) |

PAGE 190

1845.INTEGRATION M m ab Figure5.5. Geometricalinterpretationoftheproperty (5.14).Thegraphofafunction f liesbetweentwohorizontallines y = m and y = M because m f ( x ) M forall x [ a,b ].Sothearea A underthegraphof f lies betweentheareasofrectangleswiththebase b Š a and heights m and M ,i.e., m ( b Š a ) A M ( b Š a ). doesnotimplytheintegrabilityof f ( x ).Considerthefunction f ( x )= 1if x isarationalnumber Š 1if x isairrationalnumber Theabsolutevalue | f ( x ) | =1iscontinuousand,hence,isintegrable onanyboundedinterval.However, f ( x )isnotintegrable.Theproof ofthisassertionislefttothereaderasanexercise.32.8.EvaluationoftheIntegralbytheRiemannSum.Iftheintegralexists( f isintegrable),thenitcanbeevaluatedasthelimitoftheRiemannsum(5.6).Thelimitisindependentofthechoiceofsample points.Thefollowingchoicesareoftenusedinpractice: x k= xk Š 1(theleft-pointrule) x k= xk(theright-pointrule) x k=( xk Š 1+ xk) / 2(themidpointrule) incombinationwiththebasicpropertiesoftheintegral.Theevaluation oftheRiemannsumisrathertechnical.Formulaslike(5.1),(5.4),and (5.15)nk =1k = n ( n Š 1) 2 ,nk =1k3= n ( n Š 1) 2 2

PAGE 191

32.THEDEFINITEINTEGRAL185 canbehelpful.However,theRiemannsumismostlyusedtocalculate theintegral approximately withsomedesignatedaccuracybymeansof computersimulations,similarlytoapproximatecalculationsofthearea discussedintheprevioussection. Example 5.2 Findthede“niteintegralof f ( x )= eŠ 2 xŠ 2 x2+4 x3from 0 to 1 Solution:1 .Thefunctioniscontinuouson[0 1]andhenceintegrable; thatis,Equation(5.6)appliesforanychoiceof x k.Theleft-pointrule willbeused. 2 .Bythelinearityoftheintegral, 1 0f ( x ) dx = 1 0eŠ 2 xdx Š 2 1 0x2dx +4 1 0x3dx. The“rstintegralis(1 Š eŠ 2) / 2byExample5.1(wheretheareaunder thegraphof eŠ 2 xin[0 1]wascalculated).Theareaunderthegraph x2in[0 1]canbefoundatthebeginningoftheprevioussectionand isequalto1 / 3.Theareaunderthegraphof x3canbefoundwiththe helpofthesecondrelationin(5.15).Let x =1 /n and xk=( k Š 1) /n (theleft-pointrule),thentheRiemannsum(5.6)becomes 1 0x3dx =limn 1 n4 nk =1k3=limn 1 n4n2( n Š 1)2 4 = 1 4 3 .Thus, 1 0f ( x ) dx = 1 Š eŠ 2 2 Š 2 3 + 1 4 = 1 Š 6 eŠ 2 12 32.9.Exercises.(1)Let f ( x )=sin(1 /x )if x =0and f (0)= f0.Givenanynumber > 0,“ndthesupremumandin“mumof f on[ Š ]. (2)Findtheupperandlowersumsforthefunction f ( x )=1if x 0and f ( x )= Š 2if x< 0ontheinterval[ Š 1 1].Usethem toshowthat f isintegrable.Findthevalueoftheintegralas thelimitoftheloweranduppersums. (3)Findtheupperandlowersumsforthefunction f ( x )=1if x =0and f ( x )= f0 =1if x =0onaninterval[ a,b ].Use themtoshowthat f isintegrableonany[ a,b ].Findthevalue oftheintegralasthelimitoftheloweranduppersums. (4)Supposethatthelengthofthelargestpartitioninterval,n= maxk xk,decreasesasthenumber n ofpartitionintervals

PAGE 192

1865.INTEGRATION increases(i.e.,n +1< n).Showthat AL n AL n +1and AU n AU n +1.Deducefromthispropertythat AL n b afdx AU nforany n andforanyintegrable f on[ a,b ]. (5)Usetheinequalityfromthepreviousexercisetoprove Theorem5.2. (6)Let a 0 ,f ( x )= x,x a/ 2 Š 2 a + x,x>a/ 2 (iv) a Š a a2Š x2dx (v) 1 / 2 Š 1 / 2 1 Š x2Š| x | dx (vi) a 0 a2Š x2Š1 2 a2Š (2 x Š a )2 dx (10)UseRiemannsumsforequispacedpartitionstoevaluateeach ofthefollowingde“niteintegralscorrectuptothreedecimal places:

PAGE 193

32.THEDEFINITEINTEGRAL187 (i) 2 Š 1x1 / 3dx (ii) 2 0sin( x2) dx (iii) 1 Š 1exp(sin x ) dx (iv) 3 1x ln xdx (11)Let f becontinuouson[ a,b ].Then g ( x )= | f ( x ) | isintegrable on[ a,b ].Why?Showthat b afdx b a| f ( x ) | dx. Hint: CompareRiemannsumsfor f ( x )and | f ( x ) | (12)Let f haveaboundedderivativeon[ a,b ],thatis, | f( x ) | M1forall x [ a,b ].Consideranequispacedpartitionof[ a,b ]with x =( b Š a ) /n .Foreverypartitioninterval Ik=[ xk Š 1,xk], xk= a + kx x k =0 1 ,...,n ,showthatthereis xk Iksuchthat Ak= f ( xk) x istheareaunderthegraphof f on Ik.Let x k IkbesamplepointsintheRiemannsumfor f Usethemeanvaluetheoremtoprovethat | f ( x k) x Š Ak| M1 x2forany k .Deducefromthisinequalitythat b afdx Š Rn( f ) M1( b Š a )2 n (13)(TheTrapezoidalRule).Let f haveaboundedsecondderivativeon[ a,b ],thatis, | f( x ) | M2forall x [ a,b ].Bythe meanvaluetheorem,thereis c such f ( b ) Š f ( a )= f( c )( b Š a ). De“nethefunction g ( x )= f ( a )+ f( c )( x Š a ).Thegraph y = g ( x )isthesecantlinethroughthepoints( a,f ( a ))and ( b,f ( b )).Then T = b ag ( x ) dx =( f ( a )+ f ( b ))( b Š a ) / 2isthe areaofthetrapezoidboundedbytheline y = g ( x )on[ a,b ]. Usethemeanvaluetheoremforthederivative ftoprovethat | f ( x ) Š g ( x ) | M2( b Š a )2. Usethisinequalitytoshowthat b afdx Š b agdx M2( b Š a )3. Thetrapezoidalruletocalculate b afdx usesthepiecewise linearapproximationof f by g oneachpartitioninterval Ik=

PAGE 194

1885.INTEGRATION [ xk Š 1,xk]oflength xk: b afdx Tn( f )=nk =1xkxk Š 1gdx =nk =1 1 2( f ( xk Š 1)+ f ( xk)) xk. Provethat,foranequispacedpartition, b afdx Š Tn( f ) M2( b Š a )3 n2. Bycomparingthisresultwiththatinexercise12,onecansee thattheerrorofthetrapezoidalruledecreasesfasterthanthat intheRiemannsumapproximationasthenumberofpartition intervalsincreases.Soitisabetterwaytoapproximatethe integral.Oneshouldkeepinmind,however,thattheintegrand hastohaveabounded second derivativeforsuchasuperiority. (14)Evaluate 2 0sin( x2) dx correctuptothreedecimalplacesusing theRiemannsumandtrapezoidalapproximations.Howmany partitionintervalsarerequiredtoachievethisaccuracyineach oftheapproximations? 33.TheFundamentalTheoremofCalculus Inthissection,therelationbetweenthede“niteintegralofafunctionanditsantiderivativewillbeestablished.Thisrelationprovides apowerfulmethodforcalculatingthede“niteintegralthatavoidsthe useofRiemannsums.33.1.IntegrationandDifferentiation.Considerthede“niteintegralof f ( t )= t from0to x forsome x> 0.Thisintegralrepresentsthearea underthegraphof f ( t )= t intheinterval[0 ,x ],whichistheareaofa righttriangle: A ( x )= x 0tdt = x2 2 Thearea A ( x )canbeviewedasafunctionofthevariable x ,which isthelengthofthetrianglecatheti.Thisfunctionhasaninteresting property: A( x )= x = f ( x ) Inotherwords,thederivativeofthede“niteintegralwithrespecttoits upperlimitequalsthevalueoftheintegrandattheupperlimit.Recall thatif v ( t ) 0isthespeedofamovingobject,thenthedistance

PAGE 195

33.THEFUNDAMENTALTHEOREMOFCALCULUS189 traveledbytheobjectintime T isgivenbytheareaunderthegraph of v ( t ): D ( T )= T 0v ( t ) dt. Ontheotherhand,thespeedistherateofchangeof D ( T ),andthereforethereshouldbe D( T )= v ( T );thatis,thederivativeoftheintegral withrespecttoitsupperlimitisagainthevalueoftheintegrandat theupperlimit.Howgeneralisthisproperty?Doesitholdforall integrablefunctions?Thefollowingtheoremanswersthesequestions. Theorem 5.5 If f iscontinuouson [ a,b ] ,thenthefunctionde“ned by g ( x )= x af ( t ) dt,a x b, iscontinuouson [ a,b ] anddierentiableon ( a,b ) ,and g( x )= f ( x ) Proof. Bythede“nitionofthederivative,onehastoprovethat (5.16)limh 0g ( x + h ) Š g ( x ) h = f ( x ) for a 0.Then m f ( t ) M for x t x + h and,bytheproperty(5.14), (5.17) mh = f ( u ) h x + h xf ( t ) dt f ( v ) h = Mh.

PAGE 196

1905.INTEGRATION Since h> 0,bydividingthisinequalityby h ,onecaninferthat (5.18) f ( u ) 1 h x + h xf ( t ) dt f ( v ) forsome u and v in[ x,x + h ].Inequality(5.18)canbeestablished for h< 0inasimilarmanner.Indeed,inequality(5.17)holdsforthe integral x x + hf ( t ) dt .Afterdividingitby Š h> 0,inequality(5.18) isobtainedbutwiththeminussignattheintegral.Bytheproperty (5.9),thesignisreversed,yielding(5.18).Thus, f ( u ) g ( x + h ) Š g ( x ) h f ( v ) Since u and v lieintheinterval[ x,x + h ], limh 0f ( u )= f ( x ) limh 0f ( v )= f ( x ) Thentherelation(5.16)followsfromthesqueezeprinciple: f ( x )=limh 0f ( u ) limh 0g ( x + h ) Š g ( x ) h limh 0f ( v )= f ( x ) Thistheorembasicallystatesthatifacontinuousfunctionis“rst integratedandthendierentiated,thenitremainsunchanged: (5.19) d dx x af ( t ) dt = f ( x ) ,a 1.But g(1)does notexist. Example 5.3 Let g ( x )= b xeŠ t2dt .Find g( x ) Solution: Thefunction eŠ t2isacontinuousfunctioneverywhereasa compositionoftwocontinuousfunctions,theexponentialandpower

PAGE 197

33.THEFUNDAMENTALTHEOREMOFCALCULUS191 functions.Bytheproperty(5.9), g ( x )= Š x beŠ t2dt .Therefore, g( x )= Š eŠ x2by(5.19). Thisexampleillustratesthegeneralproperty: d dx b xf ( t ) dt = Š d dx x bf ( t ) dt = Š f ( x ) foracontinuous f .33.2.TheDeniteIntegralandAntiderivative.Thefollowingtheorem establishestherelationbetweenthede“niteintegralofafunctionand itsantiderivative. Theorem 5.6(TheFundamentalTheoremofCalculus) If f is continuouson [ a,b ] ,then b af ( x ) dx = F ( b ) Š F ( a ) where F isanyantiderivativeof f ,thatis,afunctionsuchthat F= f Proof. Let g ( x )= x af ( t ) dt .By(5.19),thefunction g ( x )isan antiderivativeof f ( x )inanopeninterval( a,b ).If F isanyother antiderivativeof f ,then F and g maydieronlybyaconstant, F ( x )= g ( x )+ C,a
PAGE 198

1925.INTEGRATION Example 5.4 Evaluate 1 0(1+ x2)Š 1dx Solution: Anantiderivativeof(1+ x2)Š 1is F ( x )=tanŠ 1x .Therefore, 1 01 1+ x2dx =tanŠ 1x 1 0=tanŠ 1(1) Š tanŠ 1(0)= 4 Š 0= 4 Example 5.5 Evaluate 4 1(1+ x ) / xdx Solution: Bythelinearityoftheintegral, 4 11+ x x dx = 4 1( xŠ 1 / 2+ x1 / 2) dx = 4 1xŠ 1 / 2dx + 4 1x1 / 2dx. Anantiderivativeof xnis xn +1/ ( n +1)foranyreal n =1.Bytaking n = Š 1 / 2and n =1 / 2,anantiderivativeisobtained: F ( x )=2 x1 / 2+ 2 x3 / 2/ 3.Hence, 1 01+ x x dx = 2 x1 / 2+ 2 x3 / 2 3 1 0= 4+ 16 3 Š 2+ 2 3 = 20 3 Ifanobjectmovesalongastraightline,itspositionrelativetoa “xedpointontheline(theorigin)maybede“nedbyasinglecoordinate x whichisafunctionoftime.Thevelocity v ( t )= x( t )ispositiveif theparticlemovesinthedirectioninwhich x increasesandisnegative ifitmovesintheoppositedirection.Theaccelerationis a ( t )= v( t )= x( t ).Alawaccordingtowhichtheaccelerationchangeswithtimeis usuallyestablishedbythelawsofphysics.Thenapracticalquestion isto“ndtheposition x ( t ).Since x ( t )isasecondantiderivativeof theacceleration,itisnotuniqueandtwo(initial)conditionsmustbe imposedtogetauniquesolution. Example 5.6 Aparticlemovesalongthe x axiswiththeacceleration a ( t )=2 Š 6 t .Findthepositionoftheparticleatthetime t =3 ifitspositionandvelocityat t =1 were x (1)=1 and v (1)=2 Solution: Since v( t )= a ( t ),thevelocityistheantiderivativeofthe accelerationsubjecttothecondition v (1)=2.Hence,bytheproperty (5.10), v ( t )= v (1)+ t 1a ( s ) ds =2+(2 s Š 3 s2) t 1=2 t Š 3 t2+3

PAGE 199

33.THEFUNDAMENTALTHEOREMOFCALCULUS193 Since x( t )= v ( t ),theposition x ( t )istheantiderivativeofthevelocitysubjecttothecondition x (1)=1.Byproperty(5.10)suchan antiderivativereads x ( t )= x (1)+ t 1v ( s ) ds =1+( s2Š s3+3 s ) t 1= t2Š t3+3 t Š 2 Therefore x (3)= Š 11. 33.3.Exercises.(1)Findthederivativeofeachofthefollowingfunctions: (i) f ( x )= x 1(1+ t6)Š 1dt (ii) f ( x )= x20sin( t2) dt (iii) f ( x )= x Š xcos( et) dt (iv) f ( x )= x sin xet2dt (2)Let f ( x )beapiecewiseconstantfunction: f ( x )= 0 ,x< 0and x> 6 1 0 x< 2 2 2 x< 4 Š 3 4 x 6 Usethegeometricalinterpretationofthede“niteintegralto drawthegraphsof g ( x )= x 1f ( t ) dt and h ( x )= x 4f ( t ) dt (3)Foraparticlemovingdownaroughinclinedplane,thevelocity is v ( t )= t .Whereistheparticleattheendof2seconds? (4)Findthelocationofaparticlemovingalongalineattheendof 2secondsiftheaccelerationoftheparticleis a ( t )=6 t Š 12 t2andifitspositionandvelocityattheendof1secondwere s (1)=5and v (1)=10. (5)Aspacecrafthadaconstantvelocityof v0.Thenitsengines were“redforatime T1,thenstoppedforatime T ,andthen “redagainforatime T2.Ifduringthetimeintervals T1and T2,theenginescreatedconstantaccelerations a1and a2,respectively,whatisthe“nalvelocityofthespacecraft? (6)Findtheareaoftheplanarregionboundedabovebythe parabola y =3 Š x2andbelowbytheparabola y =1+ x2. (7)Findtheareaoftheplanarregionboundedabovebythe parabola y =2 Š x2andbelowbytheline y = x (8)Evaluatetheintegralsusingthetableofantiderivatives: (i) 8 Š 13 xdx (ii) 2 1( x +1 /x ) dx

PAGE 200

1945.INTEGRATION (iii) 3 Š 1 / 3(1+ x2)Š 1dx (iv) 1 / 2 Š 1 / 2(1 Š x2)Š 1 / 2dx (v) 1 Š 1(1+ x2)Š 1 / 2dx (vi) 2 0| 1 Š x | dx (vii) 2 Š 1eŠ 2 xdx (viii) 0sin xdx (ix) 0cos2xdx (x) 1 06( x Š4 x ) dx (xi) 3 1x2(4+ x2)Š 1dx (xii) 1 0(1+ x + x2+ x3+ + xn) dx (xiii) 2 0(1+cos x +cos(2 x )+ +cos( nx )) dx (xiv) 1 Š 1( x2Š 2 x cos +1)Š 1dx Hint: To“ndanantiderivative,completethesquares. 34.IndeniteIntegralsandtheNetChange Ashasbeenshownintheprevioussection,thederivativeofthe de“niteintegralofacontinuousfunction f withrespecttotheupper limitequalsthevalueof f attheupperlimit.Sointegrationanddierentiationappearasoperationsinversetooneanother.Tofurtherstress thisrelationbetweentheintegrationanddierentiation,thenotionof aninde“niteintegralisintroduced. Definition 5.6(Inde“niteIntegral) Thefunction F iscalledan inde“niteintegral of f andisdenotedby F ( x )= f ( x ) dx if F( x )= f ( x ) Itfollowsfromthisde“nitionthataninde“niteintegralisnothingbutthegeneralantiderivativeof f .Thereasonforintroducing theintegralsymbolintotheantiderivativenotationisthefundamental theoremofcalculus: b af ( x ) dx = F ( b ) Š F ( a ) where F isanyantiderivativeof f .Sinceallantiderivativesdieronly byaconstant,whichisalwayscancelledoutinthedierence F ( b ) Š F ( a ),thede“niteintegralisthedierenceinvaluesoftheinde“nite integralattheupperandlowerlimitsofthede“niteintegral.

PAGE 201

34.INDEFINITEINTEGRALSANDTHENETCHANGE195 Theinde“niteintegralhasthesamepropertiesastheantiderivative. Itislinear: (5.21) c1f ( x )+ c2g ( x ) dx = c1 f ( x ) dx + c2 g ( x ) dx foranyconstants c1and c2andanyfunctions f and g Usingthetableofantiderivativesofbasicfunctions,onecanmake atableofinde“niteintegralsofbasicfunctions.Let C beanarbitrary constant.Thefollowingtablecanbeveri“edbydierentiation. Tableofbasicindeniteintegrals xndx = xn +1 n +1 + C,n =1 dx x =ln | x | + C,x =0 sin( ax ) dx = Š cos( ax ) a + C cos( ax ) dx = sin( ax ) a + C exdx = ex+ C axdx = ax ln a + C,a> 0 ,a =1 dx 1+ x2= tanŠ 1x + C Š cotŠ 1x + C dx 1 Š x2= sinŠ 1x + C Š cosŠ 1x + C sec2( ax ) dx = tan( ax ) a + C csc2( ax ) dx = Š cot( ax ) a + C sec x tan xdx =sec x + C csc x cot xdx = Š csc x + C dx 1 Š x2= 1 2 ln 1+ x 1 Š x + C dx x2 1 =ln x + x2 1 + C Recallthatthegeneralantiderivativeona giveninterval isobtained fromaparticularantiderivativebyaddinganarbitraryconstant.This doesnotholdforadomainbeingaunionoftwoormore disjoint intervals(reviewthepropertiesofantiderivatives).So,inthepreceding table,theconventionisusedthatthegivenexpressionsforinde“nite integralsarevalidonlyina singleinterval ofcontinuityoftheintegrand. Example 5.7 Findageneralinde“niteintegralfor xŠ 3. Solution: Thefunction xŠ 3isnotde“nedat x =0.Soitsdomain istheunionoftwo disjoint intervals( Š 0)and(0 ).Bythe“rst equalityintheprecedingtable( n = Š 3), xŠ 3dx = Š xŠ 2 2 + C1,x> 0; xŠ 3dx = Š xŠ 2 2 + C2,x< 0 ,

PAGE 202

1965.INTEGRATION where C1and C2are arbitrary constants. Example 5.8 Evaluate 1 0[3 x2Š x +4(1+ x2)Š 1] dx Solution: Bythelinearityoftheinde“niteintegral(5.21),aninde“nite integraloftheintegrandis x3Š x2/ 2+4tanŠ 1x .Anarbitraryconstant intheinde“niteintegralmaybeomittedherebecause,asalreadynoted, itisalwayscancelledoutinthede“niteintegral.Therefore, 1 0 3 x2Š x + 4 1+ x2 dx = x3Š x2 2 +4tanŠ 1x 1 0= 1 2 Š wheretanŠ 1(1)= / 4hasbeenused. 34.1.TheNetChangeTheorem.Put f ( x )= F( x )inthefundamental theoremofcalculus(5.20).Theresultobtainedisknownasthenet changetheorem. Theorem 5.7 Theintegralofacontinuousrateofchangeisthe netchange: b aF( x ) dx = F ( b ) Š F ( a ) Thecontinuityof F( x )iscrucialfortheaboverelationtohold.A formal applicationofthetheoremwithoutcheckingthecontinuitymay leadtoincorrectresults.Forexample,thefollowinglineofcalculations is false : 1 Š 1d dx Š 1 x dx = Š 1 x 1 Š 1= Š 1 Š 1= Š 2 Indeed,thederivative F( x )=( Š 1 /x )=1 /x2isnotde“nedat x =0. Itisnotpossibletoassignanynumericalvalueto F(0)tomake F( x ) continuousat x =0becauselimx 0F( x )= whichisnotanumber. Furthermore, F( x ) > 0isstrictlypositiveinanyintervaland,bythe property(5.12),soshouldbethede“niteintegralifitexists.Hence, theaboveresultcannotpossiblybetrue. Therate F( x )maybepositiveandnegativeintheinterval[ a,b ]so thatthequantity y = F ( x )may increaseanddecrease .Thedierence F ( b ) Š F ( a )representsthe net changeof y when x changesfrom a to b .Thenetchangevanishesif F ( b ) Š F ( a )=0.Thisdoesnotmean thatthequantity y doesnotchangeatall,butratherthismightmean, forexample,thatthequantity y increasesfromthevalue F ( a ),then, atsome c in[ a,b ],itbeginstodecrease,returningtoitsinitialvalue when x = b sothatitsnetchangevanishes.

PAGE 203

34.INDEFINITEINTEGRALSANDTHENETCHANGE197 Ananalogywithanobjectmovingalongastraightlinecanbe madetoillustratethenetchange.Let x ( t )beapositionfunctionof theobjectrelativetosomepointontheline.Then x( t )= v ( t )isits velocity(notethatthevelocitycanbenegativesothattheobjectcan movebackandforth).Thenetchangeofthepositionoverthetime interval[ t1,t2]is t2t1v ( t ) dt = x ( t2) Š x ( t1) Example 5.9 Supposeanobjecttravelsalongastraightlinewith avelocityof v ( t )=1 Š 2 t .Findthenetchangeofitspositionoverthe timeinterval [0 1] andthetotaldistancetraveledbytheobjectoverthe sametimeinterval. Solution:1 .Theinde“niteintegralof v ( t )is x ( t )= t Š t2+ C .So thenetchangeoftheobjectpositionis 1 0v ( t ) dt = 1 0x( t ) dt = x (1) Š x (0)=0 2 .Notethatthevelocitychangesitssignat t =1 / 2.So,inthe interval[0 1 / 2],itispositive(i.e.,theobjectmovestotherightfrom itsinitialposition),thenthevelocitybecomesnegativein[1 / 2 1](i.e., theobjectgoesbacktotheinitialpoint).To“ndthedistancetraveled bytheobject,theabsolutevalue | v ( t ) | mustbeintegratedoverthe interval[0 1].Thinkof | v ( t ) | asthespeedshownonthespeedometer ofyourcar;itisalwaysnonnegativeregardlessofthedirectionin whichthecarismoving. 1 0| 1 Š 2 t | dt = 1 / 2 0(1 Š 2 t ) Š 1 1 / 2(1 Š 2 t ) dt =[ x (1 / 2) Š x (0)] Š [ x (1) Š x (1 / 2)]=1 / 2 wherethede“nition | v | = v if v> 0and | v | = Š v if v< 0hasbeen used. Otherexamplesofthenetchangeincludesthevolume V ( t )ofwater inareservoirbetweentwomomentsoftime t2t1V( t ) dt = V ( t2) Š V ( t1) ,

PAGE 204

1985.INTEGRATION where V( t )istherateofchangeofthevolume;thenetchangeofthe populationgrowth t2t1n( t ) dt = n ( t2) Š n ( t1) where n( t )isthegrowthrate;therelationbetweenthecostandmarginalcostfunctions: t2t1C( t ) dt = C ( t2) Š C ( t1); andsimilarlyformanyotherquantities.34.2.Exercises.(1)Findtheinde“niteintegrals.Assumethat x liesinasingle intervalofcontinuityoftheintegrand. (i) 6 x5dx (ii) xŠ 3dx (iii) ( x2 / 3Š xŠ 2 / 3) dx (iv) ( x2Š 4)2dx (v) (1+ x )3dx (vi) ( x + a )2/ xdx (vii) x/ x2+1 dx (viii) (1+ xŠ 1) x xdx (ix) (1+cos x +cos(2 x )) dx (x) ( x +sin(4 x )) dx (xi) (1+2 x ) / 1 Š x2dx (xii) (1 Š x + x3) /x2dx (xiii) x2/ ( a2+ x2) dx (xiv) ( x2+3) / ( x2Š 1) dx (xv) xŠ 2 x4+ xŠ 4+2 dx (xvi) [ 1+ x2+ 1 Š x2] / 1 Š x4dx (xvii) [ x2+1 Š x2Š 1] / x4Š 1 dx (xviii) (2x+3x)2dx (xix) (2x +1Š 5x Š 1) / 10xdx (xx) ( e3 x+1) / ( ex+1) dx (xxi) 1 Š sin(2 x ) dx ,0 x Hint: Usethefundamentaltrigonometricidentity. (xxii) cot2xdx (xxiii) tan2xdx

PAGE 205

34.INDEFINITEINTEGRALSANDTHENETCHANGE199 (2)Provethattheexistenceoftheinde“niteintegralof f ( x )impliestheexistenceoftheinde“niteintegralof f ( ax + b ), a =0, and f ( x ) dx = F ( x )+ C = f ( ax + b ) dx = 1 a F ( ax + b )+ C (3)Usetheresultofthepreviousexerciseto“nd (i) ( x + a )Š 1dx (ii) (2 x Š 3)10dx (iii) 3 1 Š 3 xdx (iv) (5 x Š 2)5 / 2dx (v) (2+3 x2)Š 1dx (vi) (3 Š 2 x2)Š 1dx (vii) (4 Š 3 x2)Š 1 / 2dx (viii) (2 x2Š 5)Š 1 / 2dx (ix) csc2(2 x + / 4) dx (x) (1+cos x )Š 1dx (xi) (1 Š sin x )Š 1dx (xii) (1+sin x )Š 1dx Hint: Usesin x =2sin( x/ 2)cos( x/ 2)andthefundamental trigonometricidentity. (4)Findageneralinde“niteintegral: (i) xŠ 1dx (ii) x1 /pdx p> 1 (iii) sec2xdx (iv) (1 Š x2)Š 1dx (v) ( x2Š 1)Š 1dx (5)Explainwhyaformalapplicationofthefundamentaltheorem ofcalculuswithagivenantiderivative F ( x )leadstoincorrect resultsif (i) 1 Š 1dx x ,F ( x )=ln | x | (ii) 1 Š 1d dx tanŠ 1 1 x dx,F ( x )=tanŠ 1 1 x (6)Evaluate 1 Š 1d dx 1 1+2x dx and 1 Š 1d dx 1 1+21 /x dx.

PAGE 206

2005.INTEGRATION (7)Aparticletravelswithvelocity v ( t )=sin( t/ 2).Findthenet displacementoftheparticleoverthetimeinterval[0 2 ]and thedistancetraveledbytheparticle. (8)Abacteriapopulationgrowsatanexponentialrate n( t )= n0et,where n0istheinitialpopulationand isaconstant.If inthetime T thepopulationhasdoubled,“ndtheconstant Whatisthepopulationat t =10 T ascomparedtotheinitial population? (9)Thedecayrateofaradioactiveelementisproportionaltothe totalamountoftheelementateachmomentoftime.Find thelawofdecayoftheradiumisotopeRa-226ifataninitial momentoftimetherewere n0gramsofRa-226andin1600 yearsitsquantityhaddecreasedbytwotimes.If1gramof Ra-226isdepositedintoaradioactivestorage,howmuchofit willremainin800years? 35.TheSubstitutionRule35.1.IndeniteIntegrals.Aninde“niteintegralofthederivative F( x ) isthefunction F ( x )itself,provided F( x )iscontinuous.Let u = F ( x ), where u isanewvariablede“nedasadierentiablefunctionof x Considerthedierential du = F( x ) dx .Thenthefollowingequalities hold: F( x ) dx = F ( x )+ C = u + C = du, where C isanarbitraryconstantandthelastequalityfollowsfromthe factthataninde“niteintegralof f ( u )=1is u .Sowecanconclude that F( x ) dx = du ,providedthevariables u and x arerelatedas u = F ( x ).Thisalsoshowsthat itispermissibletooperatewith dx and du aftertheintegralsignasiftheyweredierentials .Thisobservationleadstoaneattechnicaltricktocalculateinde“niteintegrals. Forexample, 1 x +1 dx = d 2 x +1 =2 x +1+ C, wherethesubstitution u =2 x +1hasbeenused.Thistrickcanbe generalized. Let F ( u )beaninde“niteintegralofacontinuousfunction f ( u )on aninterval I .Let u = g ( x ),where g isdierentiableanditsrangeis theinterval I .Bythechainrule, F ( g ( x )) = F( g ( x )) g( x )= f ( g ( x )) g( x ) .

PAGE 207

35.THESUBSTITUTIONRULE201 Inotherwords, F ( g ( x ))+ C isaninde“niteintegralof f ( g ( x )) g( x ).On aninterval,themostgeneralinde“niteintegralof f ( u )is f ( u ) du = F ( u )+ C .Therefore, F ( g ( x ))and f ( u ) du candieratmostbyan additiveconstant.Thisprovesthefollowingtheorem. Theorem 5.8 (TheSubstitutionRule) .If u = g ( x ) isadifferentiablefunctionwhoserangeisaninterval I and f iscontinuous on I ,then (5.22) f ( g ( x )) g( x ) dx = f ( g ( x )) dg ( x )= f ( u ) du. Thesubstitutionruleisoftenreferredtoasa changeoftheintegrationvariable .Itisapowerfulmethodtocalculateinde“niteintegrals. Example 5.10 Find x sin( x2+1) dx Solution: x sin( x2+1) dx = sin( x2+1) 1 2 d ( x2+1)= 1 2 sin udu = Š 1 2 cos u + C = Š 1 2 cos( x2+1)+ C, wherethesubstitution u = x2+1hasbeenused. Example 5.11 Find tan xdx Solution: tan xdx = sin x cos x dx = Š d (cos x ) cos x = Š du u = Š ln | u | + C = Š ln | cos x | + C =ln | sec x | + C, wherethesubstitution u =cos x andthelogarithmpropertyln(1 /a )= Š ln a havebeenused.Hereitshouldbenotedthatthecalculations arevalidonlyinasingleintervalofcontinuityoftan x ,e.g., Š / 2 < x
PAGE 208

2025.INTEGRATION Therefore dx exŠ 1 = 1 u du u +1 =2 d u u +1 =2 dv v2+1 =2tanŠ 1v + C =2tanŠ 1( u )+ C =2tanŠ 1( exŠ 1)+ C wherethesecondsubstitution v = u ,2 dv = du/ u ,hasbeenmade. 35.2.DeniteIntegrals.Thesubstitutionrulecanbeusedtoevaluate de“niteintegralsbymeansofthefundamentaltheoremofcalculus. Example 5.13 Evaluate 2 0xex2dx Solution: First,“ndaninde“niteintegral: F ( x )= xex2dx = 1 2 ex2dx2= 1 2 eudu = 1 2 eu+ C = 1 2 ex2+ C. where u = x2.Bythefundamentaltheoremofcalculus, 2 0xex2dx = F (2) Š F (0)= 1 2 ( e4Š 1) Notethat,whenevaluatingtheintegral,theoriginalvariable x has beenrestoredintheinde“niteintegralinordertoapplythefundamentaltheoremofcalculus.Thefundamentaltheoremofcalculuscanalso beapplieddirectlyinthenewvariable u ,providedtherangeof u is properlychanged.Indeed,inthepreviousexample,theanswercould havebeenrecoveredfromtheinde“niteintegral1 2eu+ C if u = x2rangesfrom0=02to4=22as x rangesfrom0to2.Thisisespecially usefulwhenacalculationofade“niteintegralrequiresseveralchanges oftheintegrationvariable. Theorem 5.9 (TheSubstitutionRuleforDe“niteIntegrals) .If giscontinuouson [ a,b ] and f iscontinuousontherangeof u = g ( x ) then (5.23) b af ( g ( x )) g( x ) dx = g ( b ) g ( a )f ( u ) du. Proof. Let F beanantiderivativeof f .Then F ( g ( x ))isanantiderivativeof( F ( g ( x )))= F( g ( x )) g( x )= f ( g ( x )) g( x ).Bythefundamentaltheoremofcalculus, b af ( g ( x )) g( x ) dx = F ( g ( x )) b a= F ( g ( b )) Š F ( g ( a )) .

PAGE 209

35.THESUBSTITUTIONRULE203 Ontheotherhand,since F ( u )isanantiderivativeof f ( u ),thefundamentaltheoremofcalculusyields g ( b ) g ( a )f ( u ) du = F ( u ) g ( b ) g ( a )= F ( g ( b )) Š F ( g ( a )) Sincetheright-handsidesoftheseequalitiescoincide,somusttheir left-handsides,whichimplies(5.23). Example 5.14 Evaluate e 1ln x/xdx Solution: Theintegrandcanbetransformedas ln x x dx =ln xd ln x. Sothesubstitution u =ln x canbemade.Therangeofthenew integrationvariable u isdeterminedbytherangeoftheoldone: u =0 when x =1and u =1when x = e .Thus, e 1ln x x dx = 1 0udu = u2 2 1 0= 1 2 35.3.Symmetry.Thecalculationofade“niteintegraloverasymmetricintervalcanbesimpli“ediftheintegrandpossessessymmetryproperties. Theorem 5.10 Suppose f iscontinuousonasymmetricinterval [ Š a,a ] .Then a Š af ( x ) dx =2 a 0f ( x ) dx if f ( Š x )= f ( x )( f iseven) (5.24) a Š af ( x ) dx =0if f ( Š x )= Š f ( x )( f isodd) (5.25) Proof. Theintegralcanbesplitintotwointegrals: a Š af ( x ) dx = 0 Š a+ a 0 f ( x ) dx = Š Š a 0f ( x ) dx + a 0f ( x ) dx. Inthe“rstintegralontheveryright-handside,thesubstitution u = Š x ismadesothat u =0when x =0and u = a when x = Š a and dx = Š du .Hence, ŠŠ a0f ( x ) dx =a0f ( Š u ) du

PAGE 210

2045.INTEGRATION A a a A I 0 Figure5.6. Illustrationoftheproperty(5.25).Afunctionisoddif f ( Š x )= Š f ( x ).Itsintegralovera symmetric interval[ Š a,a ]vanishes.Thearea A underthe graphof f andabovetheinterval[0 ,a ]isthesameasthe areaabovethegraphof f andbelowtheinterval[ Š a, 0] becauseoftheskewsymmetryofthefunction and the symmetryoftheinterval[ Š a,a ]relativetothere”ection x Š x .BythepropertydepictedinFigure5.4,the integralof f over[ Š a,a ]is A +( Š A )=0. and a Š af ( x ) dx = a 0f ( Š u ) du + a 0f ( x ) dx. Now,if f iseven,then f ( Š u )= f ( u )and(5.24)follows.If f isodd, then f ( Š u )= Š f ( u )and(5.25)follows. Thegeometricalinterpretationofthistheoremistransparent(see Fig5.6).Suppose f ( x ) 0for0 x a .Theintegral a 0f ( x ) dx = A istheareaunderthegraphof f on[0 ,a ].If f iseven,then,by symmetry,thegraphof f on[ Š a, 0]isobtainedfromthaton[0 ,a ] byare”ectionaboutthe y axis.Therefore,thearea 0 Š af ( x ) dx must coincidewith A .If f isodd,thenitsgraphon[ Š a, 0]isobtainedbythe mirrorre”ectionabouttheoriginsothatthearea A appearsbeneath the x axis.Hence, 0 Š af ( x ) dx = Š A Example 5.15 Evaluate Š sin( x3) dx .

PAGE 211

35.THESUBSTITUTIONRULE205 Solution: Unfortunately,anantiderivativeofsin( x3)cannotbeexpressedinelementaryfunctions,andthefundamentaltheoremofcalculuscannotbeused.Onecanalwaysevaluatetheintegralbytaking thelimitofthesequenceofRiemannsums.Analternativesolution isduetoasimplesymmetryargument.Notethatsin( x3)isanodd function,sin(( Š x )3)=sin( Š x3)= Š sin( x3).Theintegrationinterval isalso symmetric ,[ Š ].Thus,byproperty(5.25), Š sin( x3) dx =0 Remark .Inthepreviousexample,takeapartitionof[ Š ]by points xk= k x k = Š n, Š n +1 ,..., Š 1 0 1 ,...,n Š 1 ,n ,where x = /n .ConsidertheRiemannsumwithsamplepointsbeingthe midpoints.Theyhavethepropertythat x Š k= Š x k.ItisthenstraightforwardtoshowthattheRiemannsumvanishesbecausesin( x 3 Š k)= sin(( Š x k)3)= Š sin( x 3 k)for k =1 2 ,...,n (thetermscorresponding tonegative x kcanceloutthetermscorrespondingtopositive x kinthe Riemannsum).35.4.Exercises.(1)Usethesuggestedsubstitutionto“ndtheinde“niteintegrals: (i) x3( x4+1)1 / 3dx,u = x4+1 (ii) sin( x ) / xdx,u = x (iii) sin xecos xdx,u =cos x (iv) x2 1 Š x2dx,u =sin x (v) (ln x )3/xdx,u =ln x (vi) (tan x )nsec2xdx n = Š 1, u =tan x (vii) (cot x )ncsc2xdx n = Š 1, u =cot x (viii) (sinŠ 1x )2(1 Š x2)Š 1 / 2dx u =sinŠ 1x (ix) ex( e2 x+1)Š 1dx u = ex(x) e2 x(1+ ex)Š 1dx u =1+ ex(xi) sin(2 x )(1+cos2x )pdx u =cos2x (xii) ex+1 dx u = ex+1 Hint: SeeExample5.12toproceed. (xiii) 1 Š exdx u =1 Š exHint: SeeExample5.12toproceed. (2)Useasubstitutionto“ndtheinde“niteintegrals: (i) x 1+2 xdx (ii) x2/ 2 Š 3 xdx (iii) x3 1+ xdx

PAGE 212

2065.INTEGRATION (iv) eŠ x/ xdx (v) x/ ( x4+2 x2+2) dx Hint :Completethesquaresinthedenominator. (vi) x tanŠ 1x (1+ x2)Š 2dx (vii) cosŠ 1x/ 1 Š x2dx (viii) 1 Š x2sinŠ 1xdx (ix) (1+cos2x +cos4x )sin xdx (x) sin3x/ (1+cos2x ) dx (xi) [ 1+ ex+ 1 Š ex]Š 1dx Hint: Transformtheintegrandtomakethedierenceofperfectsquaresinthedenominator: a2Š b2=( a + b )( a Š b ). (3)Useachangeofvariablesand/orsymmetrytoevaluatethe de“niteintegrals: (i) 1 0x 2+ x2dx (ii) 1 0tanŠ 1x/ (1+ x2) dx (iii) 0x cos( x2) dx (iv) 1 Š 1x3ex4dx (v) 2 Š 2x ( ex2Š eŠ x2) dx (vi) 2 Š 2( ex3Š eŠ x3) dx (vii) a Š ag ( x ) dx,g ( x )= x 0cos( t2) dt (viii) 2 1(2 x +1) x2+ x +3 dx (ix) / 2 0sin(2 x )3 1+cos2xdx (x) / 4 0(tan x )psec2xdx,p> 0 (xi) / 6 Š / 6tan3(3 x )sin5(2 x ) dx (xii) a 0 1 Š eŠ xdx a> 0 (xiii) 3 Š 3x3/ (1+ x6) dx



PAGE 1

ConceptsinCalculusIUNIVERSITYPRESSOFFLORIDAFloridaA&MUniversity,Tallahassee FloridaAtlanticUniversity,BocaRaton FloridaGulfCoastUniversity,Ft.Myers FloridaInternationalUniversity,Miami FloridaStateUniversity,Tallahassee NewCollegeofFlorida,Sarasota UniversityofCentralFlorida,Orlando UniversityofFlorida,Gainesville UniversityofNorthFlorida,Jacksonville UniversityofSouthFlorida,Tampa UniversityofWestFlorida,Pensacola OrangeGroveTexts Plus

PAGE 3

ConceptsinCalculusI MiklosBonaandSergeiShabanov UniversityofFloridaDepartmentof MathematicsU niversity P ressof F lorida Gainesville € Tallahassee € Tampa € BocaRaton Pensacola € Orlando € Miami € Jacksonville € Ft.Myers € Sarasota

PAGE 4

Copyright2011bytheUniversityofFloridaBoardofTrusteesonbehalfoftheUniversityof FloridaDepartmentofMathematics Thisworkislicensedunderamodi“edCreativeCommonsAttribution-Noncommercial-No DerivativeWorks3.0UnportedLicense.Toviewacopyofthislicense,visithttp:// creativecommons.org/licenses/by-nc-nd/3.0/.Youarefreetoelectronicallycopy,distribute,and transmitthisworkifyouattributeauthorship. However,allprintingrightsarereservedbythe UniversityPressofFlorida(http://www.upf.com).PleasecontactUPFforinformationabout howtoobtaincopiesoftheworkforprintdistribution .Youmustattributetheworkinthe mannerspeci“edbytheauthororlicensor(butnotinanywaythatsuggeststhattheyendorse youoryouruseofthework).Foranyreuseordistribution,youmustmakecleartoothersthe licensetermsofthiswork.Anyoftheaboveconditionscanbewaivedifyougetpermissionfrom theUniversityPressofFlorida.Nothinginthislicenseimpairsorrestrictstheauthorsmoral rights. ISBN978-1-61610-160-2 OrangeGroveTexts Plus isanimprintoftheUniversityPressofFlorida,whichisthescholarly publishingagencyfortheStateUniversitySystemofFlorida,comprisingFloridaA&M University,FloridaAtlanticUniversity,FloridaGulfCoastUniversity,FloridaInternational University,FloridaStateUniversity,NewCollegeofFlorida,UniversityofCentralFlorida, UniversityofFlorida,UniversityofNorthFlorida,UniversityofSouthFlorida,andUniversityof WestFlorida. UniversityPressofFlorida 15Northwest15thStreet Gainesville,FL32611-2079 http://www.upf.com

PAGE 5

Contents Chapter1.Functions 1 1.Functions1 2.ClassesofFunctions4 3.OperationsonFunctions7 4.ViewingtheGraphsofFunctions10 5.InverseFunctions13 6.TheVelocityProblemandtheTangentProblem18 Chapter2.LimitsandDerivatives 23 7.TheLimitofaFunction23 8.LimitLaws31 9.ContinuousFunctions36 10.LimitsatIn“nity40 11.Derivatives45 12.TheDerivativeasaFunction48 Chapter3.RulesofDifferentiation 53 13.DerivativesofPolynomialandExponentialFunctions53 14.TheProductandQuotientRules57 15.DerivativesofTrigonometricFunctions59 16.TheChainRule62 17.ImplicitDierentiation66 18.DerivativesofLogarithmicFunctions68 19.ApplicationsofRatesofChange71 20.RelatedRates75 21.LinearApproximationsandDierentials81 Chapter4.ApplicationsofDifferentiation 89 22.MinimumandMaximumValues89 23.TheMeanValueTheorem96 24.TheFirstandSecondDerivativeTests104 25.TaylorPolynomialsandtheLocalBehaviorofaFunction111 26.LHospitalsRule118 27.AnalyzingtheShapeofaGraph123

PAGE 6

viCONTENTS 28.OptimizationProblems129 29.NewtonsMethod135 30.Antiderivatives142 Chapter5.Integration 149 31.AreasandDistances149 32.TheDe“niteIntegral156 33.TheFundamentalTheoremofCalculus167 34.Inde“niteIntegralsandtheNetChange172 35.TheSubstitutionRule177

PAGE 7

CHAPTER1 Functions 1.Functions A function f isarulethatassociatestoeachelement x inaset D a unique element f ( x )ofanotherset R .Heretheset D iscalledthe domain of f ,whiletheset R iscalledthe range of f .Thefactthat f associatestoeachelementof D anelementof R isrepresentedbythe symbol f : D R .Insteadofsayingthat f associates f ( x )to x ,we oftensaythat f sends x to f ( x ),whichisshorter.SeeFigure1.1for anillustration. DomainRangex a b y z f Figure1.1. Domainandrange. Ifthesetsmentionedinthepreviousde“nitionaresetsofnumbers, thenitisofteneasiertodescribe f byanalgebraicexpression.Let N bethesetofallnaturalnumbers(whicharethenonnegativeintegers). Thenthefunction f : N N givenbytherule f ( x )=2 x +3is thefunctionthatsendseachnonnegativeinteger n tothenonnegative integer2 n +3.Forinstance,itsends0to3,1to5,17to37,andsoon. Inthiscase,thealgebraicdescriptionissimplerthanactuallysaying  f isthefunctionthatsends n to2 n +3.Ž Therulethatdescribes f maybesimpleorcomplicated.Itcould bethatafunctionisde“nedbycasessuchas f ( x )= 0 1 x if0 x 40 4+0 15( x Š 40)if40 80. Thisexamplecoulddescribeanincometaxcode.The“rst$40,000 ofincomeistaxedatarateof10%,incomeabove$40,000butbelow$60,000istaxedatarateof15%,andincomeabove$80,000is1

PAGE 8

21.FUNCTIONS taxedatarateof20%.Thevalueof f ( x )istheamountoftaxto bepaidafteranincomeof x thousanddollarsforanypositivereal number x Therearetimeswhentherulesthatapplyinvariouscasesare closelyconnectedtoeachother.Aclassicexampleisthe absolutevalue function ,thatis, f ( x )= | x | = x if0 x, Š x if x< 0. 3 2 10123 x 0.5 1.0 1.5 2.0 2.5 3.0 y Figure1.2. Graphof | x | Inthiscase, f ( x )= f ( Š x )forall x .Whenthathappens,wesaythat f isan evenfunction .Forinstance, g ( x )=cos x and h ( x )= x2areeven functions.Therearealsofunctionsforwhich Š f ( x )= f ( Š x )holdsfor all x .Thenwesaythat f isan oddfunction .Examplesofoddfunctions include g ( x )=sin x and h ( x )= x3. TherearetimeswhenaplainEnglishdescriptionofafunctionis simplerthananalgebraicone.Forinstance,let g bethefunction thatsendseachintegerthatisatleast2intoitslargestprimedivisorŽ issimplerthandescribingthatfunctionwithalgebraicsymbols(and symbolsofformallogic).Ifthesets D and R arenotsetsofnumbers, analgebraicdescriptionmaynotevenbepossible.Anexampleofthis iswhen D and R arebothsetsofpeopleand f ( x )isthebiological fatherofperson x .Notethatitisnotbyaccidentthatwesaidthat f ( x )isthefather(andnottheson)of x .Indeed,afunctionmustsend x toa unique f ( x ).Whileapersonhasonlyonebiologicalfather,he orshemayhaveseveralsons. Sometimestherulethatsends x to f ( x )canonlybegivenbylisting thevalueof f ( x )foreach x ,asopposedtoageneralrule.Forinstance, let D bethesetof200speci“ccitiesintheUnitedStates,let R bethe setofallnonnegativerealnumbers,andforacity x ,let f ( x )bethe

PAGE 9

1.FUNCTIONS3 amountofprecipitationthat x hadin2009.Then f isafunctionsince itsendseach x D intoanelementof R .Thisfunctionisgivenbyits listofvalues,notbyarulethatwouldspecifyhowto compute f ( x )if given x Finally,functionscanalsoberepresentedbytheir graphs .If f : D R isafunction,thenletusconsideratwo-dimensionalcoordinatesystemsuchthatthehorizontalaxiscorrespondstoelements of D ,andtheverticalaxiscorrespondstoelementsof R .Thegraphof f isthesetofallpointswithcoordinates( x,f ( x ))suchthat x D .The requirementthat f ( x )isuniqueforeach x willensurethatnovertical lineintersectsthegraphof f morethanonce.Thisiscalledthe vertical linetest .1.1.Exercises.(1)Foreachperson x ,let f ( x )denotethebirthday(day,month, andyear)of x .Is f afunction? (2)Foreachperson y ,let g ( y )denotethebiologicalmotherof y Is g afunction?Ifyes,whatisthedomainof g andwhatis therangeof g ? (3)Fortwopeople x and y ,letussaythat f ( x )= y if y isachild of x .Is f afunction? Fortheremainingexercisesinthissection,allfunctionsare de“nedonsome realnumbers (4)Let f ( x )= x + | x | .Findthedomainandtherangeof f (5)Let f ( x )=( x +1) / ( x Š 2).Findthedomainandtherange of f (6)Let g ( x )= x/ | x | .Findthedomainandtherangeof f (7)Let h ( x )=x x +3+x +3 x.Findthedomainandtherangeof f (8)Canthegraphofafunctionintersectaverticallinetwice? (9)Canthegraphofafunctionintersectahorizontalline twice? (10)Anin“nite sequence isanin“nitearrayofnumbers a1,a2,... Explainwhyin“nitesequencesare,infact,functions.What isthedomainofthesefunctions? (11)Let f ( x )=3 x +2.Findfourpointsthatareonthegraph of f .Whatcanbesaidaboutthecurvedeterminedbythose fourpoints? (12)Let f and g betwofunctionsandletusassumethatthereis exactlyonepoint( x,y )thatisonthegraphofboth f and g Whatisthealgebraicmeaningofthatfact?

PAGE 10

41.FUNCTIONS 2.ClassesofFunctions2.1.PowerFunctions.A powerfunction isafunction f givenbythe rule f ( x )= xa,where a isa“xedrealnumber.Notethat xŠ a=1 /xa, so,forinstance, xŠ 3=1 /x3.Thespecialcaseof a = Š 1,thatis,the function f ( x )=1 /x ,iscalledthe reciprocal function .Notethatthe rule g ( x )=1forallrealnumbers x alsode“nesapowerfunction,one inwhich a =0.If a =1 /n ,where n isapositiveinteger,thenthe powerfunction f givenbytherule f ( x )= xa= x1 /n=n a isalsocalleda rootfunction .2.2.Polynomials.A polynomialfunction isthesumofa“nitenumber ofconstantmultiplesofpowerfunctionswithnonnegativeintegerexponents,suchasthefunction f givenbytherule f ( x )=3 x4+2 x2+ 7 x Š 5.Thedomainofthesefunctionsisthesetofallrealnumbers. Thelargestexponentthatispresentinapolynomialfunctioniscalled the degree ofthepolynomial.Sothedegreeof f inthelastexampleis 4.Therealnumbersthatmultiplythepowerfunctionsinapolynomial arecalledthe coecients ofthepolynomial.Inthelastexample,they are3,2,7,and Š 5. Somesubclassesofpolynomialfunctionshavetheirownnamesas follows: € Polynomialsofdegree0,suchas f ( x )=6,arecalled constant functions. € Polynomialsofdegree1,suchas g ( x )=3 x Š 2,arecalled linearfunctions. € Polynomialsofdegree2,suchas h ( x )= x2Š 4 x Š 21,are called quadraticfunctions. € Polynomialsofdegree3,suchas p ( x )= x3Š x2+6 x Š 2,are called cubicfunctions.2.3.RationalFunctions.A rationalfunction istheratiooftwopolynomialfunctionssuchas R ( x )= 3 x2+4 x Š 7 x3Š 8 Thedomainofarationalfunctionisthesetofallrealnumbers,except forthenumbersthatmakethepolynomialinthedenominator0.In theprecedingexample,theonlysuchnumberis x =2.

PAGE 11

2.CLASSESOFFUNCTIONS5 2.4.TrigonometricFunctionsPeriodicity.Thereaderhassurelyencounteredthetrigonometricfunctionssin,cos,tan,cot,sec,andcscinearliercourses.Wewilldiscussthesefunctions,andtheirinverses,later inthetext.Fornow,wementiononeoftheirinterestingproperties, their periodicity .Afunction f iscalled periodic withperiod T> 0if f ( x )= f ( x + T )forall x and T isthesmallestpositiverealnumber withthisproperty. Forexample,sinandcosarebothperiodicwithperiod2 ,andtan andcotareperiodicwithperiod .SeeFigure1.3foranillustration. ThereaderwillbeaskedinExercise2.7.1abouttheperiodicityofsec andcsc. 23222322 1 0.5 0.5 1sin x 2322232210.5 0.5 1cos x 232223221 csc x 232223221 sec x 23222322 1 1 tan x 23222322 1 1 cot x Figure1.3. Trigonometricfunctions.2.5.AlgebraicFunctions.An algebraicfunction isafunctionthatcontainsonlyaddition,subtraction,multiplication,division,andtaking roots.Forinstance,powerfunctionswithintegerexponentsarealgebraicfunctions,sincetheyonlyusemultiplication,thoughpossibly manytimes.Therefore,polynomialsarealgebraicfunctionsaswellsince theyaresumsofconstantmultiplesofpowerfunctions.Thisimplies

PAGE 12

61.FUNCTIONS thatrationalfunctionsarealsoalgebraicsincetheyareobtainedby dividingapolynomial(alsoanalgebraicfunction)byanotherone. Theprecedinglistdidnotcontainallalgebraicfunctionssinceit didnotcontainanyfunctionsinwhichrootswereinvolved.Soweget additionalexamplesifweincluderoots,suchasthefunctionsgivenby therules f ( x )= x +3, g ( x )=3 x h ( x )= ( x +1) / ( x Š 1).2.6.TranscendentalFunctions.Functionsthatarenotalgebraicare called transcendentalfunctions .Theseincludetrigonometricfunctions andtheirinverses,exponentialfunctions,whicharefunctionsthatcontainavariableintheexponent,suchas f ( x )=2x,andtheirinverses, whicharecalled logarithmicfunctions .SeeFigure1.4foranillustration.Wewilldiscussthesefunctionsinlatersectionsofthischapter. Therearemanyadditionalexamples,whichdonothavetheirown names. 1 12 23 34 45 5 x 1 1 2 2 3 3 4 4 5 5 y x log2( x ) 2x Figure1.4. Logarithmicfunctions.2.7.Exercises.(1)Aresecantandcosecantperiodicfunctions?Ifyes,whatis theirperiod? (2)Are f ( x )=3 x5+7 x Š 31and g ( x )=(2 x +7) / (3 x Š 1) polynomialfunctions? (3)Are f ( x )=2xand g ( x )=sin2x powerfunctions? (4)Are1 / ( x +3), g ( x )=( x2+3 x +9) / ( x3+1),and h ( x )= (sin x ) / ( x +2)rationalfunctions?

PAGE 13

3.OPERATIONSONFUNCTIONS7 (5)Let f ( x )= x2 / 3.Is f analgebraicfunction? (6)Issin(3 x )aperiodicfunction?Ifyes,whatisitsperiod? (7)Showanexampleofaperiodicfunctionthathasperiod1. (8)Let f ( x )= xŠ 2 / 7.Is f analgebraicfunction? (9)Is g ( x )=(2 / 3)xanalgebraicfunction? (10)Showanexampleofaperiodicfunctionwithperiod (11)Issin x +tan x aperiodicfunction?Ifyes,whatisitsperiod? (12)Issin2x aperiodicfunction?Ifyes,whatisitsperiod?3.OperationsonFunctions3.1.TransformationsofaFunction.Wehaveseenthebasicmathematicalfunctionsandtheirgraphsinthelastsection.Inthissection,we willlookattheirtransformations. Itiseasytoseewhathappenstothegraphofafunctionifwe increaseordecreaseeachvalueofafunctionbyaconstant.Indeed,the graphofthefunction g givenby g ( x )= f ( x )+5forall x issimplythe graphofthefunction f translatedby“veunitstothenorth.Similarly, thegraphofthefunction h givenby h ( x )= f ( x ) Š 7isthegraphof f translatedbysevenunitstothesouth. Horizontaltranslationsarealittlebittrickier.Thereaderisinvited toverifythatif g isthefunctiongivenby g ( x )= f ( x Š 2),thenthe graphof g isthegraphof f translatedbytwounitstotheeast,thatis, inthepositivedirection.Indeed,wemustsubstitutea larger number into g togetthesamevalueasfrom f .Forinstance, g (8)= f (6). SeeFigure1.5foranillustration.Similarly,if h isthefunctiongiven by h ( x )= f ( x +3)forall x ,thenthegraphof h isthegraphof f translatedbythreeunitstothewest,thatis,inthenegativedirection. 2 101234 x 1 2 3 4 5 6 7 8 9 10 y f ( x 2) f ( x 5) f ( x ) Figure1.5. Horizontalandverticaltranslationsof f ( x ).

PAGE 14

81.FUNCTIONS Theinteractivewebsitehttp://www.math.u”.edu/ mathguy/ufcalc book/translations.htmlprovidesfurthertoolstovisualizetransformationsoffunctions. Theeectofmultiplicationanddivisiononfunctionscanbedescribedsimilarly.If f isafunctionand g isthefunctiongivenby g ( x )= c f ( x ),where c> 1isarealnumber,thenthegraphof g issimplythegraphof f stretchedŽverticallybyafactorof c .That is,eachpointonthegraphof g is c timesasfar away fromthehorizontalaxisasthecorrespondingpointonthegraphof f .Itgoes withoutsayingthatdividingby c> 1hastheoppositeeect.Thatis, if h ( x )= f ( x ) /c ,thenthegraphof h isaverticallycompressedversion ofthegraphof f .Inotherwords,eachpointonthegraphof h is c timesasclosetothehorizontalaxisasthecorrespondingpointonthe graphof f .SeeFigure1.6foranillustration. Atthispoint,thereadershouldstopandthinkaboutwhathappensif c< Š 1isanegativeconstant.Asthereaderprobably“gured out,thestretchingorcompressingeectwillnotchange(itwillonly dependon | c | ),buteachpointonthegraphwillbere”ectedthrough thehorizontalaxis. 12345 x 2 1 1 2 3 y f ( x ) 3 f ( x ) 1 2 f ( x ) f ( x ) Figure1.6. Eectsofmultiplyingafunctionbyaconstant. Thereaderisencouragedtoconsulttheinteractivewebsitehttp:// www.math.u”.edu/ mathguy/ufcalcbook/squeeze.htmlforfurther illustrations. Horizontaltransformationsinvolvingmultiplicationanddivisionare similartotheircounterpartsinvolvingadditionandsubtractioninthat theireectistheoppositeofwhatonemightthinkat“rst.If c> 1 and g isthefunctionobtainedfrom f bytherule g ( x )= f ( cx ),then thegraphof g isthegraphof f compressedhorizontallybyafactor

PAGE 15

3.OPERATIONSONFUNCTIONS9 of c .Thatis,eachpointonthegraphof g is c timesasclosetothe verticallineasthecorrespondingpointonthegraphof f .Inother words,if( x,y )isapointonthegraphof f ,then( x/c,y )isapointon thegraphof g .Ontheotherhand,if h isobtainedfrom g bytherule h ( x )= f ( x/c ),thenthegraphof h isahorizontallystretchedversion ofthegraphof f .Thatis,eachpointonthegraphof h is c timesas farfromtheverticalaxisasthecorrespondingpointonthegraphof f .Soif( x,y )isapointonthegraphof f ,then( cx,y )isapointon thegraphof h .Again,thereadershouldstopforaminuteandthink aboutthegraphsofthefunctions f ( cx )and f ( x/c )when c< Š 1isa negativeconstant.3.2.CombiningTwoFunctions.If f and g aretwofunctions,thentheir sum,dierence,andproductarede“nedwhereverboth f and g are de“ned.Thatis,thedomainof f + g f Š g ,and fg istheintersection ofthedomainsof f and g .Furthermore,( f + g )( x )= f ( x )+ g ( x ), ( f Š g )( x )= f ( x ) Š g ( x ),and( fg )( x )= f ( x ) g ( x ).Figure1.7illustrates thesumoftwofunctions.Wehavetobejustalittlebitmorecareful with f/g ,sincethisfunctionisnotde“nedwhen g ( x )=0,evenif x isinthedomainofboth f and g .Sothedomainof f/g isthe intersectionofthedomainof f andthedomainof g ,withtheexception ofthepoints x satisfying g ( x )=0.Foreachpointofthisdomain, ( f/g )( x )= f( x ) /g ( x ). 12345 x 1 1 2 3 y f ( x ) g ( x ) g ( x ) f ( x ) Figure1.7. Addingtwofunctionstogether. Iftherangeof f ispartofthedomainof g ,thenwecan compose f and g by“rstapplying f andthen g .Thefunctionweobtaininthis waysends x to g ( f ( x ))andiscalledthe composition of f and g .Itis denotedby g f .Notethatin g f ,“rst f ,andthen g isapplied.

PAGE 16

101.FUNCTIONS Example 1.1 Let R bethesetofallrealnumbers.If f and g are bothfunctionsfrom R to R and f ( x )= x2and g ( x )= x +1 ,then ( g f )( x )= g ( f ( x ))= x2+1 while ( f g )( x )= f ( g ( x ))=( x +1)2= x2+2 x +1 Notethat f g and g f are,ingeneral,dierentfunctions.3.3.Exercises.(1)Sketchthegraphof f ( x )= x2, g ( x )=( x Š 3)2,and h ( x )= (2 x +5)2. (2)Sketchthegraphof f ( x )=( x +4)2and g ( x )= x2+4. (3)Sketchthegraphof f ( x )= | x +5 | and g ( x )= | x | +5. (4)Sketchthegraphof f ( x )=sin( x/ 2)and g ( x )=(sin x ) / 2. (5)Sketchthegraphof f ( x )= | sin x | (6)Sketchthegraphof f ( x )= x and g ( x )=1 / x (7)Sketchthegraphof f ( x )= x +10and g ( x )= x +10. (8)Sketchthegraphof f ( x )=cos(2 x ), g ( x )=sin( x Š 2),and h ( x )=3tan x (9)Showexamplesfor f and g when g f isde“nedforallreal numbers,but f g isnot. (10)Showexampleswhen f g = g f (11)Sketchthegraphof g ( x )=sin x Š 4 (12)Let f ( x )=sin x and g ( x )= x2.Determine f g and g f and sketchtheirgraph.4.ViewingtheGraphsofFunctions The graph ofafunction f istheset { ( x,f ( x )) | x D ( f ) } .Itis agoodwayofvisuallydescribingwhatafunctiondoes.Today,we haveplentyofadvancedtools,suchascomputersoftwarepackagesand graphingcalculators,tostudythegraphoffunctions.Inthissection, wepointoutafewofthecommonmistakesinusingthesetools. Inordertofacilitatethediscussion,letusagreeonsometerminology.Ifthedomainof f containsaninterval I andforallrealnumbers x and xin I ,itistruethat xf ( x),thenwesaythat f is decreasing on I .Intermsofthe graphof f ,thismeansthatthegraphgoesroughlyfromthenorthwest tothesoutheast.

PAGE 17

4.VIEWINGTHEGRAPHSOFFUNCTIONS11 Ifwesimplyaskacomputerorgraphingcalculatortoplotthegraph ofafunctionwithoutspecifyingtheinterval[ x1,x2]inwhichthevalue of x canrange,wemaygetanerrormessage,orthecomputermay simplysubstitutedefaultvaluesfor x1and x2.Forexample,thesoftwarepackageMaple13usesthedefaultvalues x1= Š 10and x2=10. Theinterval[ x1,x2]isoftencalledthe viewingwindow .SeeFigure1.8 foranillustration. 10 5510 2000 2000 4000 0.2 0.2 0.4 0.6 0.8 1.0 0.2 0.4 0.6 Figure1.8. Viewing g ( x )=4 x3+9 x2+6 x +1with viewingwindow[ Š 10,10]and[ Š 1,0]. Wehavetobecareful,however,sincenotallviewingwindowsare appropriateforallfunctions,andchoosinganinappropriateviewing windowmaycausemisleadingresults. Forfunctionslike f ( x )= x g ( x )= | x | ,or h ( x )= x2+3,theviewingwindow[ Š 10 10]isappropriateasthebehaviorofthesefunctions outsidethatwindowissimilartotheirbehaviorinsidethewindow. Nowlet f ( x )=( x +10)2.Inthiscase,usingtheviewingwindow [ Š 10 10],wegetthegraphofanincreasingfunction.Thatismisleading since f isdecreasingontheinterval( Š Š 10].So,inthiscase,a viewingwindowthatstartsatapoint x1< Š 10isnecessary. Thisproblembecomesmoredicultifwearedealingwithfunctions thatchangefromincreasingtodecreasingmanytimes,perhapsinan irregularfashionandperhapsfar away fromtheorigin.Forthisreason, itisworthnotingthatif f isa polynomialfunction ofdegree n ,thenit cannotchangedirectionsmorethan n Š 1times.Ifwefoundall n Š 1 directionchanges,thenwecanbesurethatwedidnotmissanyof them.Wewillreturntothistopicinalaterchapter,whenwediscuss the derivative ofafunction. Theprecedingexampleshowedwhyselectingaviewingwindow thatistoosmallcanbemisleading.Thenextexampleshowswhya viewingwindowthatistoolargecanalsomisleadus.Plotthegraph ofthefunction g ( x )=4 x3+9 x2+6 x +1.Usingthedefaultviewing window[ Š 10 10],orsomewindowcontainingthatone,manysoftware

PAGE 18

121.FUNCTIONS packageswillshowagraphthatincreaseseverywhereanddisappearsin asmallintervaltotheleftof0.Thisshouldraiseoursuspicionthatthe programdoesnotproperlydisplaythegraphof g around0.Indeed, g isde“nedforallrealnumbers,soitsgraphshouldnotdisappear anywhere.Takingacloserlook,thatis,changingtheviewingwindow to[ Š 1 1],weseeafunctionthatisactually decreasing between x = Š 1 and x = Š 1 / 2. Trigonometricfunctions,withtheirperiodicity,areparticularly goodexamplestodemonstratewhatsoftwarepackagescanandcannotdo.Thereaderisencouragedtoplotthegraphofthefunctions sin x ,cos(2 x ),tan( x/ 4),and,“nally,sin(1 /x )andexplaintheobtained graphs.Inparticular,thereadershouldtrytoexplainwhy,forsin(1 /x ), thechoiceoftheviewingwindowisnotimportantaslongasitcontains x =0. Applicationsofgraphicalrepresentationsoffunctionsincludecountingthesolutionsofcertainequationsevenwhenwecannotexplicitly solvethoseequations,and“ndingasymptotes.A horizontalasymptote ofafunction f isahorizontalline y = a sothatthevaluesof f ( x )are neverequalto a ,butgetcloserandclosertoitas x getscloserand closertopositivein“nityornegativein“nity.A verticalasymptote of f isaline x = b sothatthefunction f isnotde“nedat x = b ,butas x getscloserandcloserto b ,thevaluesof f ( x )getcloserandcloserto in“nity,ornegativein“nity.Forinstance,thefunction f ( x )=1 /x has ahorizontalasymptoteat y =0,andaverticalasymptoteat x =0. Wewillmakethesenotionsmorepreciseinthenextchapter,when weintroducetheconceptof limits .Fornow,wecanuseagraphing softwarepackageto“ndasymptotes,asyouwillbeaskedtodointhe exercises.4.1.Exercises.Inthefollowingexercises,useagraphingsoftware packagewiththeappropriateviewingwindowto“ndthenumberof solutions(amongrealnumbers)forthegivenequation.Also“ndthe intervalsonwhichtheleft-handsideisincreasingandonwhichthelefthandsideisdecreasing.Approximatetheendpointsoftheseintervals toonedecimal. (1) x4Š x +1=0. (2) x4Š 1=0. (3) x3Š 6 x +1=0. Inthefollowingexercises,useagraphingsoftwarepackagewiththe appropriateviewingwindowto“ndthenumberofsolutions(among realnumbers)forthegivenequation.

PAGE 19

5.INVERSEFUNCTIONS13 (4) x3Š x2Š 1=0. (5) x2Š x Š 7= x3Š 1. (6) x2=sin x (7) x/ 2=cos x Inthefollowingexercises,useagraphingsoftwarepackagetodecideif thegivenfunctionhasaverticalorhorizontalasymptote. (8) f ( x )=( x +3) / ( x +2). (9) g ( x )=1 / (2 Š x ). (10) h ( x )= x +1 /x (11) s ( x )= ( x Š 4) / ( x Š 3). (12) z ( x )=( x2+1) / (2 x2Š 3).5.InverseFunctions Theinverse fŠ 1ofafunction f : A B undoesŽwhat f did. Thatis,if f ( x )= y ,then fŠ 1( y )= x ,so f sends x to y ,while fŠ 1sends y backto x .Itgoeswithoutsayingthatthis fŠ 1willonlybe afunctionif fŠ 1( y )isunambiguous,thatis,whenthereisonlyone x A sothat f ( x )= y .Inthatcase,andonlyinthatcase,itisclear that fŠ 1( y )= x Letusnowformalizetheseconcepts. Definition 1.1 Afunction f : A B iscalled one-to-one if itsendsdierentelementsintodierentelements,thatis,if x = ximpliesthat f ( x ) = f ( x) One-to-onefunctionsarealsocalled injectivefunctions or injections Visually,nohorizontallinecanintersectthegraphofaone-to-one functionmorethanonce. Forinstance,if A and B areboththesetofrealnumbers,then f ( x )= x and g ( x )= x3arebothone-to-one,but h ( x )= x2isnot. Definition 1.2 Let f beaone-to-onefunctionwithdomain A and range B .Thenthe inverse of f isthefunction fŠ 1: B A givenby fŠ 1( y )= x if f ( x )= y Example 1.2 Let A and B bothbethesetofallrealnumbers.Let f : A B begivenby f ( x )=2 x +7 .Then fŠ 1( y )=( y Š 7) / 2 Solution: If f ( x )= y ,then y =2 x +7,so y Š 7=2 x and( y Š 7) / 2= x .As x = fŠ 1( y ),itfollowsthat fŠ 1( y )=( y Š 7) / 2. Theprecedingexampleshowsageneralstrategyfor“ndingtheinverse ofafunction.Writetheequation f ( x )= y ,withtheappropriate algebraicexpressionreplacing f ( x ).Thensolvefor x .Ifthereismore

PAGE 20

141.FUNCTIONS thanonesolution,then f isnotone-to-one,andsoithasnoinverse function.Ifthereisonesolution,thenthatexpressionisthevalueof fŠ 1( y ). Example 1.3 If A isthesetofpositiverealnumbers, B isthe setofrealnumbersthatarelargerthan 1 ,and f : A B isgivenby f ( x )= x2+1 ,then fŠ 1( y )= y Š 1 Solution: Wehave f ( x )= x2+1= y .So x2= y Š 1,andbecausewe knowthat x ispositiveand y> 1,wecantakethesquarerootofboth sides,leadingto x = y Š 1.Hence, fŠ 1( y )= y Š 1. Notethatthegraphsof f and fŠ 1arere”ectedimagesofeachother throughtheline y = x asillustratedinFigure1.9. Finally,wepointoutthatif f isaone-to-onefunctionwithdomain A andrange B ,then fŠ 1 f istheidentityfunctionof A and f fŠ 1istheidentityfunctionof B 0123 x 1 2 3 y x f 1( x ) f ( x ) Figure1.9. f ( x )and fŠ 1( x )aresymmetricaboutthe identityfunction x Forinstance,usingthefunctionsofExample1.3,forallpositive realnumbers x ,theidentity( fŠ 1 f )( x )= ( x2+1) Š 1= x2= x holds,andforall y> 1,theidentity( f fŠ 1)( y )=( y Š 1)2+1= y Š 1+1= y holds.5.1.LogarithmicFunctions.Ifafunctioncontainsonlyadditions,subtractions,multiplications,anddivisions,thenitsinverseisofteneasy tocompute. Power functions,thatis,functionsoftheform f ( x )= x, where isarealnumber,arenotmuchmoredicult.However,what istheinverseofan exponentialfunction ?

PAGE 21

5.INVERSEFUNCTIONS15 Let f ( x )=2x.Itiseasytosee,byplottingthegraphof f or otherwise,that f isaone-to-onefunctionwhosedomainisthesetof allrealnumbersandwhoserangeisthesetofallpositiverealnumbers. Sotheinverseof f isafunctionfromthesetofpositiverealstotheset ofallreals.But what isthatinversefunction fŠ 1?Bythede“nition ofinversefunctionsingeneral,thisisthefunctionthatsends2xto x forallpositiverealnumbers2x.Inparticular, fŠ 1(2)=1, fŠ 1(4)=2, fŠ 1(32)=5,and fŠ 1(1 / 2)= Š 1.Thatis, fŠ 1( y )tellsus towhatpower wehavetoraise2 iftheresultistobe y .Thisimportantconcepthas itsownname. Definition 1.3 Let m beapositiverealnumber.Thentheinverse ofthefunction f ( x )= mxiscalledthe logarithmicfunctionwithbase m ,andisdenotedby logm. Soif f ( x )= xm= y ,thenlogm( y )= x .Forinstance,log2(64)=6, log3(81)=4,log5(1 / 25)= Š 2,andlog0 5(16)= Š 4. Logarithmicfunctionssatisfycertainrulesthatareverysimilarto thosesatis“edbyexponentialfunctionsandcan,infact,bededuced fromthem.Theseare (I)log( xy )=log x +log y (II)log( x/y )=log x Š log y (III)log( xa)= a log x (IV)logb x =log x b. (V) alogax= x (VI)loga( ax)= x Thelasttworulessimplyexpressthefactthatthefunctions f ( x )= axand fŠ 1( y )=loga( y )areinversesofeachother,sotheircomposition isanidentityfunction. Ifweknowthelogarithmofanumberinabaseandwanttocompute itinanotherbase,wecandosousingthefollowingtheorem. Theorem 1.1 Forpositiverealnumbers a b ,and x ,wehave logax = logbx logba Proof. Startwiththeidentity x = alogax. Nowtakethelogarithmofbase b ofbothsidestoget logbx =logax logba. Nowdividebothsidesbylogba togettheidentityofthetheorem.

PAGE 22

161.FUNCTIONS Example 1.4 WecanuseTheorem1.1tocompute log16(256) from log2(256) asfollows: log16(256)= log2(256) log2(16) = 8 4 =2 Soifacalculatororcomputercanprovidethelogarithmofall positiverealnumbersin one base,wecancomputethelogarithmofany positiverealnumberinanybase.Forthisreason,manycalculatorsand computersareprogrammedtoworkprimarilywithlogarithmsofone givenbase,namelyofbase e ,where e 2 718isanirrationalnumber thatwillbeformallyde“nedinChapter2. Thelogarithmofbase e issoimportantthatithasitsownname, naturallogarithm ,anditsownnotation,ln.Soln x =logex .5.2.InversesofTrigonometricFunctions.Basictrigonometricfunctions, suchassin,cos,andtan,areveryimportantincalculus,soitisnosurprisethattheirinversefunctionsareimportantaswell.However,we havetobeprecisewhenwede“nethemsincetrigonometricfunctions are not one-to-one.Infact,theyareperiodical,ofperiod2 or ,and sotheytakeeveryvalueintheirrangein“nitelyoften. Inordertogetaroundthisdiculty,wewillrestrictourtrigonometricfunctionstojustashortinterval,inwhichtheyareone-to-one, andde“netheirinversesbasedonthatrestriction. Forinstance,considersinasafunctionwhosedomainis[ Š / 2 ,/ 2]. Inthatinterval,sinisaone-to-onefunction(sinceitisincreasing), anditsrangeistheinterval[ Š 1 1].SeeFigure1.10foranillustration. Sotheinverseofsin:[ Š / 2 ,/ 2] [ Š 1 1]isthefunctionsinŠ 1: [ Š 1 1] [ Š / 2 ,/ 2].Thatis,if y [ Š 1 1],thensinŠ 1y isthe(only) x [ Š / 2 ,/ 2]forwhichsin x = y .Forinstance,sinŠ 1(1 / 2)= / 6, whilesinŠ 1(0)=0andsinŠ 1( 2 / 2)= / 4.Figure1.11showsthe graphofsinŠ 1x 2 32 22322 x 1 1 y Figure1.10. sin x isone-to-oneontheinterval[ Š / 2 ,/ 2].

PAGE 23

5.INVERSEFUNCTIONS17 1 1 x 2 2 y Figure1.11. GraphofsinŠ 1x Theinversesoftheothertrigonometricfunctionsarede“nedsimilarly,justtheintervalstowhichwerestrictthefunctions(inorderto makethemone-to-one)canchange. Thatis,cosŠ 1istheinversefunctionofthecosfunctionthatis restrictedtotheinterval[0 ].SocosŠ 1isafunctionwithdomain [ Š 1 1]andrange[0 ].Similarly,tanŠ 1istheinversefunctionofthe tanfunctionthatisrestrictedtotheinterval( Š / 2 ,/ 2).Itsdomain isthesetofallrealnumbers,anditsrangeistheinterval( Š / 2 ,/ 2). SeeFigure1.12forillustrations. Theinversefunctionsofcot,sec,andcsc,whilenotusedoften,can alsobede“nedanalogously.5.3.Exercises.(1)Isthereafunction f de“nedonallpositiverealnumbersfor which fŠ 1= f ? (2)Ifwearegivenlogax ,howcanwecomputelog1 /ax ? (3)Forwhichvaluesof a islogaanincreasingfunction,andfor whichvaluesof a isitadecreasingfunction? (4)Whatisthegeometricconnectionbetweenthegraphsof f and fŠ 1? (5)Isittruethatif g istheinversefunctionoftheone-to-one function f ,then g isone-to-one?

PAGE 24

181.FUNCTIONS 2 2 x 1 1 ycos x 1 1 x 2 ycos1x 22 x 5 10 5 10 ytan x 5 5 x 22ytan1x Figure1.12. Graphsofcos x andtan x withtheirinverses. (6)Let f : R R bede“nedby f ( x )= | x | .Is f aone-to-one function? (7)Let f : R R bede“nedby f ( x )= x5.Is f aone-to-one function? (8)Let f : R+ R+bede“nedby f ( x )= x2.Is f aone-to-one function? (9)Is f ( x )=logax aone-to-onefunctiononthesetofallpositive realnumbers? (10)Express x intermsof y ifloga(logax )= y (11)Letusassumethat f : R R isastrictlyincreasingfunction, thatis,if x
PAGE 25

6.THEVELOCITYPROBLEMANDTHETANGENTPROBLEM19 theaveragespeedofthecarbytheformula (1.1) v = s t where t isthetimepassed, s isthedistancecoveredintime t ,and v istheaveragespeedforthegiventimeperiod.Inphysics,whenthe directioninwhichanobjectismovingistakenintoaccount,wetalk about velocity insteadof speed ,hencetheabbreviation v .Inthegiven example,alltravelwasinonedirection(west),sothereisnodanger ofconfusion,andwecanuseeitherword.Letusassumethattimeis measuredinhoursanddistanceismeasureinmiles. ThenEquation(1.1)yields v = 100mi 2hr =50 mi hr sotheaveragevelocityofthecarforthegiventwo-hourperiodis50 milesperhour. Thecarprobablydidnotcovertheentiredistanceatitsaverage velocity.Forvarioustrac-relatedorotherreasons,itsometimesmay havegonefasterorslower.Ifwewanttoknowitsaveragevelocity forthetimeperiodbetween4:00p.m.and4:10p.m.,thenweneed knowthedistanceitcoveredinthattimeperiod.Ifthatdistanceis 10miles,thenweconcludethatinthat10-minutetimeperiod,the averagevelocityofthecarwas v = 10mi 1 / 6hr =60 mi hr Ifwewantmorepreciseinformation,liketheaveragevelocityof thecarbetween4:02p.m.and4:05p.m.,wecanproceedsimilarly, decreasingthevalueofboththenumeratorandthedenominatorof thefraction s/t .However,whatifwewanttoknowthe instantaneous velocity ofthecarinagiven moment ,suchasexactlyat4:02:23p.m. (andnotinthesecondthatpassedbetween4:02:23p.m.and4:02:24 p.m.)?Inthatcase,adirectapplicationofEquation(1.1)isimpossible, becausethedenominator t isequalto0.Thenumerator s isalsoequal to0,sincethecarneedstimetocoveranydistance;ifitisgivenno time,itwillcovernodistance. Inthissection,wewillnotgiveacompletelyformalanswertothe problemofde“ninginstantaneousvelocity;wewillleavethattaskto anupcomingsection.However,wewillsaythefollowing.Theinstantaneousvelocityofacarinagivenmoment m canbeapproximatedby choosingsmallerandsmallertimeperiodscontaining m andcomputing

PAGE 26

201.FUNCTIONS theaveragespeedofthecarforthosetimeperiods.Theseaverageswill approximatetheinstantaneousvelocity.6.2.TheTangentProblem.Theproblemof“ndingtheinstantaneous velocityofamovingobjectissimplyaspecialcaseofamuchmore generalproblem,thatof“ndingtheslopeofatangentlinetoacurve atagivenpoint. Inthepreviousproblem,thedistancethecarcoveredcanbeviewed asafunctionofthetimethatpassedsincethecarstartedmoving.So s ( t )isthedistancecoveredfromthemomentwhenthecarstarted movingtothemoment t hourslater.Inordertocomputetheaverage velocityforthetimeperiodfrom t1to t2,wesimplycomputethevalue ofthefraction s ( t2) Š s ( t1) t2Š t1. Thisfractionispreciselytheslopeofthelinethatintersectsthegraph ofthefunction s atpoints( t1,s ( t1))and( t2,s ( t2)).Ifwechoose t1and t2closerandclosertogether,thenthesepointswillgetcloserand closertogetheraswell.Finally,ifweset t1= t2,thenwewillnot immediatelyknowtheslopeofthelinethattouchesthegraphof s at thepoint( t1,s ( t1))sincewewillknowonlyone,nottwo,pointofthis line.However,andthiswillbemademorepreciseinthenextsection, theslopewearelookingforwillbeapproximatedbythesequenceof slopesofthelinesthatwegotwhenwechose t1and t2closerandcloser together. Finally,wepointoutthatthereisnothingmagicalaboutthefunction s ( t )here.Wecouldconsideranyfunction f : R R ,andask whattheslopeofthetangentlinetothiscurveisatthepoint( x,f ( x )).6.3.Exercises.(1)Acartravelsonehourataspeedof60milesperhour,then twohoursataspeedof45milesperhour.Whatistheaverage speedofthecarduringthisthree-hourperiod? (2)Considerthecarofthepreviousexercise.Whatisitsaverage speedduringthe“rsttwohoursofitstrip? (3)Idroveat40milesperhourfortwohours.HowfastdoIhave todriveinmythirdhourifIwanttoreachanaveragespeed of45milesperhourformythree-hourdrive? (4)Acartravels300milesonagivenday.Duringthe“rst100 miles,thecartravelsataspeedof40milesperhour,during thesecond100miles,ittravelsataspeedof50milesperhour,

PAGE 27

6.THEVELOCITYPROBLEMANDTHETANGENTPROBLEM21 andduringthethird100miles,ittravelsataspeedof60miles perhour.Whatistheaveragespeedofthecarfortheentire 300-miletrip? (5)Timhasriddenhisbicycletoschool,coveringa5-miledistance inhalfanhour.Canweconcludethattherewasasegmentof hisrideforwhichhisaveragespeedwasmorethan10miles perhour? (6)Considerthefunction f ( x )= x2.Canyou“ndtwopoints P and Q onthegraphof f suchthattheslopeoftheline PQ is between0and0.01? (7)Let f ( x )= x andlet P =(1 1).Findtheslopeofthe threelinesthatconnect P tothepoints(4 2),(2 25 1 5),and (1 44 1 2). (8)Let f beasinthepreviousexercise.Findtheslopeofthethree linesconnecting P =(1 1)tothepoints(0 25 0 5),(0 64 0 8), and(0 81 0 9). (9)Let g ( x )= exandlet P =(0 1).Findtheslopeofthethree linesconnecting P tothepoints( Š 1 ,eŠ 1),(1 ,e ),and(ln2 2). (10)Considerthefunction f ( x )= x2.Let P =(1 1).Canyou “ndapoint Q onthegraphof g suchthattheslopeoftheline PQ is2? (11)Considerthefunction g ( x )= x3.Let P =(1 1).Canyou“nd apoint Q onthegraphof g suchthattheslopeoftheline PQ isbetween1and1.01? (12)Considerthefunction f ( x )=1 /x .Choosetwopoints P and Q ofthegraphof f suchthat P = Q andthe x coordinatesof P and Q aresmallandpositive.Whatcanbesaidaboutthe slopeoftheline PQ ?

PAGE 29

CHAPTER2 LimitsandDerivatives 7.TheLimitofaFunction7.1.Two-SidedLimits.Considerthefunctiongivenbytherule f ( x )= 1 / (1+ x ).Letuscomputethevaluesof f ( x )forvariousrealnumbers x thatarecloseto0.We“ndthat € f (1)=1 / 2, € f (1 / 2)=2 / 3, € f (1 / 3)=3 / 4,and,ingeneral, € f (1 /n )= n/ ( n +1). Similarly,fornegativevaluesof x ,weget € f ( Š 1 / 2)=2, € f ( Š 1 / 3)=3 / 2, € f ( Š 1 / 4)=4 / 3,and,ingeneral, € f ( Š 1 /n )= n/ ( n Š 1). Whatweseeisthatif x getscloseto0(fromeitherside),then f ( x )getscloseto f (0)=1.Infact,wecanget f ( x )tobeasclose to f (0)=1aswewant;allweneedtodoistochoose xsuciently closeto0.Indeed,lookingatthepreviousexamples,weconcludethat if0
PAGE 30

242.LIMITSANDDERIVATIVES So,if f isthestartingexampleofthissection,thenlimx 0f ( x )=1. Notethatthede“nitionoflimx af ( x )requiresthat f ( x )stayclose to L when x iscloseto a regardlessofwhichof x or a islarger .That is, f ( x )hastobecloseto L if x isalittlebitlessthan a ,and f ( x )has tobecloseto L if x isalittlebitmorethan a ,though f ( x )doesnot havetobecloseto L if x = a Severalcommentsareinorder.First,limx ag ( x )doesnotalways exist. Example 2.1 Let g ( x )= 1 if 0 x 0 if x< 0 Thenthelimitof g at a =0 doesnotexist.Indeed,nomatterhow smallaninterval I wetakearoundthepoint a =0 ,thatinterval I willcontainsomepositiveandsomenegativerealnumbers.Hence,the valuesof g ( x ) willsometimesequal 1 andsometimesequal 0 for x I nomatterhowsmall I is .Thereisnonumber L suchthat both0 and 1 arearbitrarilyclosetoit…infactthereisnonumbersuchthatboth 0 and 1 arebothcloserthan 0 5 toit.So limx 0g ( x ) doesnotexist. Second,iflimx 0f ( x )exists,itis unique ;thatis, f cannothave twodierentlimitsatanygivenpoint a .Letusillustratethisusing theintroductoryexampleofthissection,thefunction f ( x )=1 / (1+ x ). Wehaveseenthatlimx 0f ( x )=1.Indeed,wesawthatthevalues of f ( x )cangetarbitrarilycloseto1iftherealnumbers x arechosen fromasuitablysmallintervalaround0.Atthispoint,onecouldaskthe followingquestion. If1satis“estherequirementstobethe limx 0f ( x ) whydoes1.0001not?Afterall,whatiscloseto1isalsocloseto1.0001. Inordertoanswerthisquestion,wemusthaveagoodunderstandingofthede“nitionoflimits.Thatde“nitionsaysthatiflimx 0f ( x )= L ,thenthevaluesof f ( x )willget arbitrarily closeto f (0)if x ischosen fromasuitablysmallintervalaround0.Thekeywordintheprevious sentenceis arbitrarily. While1.0001iscloseto1,itisnot arbitrarily closeto1;itisexactly0.0001 away.And thatisaproblem,sincewe haveseenatthebeginningofthischapterthat,as x approaches0,the valuesof f ( x )willgetarbitrarilycloseto1.Inparticular,if x isclose enoughto0,then f ( x )willbecloserthan1 106to1,butthenitcannot alsobecloserthan1 106to1.0001. Ananalogousargumentshowsthatnofunctioncanhavetwodifferentlimitsatanyonepoint. Sometimesitcanhappenthat h isnotevende“nedin a ,but limx 0h ( x )stillexists.Notethatthefactthat h ( a )isnotde“ned

PAGE 31

7.THELIMITOFAFUNCTION25 321012345 1 2 3 4 5 6 7 8 Figure2.1. h ( x )=x2Š 9 x Š 3. isnotaproblemsincethede“nitionoflimitsspeci“callystatesthat x shouldnotbeequalto a anyway. Example 2.2 Let h ( x )=( x2Š 9) / ( x Š 3) .Then h isde“nedfor allrealnumbersexcept x =3 .Still, limx 3h ( x )=6 .Inparticular, limx 3h ( x ) exists. SeeFigure2.1foranillustration. Solution: If x =3,then f ( x )= x2Š 9 x Š 3 = ( x +3)( x Š 3) x Š 3 = x +3 Soifwewant f ( x )= x +3tobecloserto6thanagivendistance a thenallwehavetodoistochoose x suchthat | x Š 3 |

PAGE 32

262.LIMITSANDDERIVATIVES However,wehavenotyetlearnedthetechniquestorigorously provethis.Plottingthegraphofthefunctionorproducingmore numericaldatashouldnotbeconsideredasacompleteanswer,since, as x approaches0,eventually x andsin x willgetsosmallthatthe computerwillnolongermanipulatethem,ortheirratio,accurately. Figure2.2. Viewing(sin x ) /x ontheTI-89graphing calculatorwithviewingwindow[ Š 5 5] [ Š 0 5 1 5]. Finally,wepointoutthatinthede“nitionofthelimit,therequirementthat f ( x )getcloseto L andstay closeto L isimportant.Consider thefunction f ( x )=sin(1 /x )around x =0.As x approaches0,the valueof1 /x willincreaseveryfast,andsoitwillequalamultipleof manytimes.Allthosetimes, f (0)=0willhold,so f ( x )willbeas closeto0aspossible.However,limx 0f ( x )doesnotexist,since f ( x ) willtakeallothervaluesintheinterval[ Š 1 1]in“nitelyoftenaswell as x approaches0.Sothevalueof f ( x )will notstay arbitrarilyclose to0,nomatterhowclose x isto0.SeeFigure2.3foranillustration. 1.0 0.50.51.0 x 1.0 0.5 0.5 1.0 y Figure2.3. f ( x )=sin(1 /x ).

PAGE 33

7.THELIMITOFAFUNCTION27 7.2.ThePreciseDe“nitionofLimits.Itistimeforustogiveaprecise mathematicalde“nitionoflimits.Theadvantageofthisformalde“nitionisthatwecan“nallydo away withthewords arbitrarilyclose and sucientlyclose .Thepricetopayforthatisthatwehavetousemore notation. Definition 2.2 Let f beafunctionde“nedonsomeopeninterval thatcontainstherealnumber a ,withthepossibleexceptionof a itself. Thenwesaythatthelimitof f at a is L ,denotedby limx af ( x )= L if,forall > 0 ,thereexists > 0 suchthatif | x Š a | < ,then | f ( x ) Š L | < SeeFigure2.4foranillustration. a L 112345 5 5 10 15 Figure2.4. As x approaches a f ( x )approaches L Example 2.3 Wehave limx 02 x sin x =0 Solution: Let beanypositiverealnumber.Thenlet = / 2.We knowthat | sin x | 1forall x .Soif | x Š 0 | = | x | < = / 2,then | f ( x ) Š 0 | = | f ( x ) | = | 2 x sin x || 2 x | < 2 = ,asrequired. 7.3.One-SidedLimits.Therearefunctionsthatbehaveinacertain wayuptoapoint a ,andthenbehaveverydierentlyafterthat.We haveseensuchafunctioninExample2.1.Thefunction g ofthat examplesatis“ed g ( x )=0fornegativevaluesof x ,and g ( x )=1for positivevaluesof x .Wehaveseenthatlimx 0g ( x )doesnotexist, sincenorealnumber L isarbitrarilyclosetoboth0and1. Nevertheless,thereareweaker, one-sided notionsoflimitsthatare relevantinthisexample.

PAGE 34

282.LIMITSANDDERIVATIVES Definition 2.3 Let f : R R beafunctionandlet a beareal number.Wesaythatthe left-handlimit of f in a istherealnumber L ifthevaluesof f ( x ) getarbitrarilycloseto L andstayarbitrarilyclose to L when x issuitablycloseto a and x
a Thefactthat L istheright-handlimitof f in a isdenotedby limx a+f ( x )= L. Forinstance,if g isthefunctionde“nedinExample2.1,then limx 0+g ( x )=1 Indeed,ifwechoose x closeto0 butmorethan0 ,then g ( x )=1,so g ( x )isarbitrarilyclose(infact,equal)to1. Atthispoint,thereadershouldcomparethede“nitionsoflimit, left-handlimit,andright-handlimit.Thede“nitionoflimit(De“nition2.1)imposesthestrongestrequirementsonthevaluesof f .Indeed, thevaluesof f ( x )havetobecloseto L when x iscloseto a and xa .Thede“nitionsofthelefthandandright-handlimitsimposeweakerrequirementsinthateach de“nitiononlyrequiresthat f ( x )becloseto L when x isona given side of a andcloseto a Itthenfollows„andthereadershouldspendaminuteverifyingit„ thatiflimx af ( x )= L ,thenlimx aŠf ( x )= L andlimx a+f ( x )= L Conversely,ifboththeleft-handlimitandtheright-handlimitof f in a isequalto L ,thenthelimitof f in a existsandisequalto L Atthispoint,thereadershouldcheckhisorherunderstandingof thematerialbyconsideringthefunction h ( x )= x | x |

PAGE 35

7.THELIMITOFAFUNCTION29 as x approaches0anddecidingifthelimitslimx 0h ( x ),limx 0Šh ( x ), andlimx 0+h ( x ),exist.ItmayhelptoconsultFigure2.5. 2 112 2 1 1 2 Figure2.5. Graphof h ( x )= x/ | x | .7.4.In“niteLimits.Inourde“nitionsoflimitsinthissection,thelimit L wasalwaysarealnumber.Inthissection,weextendthosede“nitions tothecasesof in“nite limits.If L = ,thenthevaluesof f haveto getarbitrarilycloseto ;thatis,theyhavetogetaslargeaswewant. Thisisthecontentofthefollowingde“nition. Definition 2.5 Let f : R R beafunction.Wesaythatthe limitof f in a is ifwecanget f ( x ) arbitrarilylargeandkeepit arbitrarilylargeifwechoose x suitablycloseto a withoutbeingequal to a Similarly,if g : R R isafunction,wesaythatthelimitof g in a is Š ifwecanmake g ( x ) anegativenumberwithanarbitrarilylarge absolutevalueandkeep g ( x ) thatwayifwechoose x suitablycloseto a withoutbeingequalto a Thefactthatthelimitof f in a is isdenotedby limx af ( x )= Example 2.4 Let f ( x )=1 /x2.Then limx 0f ( x )= Solution: Ifwewant f ( x )tobelargerthananarbitrarypositive realnumber N ,allweneedtodoistochoose x fromtheinterval ( Š 1 /N, 1 /N ).Then x2< 1 /N willhold,implyingthat f ( x )= 1 /x2>N Similarly,if g ( x )= Š 1 /x4,thenlimx 0g ( x )= Š .Notethat ifthelimitofafunctionatagivenpoint a is or Š ,then,as x approaches a ,thegraphofthefunctionwillapproachaverticalline

PAGE 36

302.LIMITSANDDERIVATIVES intersectingthehorizontalaxisat x = a .Thisphenomenonisreferred tobysayingthat f hasa verticalasymptote at a .7.4.1.ThePreciseDe“nitionofIn“niteLimits.Theformalde“nitionof in“nitelimitsissimilartothatof“nitelimits.Thedierenceliesin thefactthatitisnotthesametobecloseto ortobeclosetoareal number. Definition 2.6 Let f : R R beafunction.Wesaythatthe limitof f in a is if,forallpositiverealnumbers N ,thereexists > 0 suchthatif | x Š a | < ,then f ( x ) >N Similarly,let g : R R beafunction.Wesaythatthelimitof g in a is Š ifforallnegativerealnumbers M ,thereexists > 0 such thatif | x Š a | < ,then g ( x ) a Example 2.5 Let f ( x )=1 /x .Then f isnotde“nedin0.Furthermore, limx 0Š= Š and limx 0+= .Asthetwoone-sided limitsaredierent, limx 0doesnotexist. Solution: Wecanmake f ( x )=1 xsmallerthananygivennegative number M bychoosing x fromtheinterval(1 /M, 0).Wecanmake x largerthananypositivenumber P bychoosing x fromtheinterval (0 ,P ). 7.5.Exercises.(1)Findlimx 3 x2Š 4 x +3 x Š 3. (2)Doeslimx 3 x2Š 4 x +7 x Š 3exist? (3)Findlimx 0cos x (4)Findlimx 0 x2 | x |. (5)Let f ( x )= x beequaltothelargestintegerthatisatmost aslargeas x .So f (3 7)=3.Notethat f isoftencalledthe ”oorfunction or integerpartfunction .Findthevalues a for

PAGE 37

8.LIMITLAWS31 whichlimx af ( x )exists.If a issuchthat f hasnotwo-sided limitsat a ,decideif f hasone-sidedlimitsat a (6)Let g ( x )= x beequaltothesmallestintegerthatisat leastaslargeas x .So g (3 7)=4.Notethat g isoftencalled the ceilingfunction .Findthevalues a forwhichlimx ag ( x ) exists.If a issuchthat g hasnotwo-sidedlimitsat a ,decide if g hasone-sidedlimitsat a (7)Doeslimx / 2tan x exist? (8)Doeslimx 0 1 | x |exist? (9)Giveanexampleofafunction f suchthatlimx 0Šf ( x )=0 andlimx 0+f ( x )= (10)Doeslimx 01 x3+1 x2 exist? (11)Doeslimx 01 x4+1 x2 exist? (12)Giveanexampleofafunction f suchthatlimx 1Šf ( x )= limx 1+f ( x )= Š ,and f (1)isarealnumber.8.LimitLaws8.1.BasicLimitLaws.If f and g aretwofunctionsandweknowthe limitofeachofthematagivenpoint a ,thenwecaneasilycompute thelimitat a oftheirsum,dierence,product,constantmultiple,and quotient.Therulesthatprovidethislimitaregivenbelow,andthey areverysimilartothewaysinwhichthesum,dierence,product, constantmultiple,andquotientoftwofunctionsarede“ned.Indeed, (I) limx a( f + g )( x )=limx af ( x )+limx ag ( x ) (II) limx a( f Š g )( x )=limx af ( x ) Š limx ag ( x ) (III) limx a( f g )( x )=limx af ( x ) limx ag ( x ) (IV) limx a( c f )( x )= c limx af ( x ) where c isarealnumber,and (V) limx a f g ( x )= limx af ( x ) limx ag ( x ) iflimx ag ( x ) =0.

PAGE 38

322.LIMITSANDDERIVATIVES Itisnotdiculttobelievethattheserulesarevalid.Forinstance, if f ( x )getsarbitrarilycloseto L as x approaches a and g ( x )gets arbitrarilycloseto Las x approaches a ,then,as x approaches a ,the valueof f ( x )+ g ( x ),thatis,thevalueof( f + g )( x ),willgetarbitrarily closeto L + L.Thisintuitiveargumentcanbemadeformalusingthe precisede“nitionoflimits. Example 2.6 Let f ( x )= | x | andlet g ( x )= x2.Findthelimitsof f + g f Š g fg 3 f +2 g ,and f/g at a =2 Solution: Basedonthe“velimitlawsgivenearlier,itmakessenseto “rstcomputethelimitsof f and g at2.Thereaderisinvitedtoverify that limx 2f ( x )=limx 2| x | =limx 2x =2 and limx 2g ( x )=limx 2x2=limx 2x limx 2x =2 2=4 whereweusedthefactthat g ( x )= x2= x x ,solawIIIcanbeapplied tocomputethelimitof g at2. Nowitissimplyamatterofbasicalgebratocomputethe“velimits thatwehavebeenaskedto“nd.Indeed,applyingthe“velimitlaws, wegetthat (I)limx 2( f + g )( x )=limx 2f ( x )+limx 2g ( x )=2+4=6, (II)limx 2( f Š g )( x )=limx 2f ( x ) Š limx 2g ( x )=2 Š 4= Š 2, (III)limx 2( f g )( x )=limx 2f ( x ) limx 2g ( x )=2 4=8, (IV)limx 2(3 f +2 g )( x )=3limx 2f ( x )+2limx 2g ( x )=3 2+ 2 4=14(notethathereweappliedlimitlawIVto“rst f thento g ,andthenweappliedlawIto3 f and2 g ),and (V) limx 2 f g ( x )= limx 2f ( x ) limx 2g ( x ) = 2 4 = 1 2 8.2.FrequentlyUsedSpecialCasesofLimitLaws.Afewspecialcases oflimitlawsI…Vareusedsofrequentlythatitisworthmentioning themseparately.First,ifwerepeatedlymultiplyafunctionbyitself, wegetapowerofthatfunction.ApplyinglawIIIeachtime,weget thatforallpositiveintegers n (2.1)limx a( f ( x ))n= limx af ( x ) n. Notethatwehaveessentiallyappliedthisruleinthespecialcase of n =2whenwecomputedlimx 2x2inExample2.6.

PAGE 39

8.LIMITLAWS33 Thereaderisinvitedtoverifythatthelimitsoftheconstantfunction f ( x )= c andtheidentityfunction f ( x )= x aregivenbylimx ac = c forall a andlimx ax = a .Formalproofswillbegiveninthe nextsection. ApplyingEquation(2.1)totheidentityfunction f ( x )= x yields theequation (2.2)limx axn= an. Itturnsout(thoughitisnotobvious)thatinEquation(2.1)the exponent n canbereplacedby1 /n ;inotherwords,powerscanbe replacedbyroots,yielding (2.3)limx an f ( x )=n limx af ( x ) (Here f ( x )hastobenonnegativeif n iseven.)So,inparticular,if f ( x )= x ,then limx an x =n a.8.3.OtherUsefulFactsAboutLimits.Inthissection,wediscussafew factsaboutlimitsthatareoftenusedtocomputelimits,butareslightly dierentinnaturefromthelimitlawswediscussedsofar. First,letusrecallthatthede“nitionof L =limx af ( x )requires that f ( x )getarbitrarilycloseto L if x issucientlycloseto a but notequaltoa .Thatis,thevalueof f ( a )doesnothavetosatisfyany requirements.Infact,wecanchange f ( a )toanythingwewant,and L =limx af ( x )willnotchange.Whatmattersiswhathappensat pointsotherthan a .Hence,wecanconcludethatif f ( x )= g ( x )forall points x = a ,thenlimx af ( x )=limx ag ( x )aslongastheselimits exist.Forinstance,let f ( x )=( x2Š 4) / ( x Š 2)forallrealnumbers x =2andlet f (2)=2010.Let g ( x )= x +2forallrealnumbers.Then f ( x )= g ( x )unless x =2,andhencelimx af ( x )=limx ag ( x )=4. Thestatementthatif f ( x )= g ( x )forallpoints x = a ,then limx af ( x )=limx ag ( x )aslongastheselimitsexistcanbesigni“cantlystrengthened.SeeExercise8.4.1forapossibledirectionfor that. Second,Equation(2.2)canbeinterpretedbysayingthatthelimit ofa powerfunction f ( x )= xnatanypoint a issimplythe value of f ( a ).Nownotethat polynomials arenothingelsebutsumsofconstantmultiplesofpowerfunctionswithnonnegativeintegerexponents. Hence,usinglimitlawsIandIV,wegetthefollowingtheorem.

PAGE 40

342.LIMITSANDDERIVATIVES Theorem 2.1 Let p beapolynomialfunction.Then,foranyreal number a ,wehave limx ap ( x )= p ( a ) Nowrecallthata rationalfunction isjusttheratiooftwopolynomials.Hence,usinglimitlawV,wegetthefollowingstatementfrom Theorem2.1. Corollary 2.1 Let R ( x ) bearationalfunctionandlet a beareal numbersuchthat R ( a ) isde“ned.Then limx aR ( x )= R ( a ) Proof. If R ( x )= p ( x ) /q ( x ),where p and q arepolynomials,then by“rstapplyinglimitlawV,andthenTheorem2.1,weget limx aR ( x )=limx ap ( x ) q ( x ) = limx ap ( x ) limx aq ( x ) = p ( a ) q ( a ) = R ( a ) Sofaralltherelationshipsthatwediscussedforlimitsinvolved equations .Wewillnowdiscusstworulesthat,involve inequalities Theorem 2.2 Let f and g betwofunctionsandassumethat,for allrealnumbers x ,theinequality f ( x ) g ( x ) holds.Then (2.4)limx af ( x ) limx ag ( x ) foranyrealnumber a aslongasbothlimitsexist. Proof. If(2.4)didnothold,then Lf=limx af ( x )= D +limx ag ( x )= D + Lgwouldhold,forsomepositiverealnumber D .Thatwouldleadtoa contradiction,sinceif x issocloseto a that | f ( x ) Š Lf| < ( D/ 3),then, inparticular, f ( x ) >LfŠ ( D/ 3),so g ( x ) >LfŠ D 3 = Lg+ 2 D 3 Thisinequalitysaysthatnomatterhowclose x isto a ,thedistance between g ( x )at Lgismorethan2 D/ 3.Thiscontradictsthede“nition of Lg,sinceif Lgexists,thenthevaluesof g ( x )shouldgetarbitrarily closetoit,providedthat x issucientlycloseto a NotethatinTheorem2.2,thefactthattheinequalitiesarenot strictisimportant.SeeExercise8.4.7forarelevantquestion.

PAGE 41

8.LIMITLAWS35 Corollary 2.2(SqueezePrinciple) If f g ,and h arefunctions suchthat,forallrealnumbers x ,theinequality f ( x ) g ( x ) h ( x ) holdsand limx af ( x )=limx ah ( x )= L, then limx ag ( x ) existsand limx ag ( x )= L SeeFigure2.6foranillustrationofthisimportantprinciple. Proof. Iflimx ag ( x )exists,thenbyapplyingTheorem2.2to f and g ,itfollowsthat L limx ag ( x ),andbyapplyingTheorem2.2 to g and h ,itfollowsthatlimx ag ( x ) L .Soiflimx ag ( x )exists,it isequalto L .InExercise8.4.3youareaskedtoprovethatthislimit exists. Thesqueezeprincipleisveryusefulsinceitallowsustocompute thelimitsofrathercomplicatedfunctionsaslongaswecan squeeze thembetweentwofunctionswithidenticallimits. f ( x ) g ( x ) h ( x ) Figure2.6. Conceptofsqueezetheoremwhere f ( x ) g ( x ) h ( x ). Example 2.7 Let g ( x )= x cos(log x ) .Then limx 0g ( x )=0 Solution: Indeed,let f ( x )= Š x and h ( x )= x .Then,sincecos(log z ) isalwaysarealnumberintheinterval[ Š 1 1],theinequality f ( x ) g ( x ) h ( x )holdsforallrealnumbers x .Furthermore,limx 0f ( x )= limx 0h ( x )=0,sowecanapplyCorollary2.2toproveourclaim. WecouldnothaveusedlimitlawIIItocomputelimx 0g ( x )since limx 0cos(log x )doesnotexist.Youareaskedtoprovethisin Exercise8.4.4.

PAGE 42

362.LIMITSANDDERIVATIVES 8.4.Exercises.(1)Findlimx 23 x2+4 x +9. (2)Findlimx 3 3 x2+5 x Š 2 x +1. (3)Findlimx 2 x Š 4 x Š 2. (4)Findlimx 4 x2+2 x +5 x3+1. (5)Findlimx 2 x2Š 4 x3+8. (6)Let f ( x )and g ( x )betwofunctionsthatonlydierfora “nitenumberofvaluesofthevariable x .Isittruethat limx af ( x )=limx ag ( x )aslongastheselimitsexist?Why orwhynot? (7)Findanexampleoftwofunctions f and g suchthat f ( x ) < g ( x )forallrealnumbers x ,butthereexistsarealnumber a suchthatlimx af ( x )=limx ag ( x ). (8)Explainwhylimx ag ( x )existsiftheconditionsofCorollary2.2hold. (9)Provethatlimx 0cos(log x )doesnotexist. (10)Provethatlimx 0| x sin( x ) | =0. (11)Computelimx 0x3sin(1 /x ). (12)Computelimx 0 x4+ x5sin(ln x ).9.ContinuousFunctions Intuitivelyspeaking,afunctioniscalled continuous atapoint x = a ifitsgraphinaneighborhoodof x = a canbedrawnwithoutlifting thepencilfromthepaper,thatis,byacontinuousŽline.Theformal de“nitionofcontinuityisasfollows. Definition 2.8 Afunction f iscalled continuous at a ifthe equality limx af ( x )= f ( a ) holds. NotethatDe“nition2.8reallyrequiresthreethings.Thelimitof f at a mustexist,thefunction f mustbede“nedin a suchthat f ( a ) exists,andthevalueof f ( a )mustagreewiththelimitof f at a Ifalltheseconditionshold,thenthebehaviorof f at a isvery similartothebehaviorof f around a ;inparticular,thegraphof f can bedrawnwithoutliftingthepencilfromthepaper.Ifwehadtolift thepencilfromthepaper,thatwouldmeanthatsomekindofgapŽ wouldexistinthegraphof f ,sotherequirementsofDe“nition2.8 wouldnotbesatis“ed.

PAGE 43

9.CONTINUOUSFUNCTIONS37 Ifafunction f : R R iscontinuousatall a R ,thenitiscalled continuous .If f iscontinuousateachpointoftheopeninterval( c,d ), thenwesaythat f is continuouson ( c,d ).Finally,ifyoureallywant aformalde“nition,the neighborhood of a isaset S thatcontainsan openinterval( c,d )containing a .9.0.1.ThePreciseDe“nitionofContinuity.Astheinformalde“nition ofcontinuityisveryclosetothatoflimits,itisnotsurprisingthat theirprecisede“nitionsarealsosimilar. Definition 2.9 Let f bede“nedinanopenintervalcontaining a Wesaythat f iscontinuousin a if,forall > 0 ,thereexists > 0 suchthatif | x Š a | < ,then | f ( x ) Š f ( a ) | < .9.1.ExamplesofContinuousFunctions.Letusconsidersomeofthe mostfrequentlyusedcontinuousfunctions. Example 2.8 Polynomialfunctionsarecontinuous. Solution: ThisisadirectconsequenceofTheorem2.1,whichwe discussedinthelastsection.Theorem2.1statedthatthelimitof apolynomialfunctionat a isequaltothe value ofthepolynomialat a ,whichispreciselywhatthede“nitionofcontinuityrequires. Therearemanyclassesoffunctionsthatarecontinuousatevery pointwheretheyare de“ned .Iftheyarenotde“nedsomewhere,then, ofcourse,theycannotbecontinuousthere. Example 2.9 Thefollowingareexamplesoffunctionsthatare continuousineverypointwheretheyarede“ned. (I) Rationalfunctions (II) Exponentialfunctions (III) Trigonometricfunctions (IV) Logarithmicfunctions (V) Inversetrigonometricfunctions Thereaderisinvitedtorecallthegraphsofeachofthesefunctions andverifythattheyconsistofcontinuouslinesaslongastheyare de“ned.9.2.FunctionsThatAreNotContinuous.Itistimetostopforamoment andthinkaboutfunctionsthatarenotcontinuousatagivenpoint a Therecanbethreereasonsforthis.First,itcouldbethat f ( a )isnot de“ned,forinstance,when f isarationalfunctionwhosedenominator

PAGE 44

382.LIMITSANDDERIVATIVES becomes0when x = a .Oritcouldbethat g isde“nedat a ,but limx ag ( x )doesnotexistat a .Anexampleofthisisthefunction de“nedby g ( x )=1if x 0and g ( x )=0if x< 0.Aswehaveseen before,thelimitofthisfunctiondoesnotexistin a =0,evenif g (0)is de“ned.So g isnotcontinuousat0.Finally,itcouldhappenthat h is de“nedin a andthelimitof h at a exists,but h ( a )isnotequaltothis limit.Thathappens,forexample,if h ( x )=( x +3) / ( x2Š 9)if | x | =3 and h ( x )=1if | x | =3.Let a = Š 3.Then h ( a )=1 =limx ah ( x )= Š 1 6 Theinterestedreaderisinvitedtothinkaboutthefollowing example. Excursion 2.1 Thefollowingfunctionisnotcontinuousanywhere.Let f ( x )=1 if x isrationalandlet f ( x )=0 if x isirrational.9.3.NewContinuousFunctionsfromOld.Itfollowsfromthelimitlaws thatseveraltransformationspreservethecontinuouspropertyoffunctions. Theorem 2.3 Let f and g betwofunctionsthatarecontinuous at a andlet c bearealnumber.Thenallofthefollowingarealso continuousfunctionsat a : (I) f + g (II) f Š g (III) f g (IV) cf ,and (V) f/g aslongas g ( a ) =0 Example 2.10 Itfollowsfromsuccessiveapplicationsoftheprevioustheoremthat h ( x )= ex sin x +3ln x Š x iscontinuousatall positiverealnumbers a Thefollowingimportanttheoremalsoholds,thoughitisnota directconsequenceofourlimitlaws. Theorem 2.4 Let f and g betwofunctionssuchthat f iscontinuousat a and g iscontinuousat f ( a ) .Thenthecompositionfunction g f iscontinuousat a Thistheoremisimportantsinceitenablesustoprovethecontinuity offunctionsthatwouldotherwisebecumbersometohandle. Example 2.11 Thefunction h ( x )= 2+sin x iscontinuousat allrealnumbers a .

PAGE 45

9.CONTINUOUSFUNCTIONS39 Solution: Let f ( x )=2+sin x andlet g ( x )= x .Then f iscontinuouseverywhere,and g iscontinuousatallpositiverealnumbers.As f ( x )isalwaysapositiverealnumber,thestatementfollows. 9.4.One-SidedContinuity.Afunctionmayhappentobecontinuousin onlyonedirection,eitherfromtheleftŽorfromtheright.ŽFormally, thismeansthefollowing. Definition 2.10 Wesaythatthefunction f is left-continuous at a if f ( a )=limx aŠf ( x ) .Similarly,wesaythat f is right-continuous at a if f ( a )=limx a+f ( x ) Example 2.12 Let g bethefunctionde“nedby g ( x )=1 if x 0 and g ( x )=0 if x< 0 .Then limx 0Šg ( x )=0 =1= g (0) ,so g isnot left-continuousat 0 .Ontheotherhand, limx 0+g ( x )=1= g (0) ,so g isright-continuousat 0 Thereaderisinvitedtoverifythat f iscontinuousat a ifandonly if f isbothleft-continuousandright-continuousat a Wesaythatafunctioniscontinuousonaninterval[ a,b ]ifitiscontinuousatallpointsof( a,b ),left-continuousat a ,andright-continuous at b .9.5.IntermediateValueTheorem.Perhapsthemostimportantproperty ofcontinuousfunctionsisthattheydonot skip anyvaluesbetweentwo valuesthattheyactuallytake.Forinstance,ifatreegrowsfrom3feet to6feet,thenthereisatimeinbetweenwhenthetreeisexactly4.47 feettall.Theintuitivereasonforthisisthatiftherewereavaluein betweenthatisnottakenbythefunction,thentherewouldbeagap inthegraphofthefunction,contradictingtherequirementthatthe functionbecontinuous.Thisisthecontentofthenexttheorem. Theorem 2.5(IntermediateValueTheorem) Let f beafunction thatiscontinuousontheinterval [ a,b ] .Then,if f ( a )= y1and f ( b )= y2and y isarealnumberthatisbetween y1and y2,thenthereexists x [ a,b ] suchthat f ( x )= y Inotherwords, f takesallvaluesbetween y1and y2ontheinterval [ a,b ] Example 2.13 Thereisarealnumber x intheinterval [0 1] such that x + ex=2 Solution: Let f ( x )= x + ex.Then f iscontinuouseverywhere, f (0)= 1,and f (1)=1+ e> 3 71.So,bytheintermediatevaluetheorem,

PAGE 46

402.LIMITSANDDERIVATIVES wegetthat f takesallvaluesbetween1and1+ e onthatinterval, including y =2. 9.6.Exercises.(1)Is e3 x +7sin x continuous? (2)Is( x2+1)ln( x +1)continuous? (3)Isx3+2 x2+3 x +4 x2+4continuous? (4)Whereistan x continuous? (5)Whereis1 /x notcontinuous? (6)Whereis3 x +2 5 x2Š 6 x +1continuous? (7)Whereissin( x2)continuous? (8)Let f ( x )= x .Determinethesetofpoints a forwhich f iscontinuousat a .Whatcanbesaidabout f atthepoints where f isnotcontinuous? (9)Let g ( x )= x .Determinethesetofpoints a forwhich g iscontinuousat a .Whatcanbesaidabout g atthepoints where g isnotcontinuous? (10)Provethattheequation x5Š x Š 1=0hasarootintheinterval ( Š 1 2). (11)Provethattheequation x3Š 3 x Š 1=0hasatleasttworoots intheinterval( Š 1 2). (12)De“neafunction f : R R thatisnotcontinuousin any point a .10.LimitsatIn“nity10.1.FiniteLimitsatIn“nity.InSection7,wede“nedwhatitmeant forafunctiontohavealimit L atarealnumber a .Inthissection,we extendthatde“nitionandde“newhatitmeansforafunctiontohave alimit L at orat Š Definition 2.11 Let f : R R beafunctionthatisde“nedon someinterval ( b, ) .Wesaythatthe limit of f at istherealnumber L ifthevaluesof f ( x ) getarbitrarilycloseto L andstayarbitrarilyclose to L when x issuitablylarge. Thefactthatthelimitof f at is L isexpressedbythenotation limx f ( x )= L. Thisde“nitionfollowstheideaofthede“nitionoflimitsat“nite points.Indeed,inorderforlimx f ( x )= L tohold,werequirethat thevaluesof f ( x )getarbitrarilycloseto L andstayarbitrarilyclose

PAGE 47

10.LIMITSATINFINITY41 to L if x islargeenough.Here x islargeenoughŽmeansthat x is inasuitablyselectedneighborhoodof ,inotherwords,inanopen interval( c, ).Recallthatthisisanalogoustowhatwerequiredinthe “nitecase.Therewesaidthatlimx af ( x )= L if f ( x )gotarbitrarily closeto L andstayedarbitrarilycloseto L once x wassuitablyclose to a ,thatis,when x wasinasuitablyselectedneighborhoodof a Example 2.14 Let f ( x )=1 /x .Then limx f ( x )=0 Solution: Ifwewantthevalueof f ( x )tobecloserthan to0,allwe havetodoistoselect x suchthat x> 1 / holds.Once x getspast1 / thevaluesof f ( x )willstaybetween0and Thede“nitionoflimitsat Š iswhatthereaderprobablyexpects. Definition 2.12 Let f : R R beafunctionde“nedonsome interval ( Š ,b ) .Wesaythatthe limit of f at Š istherealnumber L ifthevaluesof f ( x ) getarbitrarilycloseto L andstayarbitrarily closeto L when x isanegativenumberwithasuitablylargeabsolute value. Thefactthatthelimitof f at Š is L isexpressedbythenotation limx Šf ( x )= L. Example 2.15 Let f ( x )=1 /x2.Then limx Šf ( x )=0 Solution: Ifwewanttoget f ( x )closerthan to0andkeepitthere, itsucestochoose x suchthat x< Š 1 / .Then x2> 1 / ,andhence f ( x )=1 /x2< 10.1.1.TheFormalDe“nitionofLimitsatIn“nity.Theformalde“nition oflimitsatin“nityisverysimilartothatoflimitsat“nitepoints.The onlydierenceisintheformaldescriptionofwhatitmeanstobeina neighborhoodofin“nityversuswhatitmeanstobeinaneighborhood ofarealnumber. Definition 2.13 Let f : R R beafunctionde“nedonsome interval ( b, ) .Wesaythat limx f ( x )= L if,forallpositivereal numbers ,thereexistsapositiverealnumber N suchthatif x>N then | f ( x ) Š L | < .

PAGE 48

422.LIMITSANDDERIVATIVES Theformalde“nitionoflimitsatnegativein“nityisanalogous.The onlydierenceisagainintheformaldescriptionofwhatitmeansfor x tobeinaneighborhoodof Š .Itmeanstobeinaninterval( Š ,c ). Definition 2.14 Let f : R R beafunctionde“nedonsome interval ( Š ,b ) .Wesaythat limx Šf ( x )= L if,forallpositivereal numbers ,thereexistsanegativerealnumber N suchthatif x ln M Thefollowingnotationisde“nedinananalogousway: (I)limx f ( x )= Š (II)limx Šg ( x )= (III)limx Šh ( x )= Š Eachofthesede“nitionsreferstoafactthatthevaluesofafunction getarbitrarilyfar away from0andstayarbitrarilyfar away from0(in theappropriatedirection)if x getssucientlyfar away from0(inthe appropriatedirection).Thereadershouldtesthisorherunderstandingoftheseconceptsbyverifyingthatlimx 1 Š x = Š ,while limx Šx2= ,andlimx Šx3= Š .

PAGE 49

10.LIMITSATINFINITY43 10.2.1.TheFormalDe“nitionofIn“niteLimitsatIn“nity.Bynow,the formalde“nitionofin“nitelimitsatin“nityprobablydoesnotcome asasurprise.Weareprovidingaformalde“nitionforoneofthefour possiblescenariosthatcanoccurduetochangesinsign.Theother threecasesareanalogous. Definition 2.16 Let f : R R beafunctionde“nedonsome interval ( b, ) .Wesaythat limx f ( x )= if,forallpositivereal numbers M ,thereexistsapositiverealnumber N suchthatif x>N then f ( x ) >M .10.3.ComputingLimitsatIn“nity.Thelimitlawsthatwelearnedfor limitsat“nitepointsstaytrueforlimitsatin“nityaswell,provided, ofcourse,thattheymakesense.Hereareafewexamples. Example 2.17 Wehave limx x +3 x Š 4 =1 Itwouldbe wrong toargueasfollows:Thenumeratoristhefunction f ( x )= x +3,andthedenominatoristhefunction g ( x )= x Š 4. At ,theybothhavelimit ,so,bythelimitlawforquotients,the limitoftheirquotientis1.Ž Theproblemwiththisargumentisthat is notanumber .So / isnotde“ned.Itispossiblefor f and g bothtohavelimit at ,andfor f/g tohavelimits c at ,foranygivenrealnumber c Indeed,let f ( x )= cx andlet g ( x )= x Instead,wecansolveExample2.17asfollows. Solution: limx x +3 x Š 4 =limx ( x Š 4)+7 x Š 4 =limx 1+ 7 x Š 4 =1+limx 7 x Š 4 =1+0 =1 Wewouldliketopointoutotherpitfallswhendealingwiththe applicationoflimitlawsandin“nitelimits.Thefollowingexpressions arenotde“ned: (I) +( Š )

PAGE 50

442.LIMITSANDDERIVATIVES (II) 0and Š 0 (III)1and1ŠThefollowingtheoremisveryusefulwhendealingwithlimitsat Theorem 2.6 Let r beapositiverationalnumber.Then limx 1 xr=0 If r isaninteger,thenthisstatementfollowsfromthefactthat limx 1 /x =0byapplyinglimitlawIII(forproducts) r times.If r = p/q ,where p and q arepositiveintegers,thenwecan“rstprove thetheoremfor xp,andthen,usingtherootlaw,for xp/q=q xp. Manylimitscanbecomputedwiththehelpofthistheorem. Example 2.18 Wehave limx x2+3 x +1 x3=0 Solution: Wehave x2+3 x +1 x3= x2 x3+ 3 x x3+ 1 x3, andeachofthethreesummandshaslimit0at bythepreceding theorem.Hence,bythelimitlawforsums,sodoestheirsum. Notethatthelimitwouldnotchangeifwechangedthedenominator from x3to x3+3 x2+4 x +5.Thiswouldhavedecreasedthevalueofour function,butwouldhavestillkeptitpositive.Hence,bythesqueeze principle,wecanthenconcludethat limx x2+3 x +1 x3+3 x2+4 x +5 =0 .10.4.Exercises.(1)Findlimx x +1 x2+4. (2)Findlimx 3 x2+4 x +1 x2+5. (3)Findlimx x3+2 x x2+4 x +6. (4)Findlimx Š 3 x2+4 x +1 x2Š 4. (5)Let R ( x )= p ( x ) /q ( x )bearationalfunction.Explainhow limx R ( x )dependson p ( x )and q ( x ). (6)Computelimx x +2 x Š 3+2 x2Š 3 x +1 x2Š 2 x +1. (7)Computelimx Š sin x x. (8)Computelimx x2+1 x.

PAGE 51

11.DERIVATIVES45 (9)Computelimx Š 100 x +4 x2Š 4 x +5. (10)Istherearealnumber L suchthat L =limx x3+1 1000 x2+9 x +35holds? (11)Computelimx xŠ 0 1+ xŠ 0 9. (12)Doeslimx x sin x exist?11.Derivatives11.1.TangentLines.Letusconsiderafunction,suchas f ( x )= x2,and itsgraph.Letuschooseapointonthegraph,saythepoint P =(3 9). Nowletuslookfortheslopeofthe tangentline tothegraphatthat point. Thatis,considerasequenceofpoints P1,P2,... thatareallonthe graphof f andarecloserandcloserto P .Foreachofthesepoints, drawtheline PiP .Theslopeoftheselineswillapproachacertain slope,andsothelines PiP willapproachacertainline. That lineis calledthetangentlineof f at P .SeeFigure2.7foranillustration. Definition 2.17 Let f beafunctionandlet P =( a,f ( a )) bea pointonthegraphof f .Thenthe tangentline to f at P istheline thatcontains P andhasslope (2.5)limx af ( x ) Š f ( a ) x Š a providedthatthislimitexists. ( a f ( a ))( a f ( a )) ( x f ( x )) ( x f ( x )) 2 112 3 2 1 1 2 112 3 2 1 1 Figure2.7. Noticethatas x approaches a thesecant lineapproachesthetangentline.

PAGE 52

462.LIMITSANDDERIVATIVES Theinteractivewebsitehttp://www.math.u”.edu/ mathguy/ ufcalcbook/derivative def.htmlprovidesfurtherexamplesofthis phenomenon. Notethatintheprecedingde“nition,( f ( x ) Š f ( a )) / ( x Š a )issimply theslopeofthelineconnectingthepoints P and( x,f ( x )). Example 2.19 Inourrunningexample,thatis,when f ( x )= x2and P =(3 9) ,thetangentlineisthelinethatgoesthrough P andhas slope limx 3f ( x ) Š f (3) x Š 3 =limx 3x2Š 9 x Š 3 =limx 3( x +3)=6 .11.2.Velocities.RecallthatinSection6,wementionedthattheaveragevelocityofamovingobject,suchasacar,canbecomputedby therule v = s/t .Thatis,theaveragevelocityisequaltothedistance covereddividedbythetimeneededtocoverthatdistance.However, whatcanbesaidaboutthe instantaneousvelocity ,thatis,thevelocity inagivenmoment? WecouldnotanswerthatquestioninSection6sincewedidnot havethetoolstohandlethefactthatwhenonlyagivenmomentis considered,boththenumeratorandthedenominatoroftheformula v = s/t are0.Nowthatwehavelearnedaboutlimits,wecanovercome thatdicultyasfollows. Definition 2.18 Let f ( t ) beafunctionsuchthat f ( t ) isthedistancecoveredbyamovingobjectin t unitsoftime.Thenthe instantaneousvelocity oftheobject a unitsoftimeafteritstartsmovingis v ( t )=limt af ( t ) Š f ( a ) t Š a providedthatthislimitexists. Example 2.20 Acarstartsoutbyacceleratingfor10secondsso thatthedistancecoveredinthe“rst t secondsisobtained(inmeters) bythefunction f ( t )=1 2t2if t 10 .Whatistheinstantaneousvelocity ofthecarafter4seconds? Solution: Bythede“nitionofinstantaneousvelocity,wemust compute v (4)=limt 4f ( t ) Š f (4) t Š 4 =limt 4t2Š 16 2( t Š 4) =limt 4t +4 2 =4 .

PAGE 53

11.DERIVATIVES47 So,attheendofthefourthsecond(exactly4secondsafterstarting out),thecarwillmoveatarateof4meterspersecond. 11.3.TheDerivativeofaFunction.Thefactthatthelasttwoconcepts, thetangentlineandtheinstantaneousvelocity,ledtoverysimilar de“nitionssuggeststhatthereisaverygeneralprincipleatworkand wehaveseentwospecialcasesofthatprinciple. Thisisindeedthecase. Definition 2.19 Let f beafunction.The derivative of f at a is thelimit f( a )=limx af ( x ) Š f ( a ) x Š a ifthislimitexistsandis“nite. So,inparticular, f( a )istheslopeofthetangentlineof f at a (unlessthattangentlineisvertical).Furthermore,theinstantaneous velocityattime a isthederivativeofthedistancecovered(asafunction ofthetime t neededtocoverthatdistance)at t = a Inotherwords,thederivativeisacommongeneralizationofthe conceptsoftangentlineandinstantaneousvelocity.11.4.Exercises.(1)Findtheslopeofthetangentlinetothecurve f ( x )=3 x2Š 7 atthepoint(2 5). (2)Findtheslopeofthetangentlinetothecurve f ( x )= x3at x =0. (3)Findtheslopeofthetangentlinetothecurve f ( x )= x (1 Š x ) at x =1 / 2. (4)Findtheslopeofthetangentlinetothecurve f ( x )= x2at threedierentpoints.Doyouseeapattern? (5)Findtheslopeofthetangentlinetothecurve f ( x )= x2+ x atthreedierentpoints.Doyouseeapattern? (6)Showanexampleofacurvethatdoesnothaveatangentline atsomepoint a becausethelimitde“nedin(2.5)doesnot existorisin“nite. (7)Thedistancecoveredbyacarinacertaintimeperiodisdescribedbythefunction f ( t )= tm + t2( b Š m ) 2 ,

PAGE 54

482.LIMITSANDDERIVATIVES where b and m arepositiveconstants.Letusassumethat t [0 1].Findtheinstantaneousvelocityofthecarata givenmoment t = a (8)Acarismovingataspeedof20meterspersecondwhenits driverappliesthebrakesandthecarstartsslowingdown.The carstops10secondslater.Thedistancecoveredbythecarin t secondsstartingatthemomentwhenthedriverstepsonthe brakesisgivenbythefunction f ( t )=20 t Š t2for t [0 10]. Whatisthevelocityofthecar t secondsafterthebrakesare applied? (9)Findthederivativeofthefunction f ( x )= x +5at a =7. (10)Findthederivativeofthefunction f ( x )=2 x2at a =2. (11)Findthederivativeofthefunctions g ( x )=2 x2+1at a =2 and h ( x )=2( x Š 1)2at a =3. (12)Let g ( x )= 2 x if0 x x if x< 0 Does f( a )exist?12.TheDerivativeasaFunction12.1.RatesofChange.Inthelastsection,wesawthatthederivative ofafunctionatagivenpointwasacommongeneralizationofthe conceptsoftangentlinesandinstantaneousvelocities.Wewillnow furtherelaborateonthat,inordertounderstandhowfar-reachingthe conceptofderivativesis. If f isafunctionand f ( x )= y ,thenthequantitydenotedby y dependsonthequantitydenotedby x .Thisissometimesexpressed bysayingthat x istheindependentvariableand y isthedependent variable.If x changes,thenthechangein y canbedescribedinterms ofthechangein x Inparticular,if x changesfrom x1to x2,then y = f ( x )changes from y1= f ( x1)to y2= f ( x2).Theaveragerateofchangeforthe interval( x1,x2)isthentheratio y2Š y1 x2Š x1= y x where x isthechange(orincrement)of x .Wehavetousetheword averageŽsinceweonlyhaveinformationaboutthevaluesof y atthe endpointsoftheinterval( x1,x2);wedonotknowhow f ( x )= y behaves intherestoftheinterval.Ifwewantmorepreciseinformation,such asthe instantaneousrateofchange of f ( x )= y atagivenpoint,then

PAGE 55

12.THEDERIVATIVEASAFUNCTION49 wehavetousethenotionoflimitsagain,justaswehavedonetwice inthelastsection.Thatis,atagivenpoint x = a ,wede“nethe instantaneousrateofchangeof f ( x )= y as limx2 af ( x2) Š f ( a ) x2Š a =lim x 0 y x .12.2.TheDerivativeoftheFunctionf.Recallthat,atagivenpoint a thederivativeofthefunction f isde“nedasthelimit f( a )=limx af ( x ) Š f ( a ) x Š a Notethatthisde“nitionassociatestherealnumber f( a )tothe realnumber a .Thatis, f: R R isafunction.Thefunction fis calledthe derivative of f .Theoperationthattakes f into fiscalled dierentiation .Thisexplainsthefollowingde“nition. Definition 2.20 Afunction f iscalled dierentiable at a if f( a ) exists. Wesaythat f isdierentiableontheinterval( a,b )if f isdierentiableat d forall d ( a,b ). Example 2.21 Thefunction f ( x )= x3isdierentiableinevery realnumber a ,and f( a )=3 a2. Solution: Wehave limx af ( x ) Š f ( a ) x Š a =limx ax3Š a3 x Š a =limx a( x Š a )( x2+ xa + a2) x Š a =limx a x2+ xa + a2 =3 a2. Thefunctionswehaveconsideredsofarhadonlyoneindependent variable,usuallythevariable x .Thedependentvariablewasusually denotedby y ,so y = f ( x )held.Soitwasalwaysclearthatthederivativewastaken withrespectto x .However,therearecircumstances whenthisisnotsoclear,usuallywhen f dependsonmorethanone variable.Therefore,thereareadditionalwaystodenotethefunction fsuchas €dy dx, €df dx,

PAGE 56

502.LIMITSANDDERIVATIVES €d dxf ( x ), € Dxf ( x ),or € Df ( x ).12.3.DifferentiabilityVersusContinuity.Thede“nitionsofdierentiabilityandcontinuityaresimilar.Whichoneimposesstrongerrequirementsonafunctionatagivenpoint?Thefollowingtheoremshows thatdierentiabilityisthestrongerrequirement. Theorem 2.7 If f isdierentiableat a ,then f iscontinuousat a Proof. If f isdierentiableat a ,then f( a )=limx af ( x ) Š f ( a ) x Š a ; inparticular,thelimitshownontheright-handsideexists.Multiplying bothsidesbythefunction g ( x )= x Š a ,weget f( a )( x Š a )=( x Š a )limx af ( x ) Š f ( a ) x Š a ; Now,takinglimitsat a onbothsides,weobtain (2.6) f( a ) limx a( x Š a )=limx a( f ( x ) Š f ( a )) sincewecanapplythelimitlawforproductsontheright-handsideto getthat limx a( x Š a )limx af ( x ) Š f ( a ) x Š a =limx a ( x Š a ) f ( x ) Š f ( a ) x Š a =limx a( f ( x ) Š f ( a )) Finally,notethattheleft-handsideof(2.6)isequalto0since f( a ) ( x Š a )isapolynomialthattakesvalue0when x = a .Hence,the right-handsideof(2.6)isequalto0aswell,thatis, 0=limx a( f ( x ) Š f ( a ))=(limx af ( x )) Š f ( a ) Adding f ( a )toboththefarleftandfarrightsides,wegetthat f ( a )=limx af ( x ) whichmeansthat f iscontinuousat a TheconverseofTheorem2.7isnottrue.Indeed,thefunction f ( x )= | x | iscontinuousat a =0,butitisnotdierentiable.The readerisinvitedtoprovethisbyshowingthat limx 0Š| x |Š 0 x Š 0 =limx 0Š| x | x =limx 0+| x | x =limx 0+| x |Š 0 x Š 0 ,

PAGE 57

12.THEDERIVATIVEASAFUNCTION51 andhence f(0)=limx 0| x |Š 0 x Š 0 doesnotexist. Ingeneral,thereareseveralreasonsacontinuousfunctionmayfail tobedierentiableatagivenpoint.Itcouldbethatthegraphofthe functionhasacorner,Žlikethatof | x | at0,andhencetheslopeof thetangentlinecannotbede“nedbecausetheleft-handlimitandthe right-handlimitofthelinesapproachingthepurportedtangentline arenotequal.Or,itcouldbethatthefunctionhasaverticaltangent lineatthegivenpoint.SeeExercise12.5.6foranexampleofthis.12.4.Higher-OrderDerivatives.Inupcomingchapters,itwilloftenbe usefultoconsidernotonlythederivativeofafunctionbutalsothe derivativeofthederivativeandeventhederivativeofthederivative ofthederivative.Thesefunctionsappearsooftenthattheyhavetheir ownnames. If f isadierentiablefunctiononaninterval( a,b )anditsderivative fisalsodierentiableon( a,b ),thenthederivativeof fiscalled the secondderivative of f andisdenotedby f.Similarly,if fis dierentiableon( a,b ),thenitsderivativeiscalledthe thirdderivative of f andisdenotedby f.Higher-orderderivativesarede“nedin ananalogousway,butdenotedslightlydierently.Forinstance,the seventhderivativeof f isdenotedby f(7),and,ingeneral,the n th derivativeisdenotedby f( n ). Example 2.22 WehaveseeninExample2.21thatif f ( x )= x3, then f( x )=3 x2.Therefore, f( a )=limx af( x ) Š f( a ) x Š a =limx a3( x2Š a2) x Š a =limx a3( x + a ) =6 a. So f( x )=6 x InExercise12.5.2,youareaskedtoprovethat f( x )=6forall x andinExercise12.5.3,youareaskedtocomputehigher-orderderivativesof f .

PAGE 58

522.LIMITSANDDERIVATIVES 12.5.Exercises.(1)Compute f( a )if f ( x )= 4 x +1. (2)Let f ( x )= x3.Provethat f( x )=6forallrealnumbers x (3)Let f ( x )= x3.Compute f(4)( x ).Whatcanbesaidabout higher-orderderivativesof f ? (4)Let f ( x )= x .Compute f( a )atsomepoint a> 0. (5)Let f ( x )=1 /x .Compute f( a )atsomepoint a =0. (6)Let f bede“nedontheinterval[0 2]by f ( x )= 1 Š x2if 0 x 1,and f ( x )= Š 1 Š ( x Š 2)2if1
PAGE 59

CHAPTER3 RulesofDifferentiation 13.DerivativesofPolynomialandExponentialFunctions13.1.Polynomials.Letusrecallthatpolynomialsaresumsofpower functionswithnonnegativeintegerexponents,suchasthefunction f ( x )=3 x2+4 x +6.Inthissection,wewilldeducegeneralrulesfor thederivativesofpolynomialfunctions.Westartbytheirbuilding blocks,Žpowerfunctions.Thesimplestoftheseistheclassofconstant functions. Theorem 3.1 Let c bearealnumberandlet f ( x )= c forall x Then f( a )=0 forallrealnumbers a Beforeweprovethetheorem,wepointoutthat,intuitively,itmakes perfectsense.Thederivativeofafunction f describestherateof changeof f ,butif f isaconstantfunction,then f neverchanges(it haszerochange). ProofofTheorem3.1. Wehave f( a )=limx af ( x ) Š f ( a ) x Š a =limx ac Š c x Š a =0 Notethatlimx a( c Š c ) / ( x Š a )=0since( c Š c ) / ( x Š a )=0forall values x = a Wenowturnourattentiontoamoregeneralclassofpowerfunctions,thoseoftheform f ( x )= xn,where n isapositiveinteger.Let usrecallthealgebraicidentity xnŠ an=( x Š a ) ( xn Š 1+ xn Š 2a + + xan Š 2+ an Š 1) Theorem 3.2 Let n beapositiveintegerandlet f ( x )= xn.Then f( a )= an Š 1.53

PAGE 60

543.RULESOFDIFFERENTIATION Proof. Wehave f( a )=limx af ( x ) Š f ( a ) x Š a =limx axnŠ an x Š a =limx a( x Š a ) ( xn Š 1+ xn Š 2a + + xan Š 2+ an Š 1) x Š a =limx a( xn Š 1+ xn Š 2a + + xan Š 2+ an Š 1) = nan Š 1. Notethatthisagreeswithourresultfromthelastsectionthat showedthatif f ( x )= x3,then f( x )=3 x2. ItturnsoutthatTheorem3.2holdsevenif n isnotapositive integer.Thatis,forallrealnumbers ,if f ( x )= x,then f( x )= x Š 1.Wewillseeaformalproofofthisfactlater.Intheexercises, youareaskedtoprovetwospecialcasesofthisgeneralresult.13.1.1.ThreeSimpleRules.Derivativesarelimitsofcertainfunctions, soitisnotsurprisingthat some ofthelawsgoverningtheircomputation areverysimilartolimitlaws.Thatis,ifweknowthederivativeof f and g ,thenwecaneasilycomputethederivativeof f + g f Š g ,and cf ,where c isagivenrealnumber.Therulesareasfollows. Theorem 3.3 Let f and g betwofunctionsthataredierentiable at a .Then f + g isdierentiableat a ,and ( f + g )( a )= f( a )+ g( a ) Proof. Wehave ( f + g )( a )=limx a( f + g )( x ) Š ( f + g )( a ) x Š a =limx a f ( x ) Š f ( a ) x Š a + g ( x ) Š g ( a ) x Š a = f( a )+ g( a ) Theothertworulesandtheirproofsaresosimilarthattheyare leftasexercises. Theorem 3.4 Let f and g betwofunctionsthataredierentiable at a .Then f Š g isdierentiableat a ,and ( f Š g )( a )= f( a ) Š g( a ) .

PAGE 61

13.DERIVATIVESOFPOLYNOMIALANDEXPONENTIALFUNCTIONS55 Theorem 3.5 Let f beafunctionthatisdierentiableat a and let c bearealnumber.Then cf isdierentiableat a and ( cf )( a )= cf( a ) Itisveryimportanttopointoutthatthe otherlimitlawsdonot carryovertoderivativesinthesamefashion. Thatis,ingeneral, ( fg ) = fg,and( f/g ) = f/g.Wewilllearnsomemorecomplicated rulestocomputethederivativesof fg and f/g inthenextsection. Theorems3.3to3.5enableustocomputethederivativeofany polynomialfunction. Example 3.1 Let p ( x )=3 x3+5 x2Š 6 x +8 .Find p( x ) Solution: Notethat p ( x )isjustasum(anddierence)ofconstant multiplesofpowerfunctions.Thederivativesofpowerfunctionsare computedinTheorem3.2.ThenwecanapplyTheorems3.3to3.5 toget p( x )=(3 x3)+(5 x2)Š (6 x )+(8)=3( x3)+5( x2)Š 6( x )+(8)=9 x2+10 x Š 6 13.2.ExponentialFunctions.Letusnowcomputethederivativeofthe exponentialfunction f ( x )= bx,where b issomepositiveconstant.By thede“nitionofderivatives,weget f( a )=limx af ( x ) Š f ( a ) x Š a =limx abxŠ ba x Š a =limz 0ba + zŠ ba z = balimz 0bzŠ 1 z = baf(0) Severalcommentsareinorder.First,notethesubstitution z = x Š a inthethirdline.Second,notethat baisaconstantthatdoesnot dependon z ;hence,thelimitlawforconstantmultipleswasusedin thefourthline.Third,inthespecialcasewhen a =0,thede“nition ofthederivativeyields f(0)=limz 0( bzŠ 1) /z .Weusedthisfactin thelastline.

PAGE 62

563.RULESOFDIFFERENTIATION Inotherwords, (3.1) f( x )= f(0) bx= f(0) f ( x ) Thatis,thederivativeofthefunction f isa constantmultiple of f Theconstantinquestionis f(0),thatis,limz 0( bzŠ 1) /z .Numerical experimentationsuggeststhatthelarger b is,thelargerthislimitis. Graphicalexperimentationsuggeststhisaswell.Indeed, f(0)isthe slopeofthetangentlinetothecurveof f ( x )= bxatthepoint x =0, andplotting f forvariousvaluesof b suggeststhatthelarger b is,the largerthisslopeis. Inparticular,itcanbeprovedthatthereexistsarealnumber e closeto2.71,suchthat limz 0ezŠ 1 z =1 Thereadermaywishtoconsulttheinteractivewebsitehttp:// www.math.u”.edu/ mathguy/ufcalcbook/exponent.htmlforfurtherillustrations. Definition 3.1 Let e betherealnumbersuchthat limz 0ezŠ 1 z =1 So,inthespecialcaseof b = e ,Equation(3.1)takestheform ( ex)= ex, since f(0)=limz 0ezŠ 1 z =1 Thatis,thederivativeof f ( x )= exis f ( x )= exitself.Inthenext section,wewillseewhatthisimpliesforthederivativesofexponential functionswithbasesdierentfrom e .13.3.Exercises.(1)Let f ( x )= x3+2 x2+3 x +4.Compute f( x )and f( x ). (2)Let f ( x )= x4Š 3 x +9.Compute f( x )and f( x ). (3)Provethatif f isapolynomialfunction,then f( x )isalsoa polynomialfunction. (4)Provethatif f isapolynomialfunctionofdegree d ,then f( d +1)( x )=0forallrealnumbers x (5)Let p beapolynomialfunctionofdegree d andlet k d bea nonnegativeinteger.Whatkindoffunctionis fk? (6)Provethatif f ( x )= x1 / 2and a> 0,then f( a )=1 2 a. (7)Provethatif f ( x )=1 /x and a =0,then f( a )= Š1 a2.

PAGE 63

14.THEPRODUCTANDQUOTIENTRULES57 (8)ProveTheorem3.4. (9)ProveTheorem3.5. (10)Let f ( x )=3 x3Š 4 x2+ x Š 2+4 ex.Compute f( x ). (11)Couldithappenthat f and g aretwodierentfunctions,but f( x )= g( x )forall x ? (12)Couldithappenthat f and g aretwodierentfunctions, fand garetwodierentfunctions,but f( x )= g( x )forall x ?14.TheProductandQuotientRules14.1.TheProductRule.Wementionedinthelastsectionthat,ingeneral,( fg ) = fg.Forinstance,if f ( x )=2 x +1and g ( x )= x +2,then ( fg )( x )=2 x2+5 x +2,so( fg )( x )=(2 x2+5 x +2)=4 x +5,while f( x )=2and g( x )=1,so f( x ) g( x )=2. Itturnsoutthatthereisaruletocomputethederivativeofa product;itisjustalittlebitmorecomplicatedthanthelimitlawfor products.Thisisthefocusofour“rsttheoreminthissection. Theorem 3.6 Let f and g betwofunctionsthataredierentiable at a .Then fg isdierentiableat a ,and ( fg )( a )= f ( a ) g( a )+ f( a ) g ( a ) Proof. Byde“nition,wehave (3.2)( fg )( a )=limx af ( x ) g ( x ) Š f ( a ) g ( a ) x Š a Thecrucialideaistodecomposethedierence f ( x ) g ( x ) Š f ( a ) g ( a )as ( f ( x ) g ( x ) Š f ( x ) g ( a ))+( f ( x ) g ( a ) Š f ( a ) g ( a ))inthenumeratorofthe right-handsideof(3.2). Usingthisidea,weobtainfromEquation(3.2) ( fg )( a )=limx a f ( x ) g ( x ) Š f ( x ) g ( a ) x Š a + f ( x ) g ( a ) Š f ( a ) g ( a ) x Š a =limx af ( x ) g ( x ) Š f ( x ) g ( a ) x Š a +limx af ( x ) g ( a ) Š f ( a ) g ( a ) x Š a =limx af ( x ) g ( x ) Š g ( a ) x Š a +limx ag ( x ) f ( x ) Š f ( a ) x Š a = f ( a ) g( a )+ g ( a ) f( a )

PAGE 64

583.RULESOFDIFFERENTIATION Example 3.2 Thederivativeof h ( x )= x2excanbecomputedas follows.Let f ( x )= x2and g ( x )= ex.Then h( x )=( fg )( x ) = f ( x ) g( x )+ f( x ) g ( x ) = x2( ex)+( x2)ex= x2ex+2 xex= ex( x2+2 x ) .14.2.TheQuotientRule.Theruleforthederivativeofthequotientof twofunctionsisalittlebitmorecomplicatedthanthatforthederivativeoftheproductoftwofunctions.Thoughmorecomplex,boththe ruleanditsproofbearsomesimilaritytotherulegiveninTheorem3.6. Theorem 3.7 Let f and g betwofunctionsthataredierentiable at a andletusassumethat g ( a ) =0 .Then f/g isdierentiableat a andwehave f g ( a )= g ( a ) f( a ) Š f ( a ) g( a ) g ( a )2. Proof. Byde“nition,wehave (3.3) f g ( a )=limx a f ( x ) g ( x )Šf ( a ) g ( a ) x Š a Letusmultiplyboththenumeratorandthedenominatoroftherighthandsideby g ( x ) g ( a )toget f g ( a )=limx af ( x ) g ( a ) Š f ( a ) g ( x ) ( x Š a ) g ( x ) g ( a ) Nowtransformthenumeratoroftheright-handsidebysubtracting andthenadding g ( a ) f ( a )toget f g ( a )=limx af ( x ) g ( a ) Š g ( a ) f ( a )+ g ( a ) f ( a ) Š f ( a ) g ( x ) ( x Š a ) g ( x ) g ( a ) =limx ag ( a ) g ( x ) g ( a ) f ( x ) Š f ( a ) x Š a Š limx af ( a ) g ( x ) g ( a ) g ( x ) Š g ( a ) x Š a = g ( a ) f( a ) Š f ( a ) g( a ) g ( a )2. Theorem3.7nowenablesustocomputethederivativeofrational functions.

PAGE 65

15.DERIVATIVESOFTRIGONOMETRICFUNCTIONS59 Example 3.3 Let h ( x )=( x +3) / ( x2+1) .Find h( x ) Solution: Let f ( x )= x +3andlet g ( x )= x2+1.Then f( x )=1 and g( x )=2 x .So,byTheorem3.7,wehave h( x )= g ( x ) f( x ) Š f ( x ) g( x ) g ( x )2= x2+1 Š ( x +3)2 x x4+2 x2+1 = Š x2Š 6 x +1 x4+2 x2+1 14.3.Exercises.(1)Let h ( x )= exx3.Find h( x )and h( x ). (2)Let f ( x )=(2 x +7) ex.Compute f( x ). (3)Findaruletocompute( f2)( x ). (4)Findaruletocompute(1 /f )( x ). (5)Usetheresultofthepreviousexercisetoproveaformulafor g( x )if g ( x )= xnfora negative integer n (6)Let g ( x )= eŠ x.Find g( x ). (7)Let h ( x )= x/ex.Find h( x ). (8)Let f ( x )= ex/ ( x +2).Compute f( x ). (9)Let g ( x )=( x Š 3) / ( ex+1).Compute g( x ). (10)Let f ( x )=(2 x +3) / (4 x +7).Compute f( x ).Tryto“ndtwo dierentwaysofgettingthesameanswer. (11)Let f ( x )= g ( x ) h ( x ),where g isapolynomialfunctionof x and h ( x )= eaxforsomeconstant a .Provethat f( x )and f( x )arebothequaltotheproductofapolynomialfunction andthefunction eax. (12)Provethatif f ( x )isarationalfunction,then f( x )isalsoa rationalfunction.15.DerivativesofTrigonometricFunctions Inthissection,weshowhowtocomputethederivativesoftrigonometricfunctions.First,wecompute(sin x ).Thiswillbeasomewhat lengthyprocedure,duetothefactthatthisisthe “rst trigonometric functionwewilldierentiateandwewillhavetoapplynewmethods. However,onceweknowthederivativesofsin x andcos x ,itwillbe muchsimplertodeducethederivativesofothertrigonometricfunctions,sincethosefunctionscanbeobtainedfromsinandcos,andthen thevariousdierentiationrulescanbeused. Theorem 3.8 Wehave (sin x )=cos x Proof. Recalltheidentitysin( a + b )=sin a cos b +sin b cos a .We have

PAGE 66

603.RULESOFDIFFERENTIATION (sin x )=limh 0sin( x + h ) Š sin x h =limh 0sin x cos h +sin h cos x Š sin x h =limh 0 sin x cos h Š 1 h + sin h cos x h =sin x limh 0cos h Š 1 h +cos x limh 0sin h h Notethatas, h approaches0,wecertainlyhavelimh 0sin x =sin x and limh 0cos x =cos x ,sincethesefunctionsdonotevendependon h Thereremainsthetaskofcomputingthetwonontriviallimits limh 0cos h Š 1 h andlimh 0sin h h Wewillcarryoutthistaskintwolemmas. Lemma 3.1 Wehave limh 0sin h h =1 Proof. Letusconsideracirclewithunitradiusandaregular n gonwhosecenterisatthecenter O ofthecircleandwhose n vertices areallontheunitcircle.Thentheareaofthecircleis ,andthearea ofthe n -gonis n 1 2 sin ,where =2 /n istheangle AOB ,with A and B beingadjacentverticesofour n -gon. Consideringjust1 /n ofboththecircleandthe n -gon,weseethat theareaofthetriangle AOB is(sin ) / 2,andtheareaof1 /n ofthe circleborderedbythelines AO BO ,andthearc AB is / (2 )= / 2.Sotheratioofthetwoareasis (sin ) / 2 / 2 = sin Ontheotherhand,as n getslargerandlarger, getssmallerand smaller,whiletheareaofthe n -gongetscloserandclosertothearea ofthecircle.Hence,theirratio,sin / ,willgetarbitrarilycloseto1 andstayarbitrarilycloseto1. Lemma 3.2 Theequality (3.4)limh 0cos h Š 1 h =0 holds.

PAGE 67

15.DERIVATIVESOFTRIGONOMETRICFUNCTIONS61 Proof. Wewillmanipulatetheexpression(cos h Š 1) /h sothatwe canusetheresultofLemma3.1.First,wemultiplyboththenumerator andthedenominatorbycos h +1toget cos h Š 1 h = cos2h Š 1 h (1+cos h ) = Š sin2h h (1+cos h ) Therefore,wehave limh 0cos h Š 1 h = Š limh 0sin2h h (1+cos h ) = Š limh 0sin h h sin h 1+cos h = Š limh 0sin h h limh 0sin h 1+cos h =( Š 1) 0=0 Wecannow“nishtheproofofTheorem3.8.Attheendofthe“rst displayedchainofequationsinthatproof,wesawthat (sin x )=sin x limh 0cos h Š 1 h +cos x limh 0sin h h Theprevioustwolemmasshowedthat,ontheright-handside,the“rst limitis0andthesecondlimitis1,so(sin x )=cos x asclaimed. Thefollowingtheoremcanbeprovedbyverysimilarmethods. Theorem 3.9 Theequality (cos x )= Š sin x holds. YouareaskedtoprovethistheoreminExercise15.1.1. Nowthatwehavethederivativesofsinandcos,thederivatives ofothertrigonometricfunctionscanbeobtainedbysimplyusingthe quotientrule.Thenexttheoremshowsanexampleofthis. Theorem 3.10 Wehave (tan x )=sec2x Proof. Notethattan x =sin x/ cos x ,sowecanapplythequotient rule.Thisleadsto

PAGE 68

623.RULESOFDIFFERENTIATION (tan x )= sin x cos x = cos x (sin x )Š sin x (cos x ) cos2x = cosx+sin2x cos2x = 1 cos2x =sec2x. Thederivativesoftheotherthreetrigonometricfunctionsaregiven intheexercises.15.1.Exercises.(1)Provethat(cos x )= Š sin x (2)Provethat(cot x )= Š csc2x. (3)Provethat(csc x )= Š csc x cot x. (4)Provethat(sec x )=sec x tan x. (5)Let h ( x )= excos x .Find h( x ). (6)Let h ( x )= ex/ sin x .Find h( x ). (7)Compute(sin2x ). (8)Compute(cos2x ).Trytogetthesameanswerintwodierent ways. (9)Compute(sin x tan x ). (10)Compute(tan2x ). (11)Compute(sin3x ).Youmaywanttousetheresultofexercise7. (12)Compute(cos3x ).Youmaywanttousetheresultofexercise8.16.TheChainRule16.1.TheDerivativeoftheCompositionofTwoFunctions.Inprevious sections,welearnedhowtocomputethederivativeofthesum,dierence,product,andquotientoftwofunctions.Westilldonotknow howtocomputethederivativeofthe composition offunctions,suchas h ( x )=sin(3 x ), t ( x )= x2+1,or r ( x )= e4 x.Inthissection,wewill learnarule,calledthe chainrule ,thatappliesinthesesituations. Theorem 3.11(ChainRule) Let h ( x )= f ( g ( x )) ,where g isdifferentiableat x and f isdierentiableat g ( x ) .Then h isdierentiable

PAGE 69

16.THECHAINRULE63 at x ,andwehave h( x )= f( g ( x )) g( x ) Theproofofthechainruleissomewhattechnical,sowewillpostponeituntiltheendofthissection.Nowwewilldiscusssomeexamples oftheapplicationsofthechainrule. Example 3.4 Findthederivativeof h ( x )=sin(3 x ) Solution: Let f ( x )=sin x andlet g ( x )=3 x .Then h ( x )= f ( g ( x )), so,bythechainrule,wehave h( x )= f( g ( x )) g( x )=(cos(3 x )) 3=3cos(3 x ) Example 3.5 Let h ( x )= x2+1 .Find h( x ) Solution: Let f ( x )= x andlet g ( x )= x2+1.Then h ( x )= f ( g ( x )), so,bythechainrule,wehave h( x )= f( g ( x )) g( x )= 1 2 x2+1 2 x = x x2+1 SometimesthechainruleiswrittenintheLeibniznotation,that is,as dh dx = dh dg dg dx .16.2.TwoApplicationsoftheChainRule. 16.2.1.ASimpleWayofObtaining(cos x ).Recallthatinthelastsection,ittookconsiderabletimeandeorttoprovethat(sin x )=cos x Finding(cos x )withsimilarmethodsisjustastime-consuming.On theotherhand,thechainruleenablesustocompute(cos x )faster. Recallthatcos x =sin x + 2 .Sowecanwritecos x asthecompositionoftwofunctions,namelycos x = f ( g ( x )),with f ( x )=sin x and g ( x )= x + 2.Sothechainruleapplies,andweget (cos x )= f( g ( x )) g( x ) =cos x + 2 1 =cos x cos 2 Š sin x sin 2 =0 Š sin x = Š sin x.

PAGE 70

643.RULESOFDIFFERENTIATION 16.2.2.TheDerivativesofExponentialFunctions.Recallthatwede“nedthenumber e suchthatthederivativeoftheexponentialfunction f ( x )= exwas f ( x )itself.Nowthechainruleenablesustocompute thederivativesofexponentialfunctionswith any base. Theorem 3.12 Let a beapositiverealnumberandlet h ( x )= ax. Thenwehave h( x )= axln a. Proof. Notethat h ( x )= ax=( eln a)x= ex ln a. Sowehavesucceededinwriting h asthecompositionoftwofunctions, namely h ( x )= f ( g ( x )),where f ( x )= exand g ( x )= x ln a .Therefore, thechainruleapplies,andweget h( x )= f( g ( x )) g( x )= ex ln a ln a = axln a. 16.3.ProofoftheChainRule.Itistimethatweprovedthechainrule. ProofofTheorem3.11. As g isdierentiableat x ,weknow that (3.5)limr 0 g ( x + r ) Š g ( x ) r Š g( x ) =0 Set t = g ( x + r ) Š g ( x ) r Š g( x ) Notethat t dependson r ,andas r approaches0, t approaches0. Similarly,let y = g ( x ).As f isdierentiableat y ,wehave (3.6)lims 0 f ( y + s ) Š f ( y ) s Š f( y ) =0 Set u = f ( y + s ) Š f ( y ) s Š f( y ) Again,notethat u dependson s andthat u approaches0as s approaches0. Nowweundertakeaseriesofmanipulationsoftheprecedingtwo equations.Ourgoalistoexpress f ( g ( x ))=limr 0f ( g ( x + r )) Š f ( g ( x )) r intermsof f( g ( x ))and g( x ).

PAGE 71

16.THECHAINRULE65 Rearrangingtheequationthatde“nesthevariable t thatwejust introduced,weget (3.7) g ( x + r )= g ( x )+( g( x )+ t ) r. Similarly,rearrangingtheequationthatde“nesthevariable u ,weget (3.8) f ( y + s )= f ( y )+( f( y )+ u ) s. Nowapplythefunction f tobothsidesof(3.7)toget (3.9) f ( g ( x + r ))= f ( g ( x )+( g( x )+ t ) r ) Observethat(3.8)holdsforall y and s ,so,inparticular,itholds when y = g ( x )and s =( g( x )+ t ) r .Makingthesesubstitutionsin (3.8),Equation(3.9)yields f ( g ( x + r ))= f ( g ( x )+( g( x )+ t ) r ) (3.10) = f ( g ( x ))+( f( g ( x ))+ u ) ( g( x )+ u ) r. (3.11) Wecannowexpressthequotient( f ( g ( x + r )) Š f ( g ( x ))) /r from theequalityoftheleft-handsideof(3.10)andtheexpressionin(3.11)as f ( g ( x + r )) Š f ( g ( x )) r = ( f( g ( x ))+ u )( g( x )+ t ) r r =( f( g ( x ))+ u )( g( x )+ t ) Finally,weareinapositiontocomputethederivativewewere lookingforasthelimitoftheleft-handsideas r approaches0.Weget limr 0f ( g ( x + r )) Š f ( g ( x )) r =limr 0( f( g ( x ))+ u )( g( x )+ t ) = limr 0f( g ( x ))+limr 0u limr 0g( x )+limr 0t = f( g ( x )) g( x ) sinceboth t and u approach0as r approaches0. 16.4.Exercises.(1)Let h ( x )=( x2+1)5.Find h( x ). (2)Let h ( x )=sin( x2).Find h( x ). (3)Let h ( x )=sin3x .Find h( x ).Compareyourresulttothe resultofexercise11oftheprevioussection.Whichotherexercisesoftheprevioussectioncanbesolvedbythechainrule? (4)Let f ( x )=2x+3x.Find f( x ). (5)Let h ( x )= esin x.Find h( x ). (6)Let h ( x )= ex2sin x.Find h( x ).

PAGE 72

663.RULESOFDIFFERENTIATION (7)Let h ( x )=sin(2 x ).Find h( x ).Howcouldyougetthesame resultwithoutusingthechainrule? (8)Let h ( x )=cos(2 x ).Find h( x ).Howcouldyougetthesame resultwithoutusingthechainrule? (9)Let h ( x )=2x2.Find h( x ). (10)Let h ( x )=1 / (1 Š x ).Find h( x ). (11)Let h ( x )=ln(1 / (1 Š x )).Find h( x ). (12)Let h ( x )= 1 Š x2.Find h( x ).17.ImplicitDifferentiation Inthelastseveralsections,wecomputedthederivativesofmany dierentfunctions.Althoughthesefunctionsweredierent,theyhad oneimportantfeatureincommon.Theywere explicitly given.That is,theyweregivenbyarulethatdirectlydescribedhow f ( x )= y is obtainedfrom x .17.1.TangentLinestoImplicitlyDe“nedCurves.Sometimeswehaveto dealwithcurvesthataregivenbyadierentkindofrule.Consider thecurvegivenbytheequation (3.12) x3+ y3=4 xy. Letussaythatwewanttocomputetheslopeofthetangentline tothiscurveatthepoint(2 2).Ifwecouldexpress y asafunction of x ,wecouldsimplytakethederivativeofthatfunctionat x =2. However,itisnotclearhowtowrite y explicitly intermsof x ,evenif (3.12) implicitly describesthisdependence. Itisinthesesituationsthatweresortto implicit dierentiation. Keepinmindthatwedonotneedtoexplicitlyknowhow y depends on x ,thatis,wedonotneedanexplicitexpressionforthefunction y ( x );weonlyneedtoknowthederivative dy/dx ofthatfunctionat x =2. ConsiderEquation(3.12),anddierentiatebothsideswithrespect tothevariable x toget d dx x3+ y3 = d dx (4 xy ) Nowrecallthat y = y ( x )isafunctionof x .So,whencomputing ( d/dx ) y3ontheleft-handside,weneedtousethechainrule.Onthe right-handside,weneedtousetheproductrule.Usingtheserules, weget 3 x2+3 y2dy dx =4 y +4 x dy dx .

PAGE 73

17.IMPLICITDIFFERENTIATION67 Expressing dy/dx fromthisequation,weget dy dx = (4 y Š 3 x2) (3 y2Š 4 x ) Atthepoint(2 2),theright-handsideis Š 4 / 4= Š 1,sotheslopeof thetangentlineat(2 2)is Š 1. Notethatthefactthatthetangentlineat(2 2)hasslope Š 1makes (intuitively)perfectsense,sincethecurveinquestionis symmetric in x and y .Thatis,if( x,y )isonthecurve,then( y,x )isalsoonthe curve.17.2.DerivativesofInverseTrigonometricFunctions.Oneplacewhere implicitdierentiationisaverypowerfultoolisinthecomputationof thederivativesofinversetrigonometricfunctions.RecallthattanŠ 1x = y isthefunctionthatistheinverseoftherestrictionofthefunction tan x totheinterval( Š / 2 ,/ 2).Thatis,if tanŠ 1x = y, then (3.13) x =tan y, where y ( Š / 2 ,/ 2). Ourgoalistodetermine d dx tanŠ 1x = dy dx Tothatend,letustakethederivativeofbothsidesof(3.13)with respectto x .Recallingthat d dz tan z =sec2z and y = y ( x ) weget 1=sec2y dy dx Solvingfor dy/dx andrecallingtheidentitysec2z =1+tan2z ,we obtain dy dx = 1 sec2y = 1 1+tan2y = 1 1+ x2. Inotherwords,weprovedthesuprisinglysimpleformula (tanŠ 1x )= 1 1+ x2. Thisformulaisinterestingfortworeasons.First,itissurprisinglysimple.Second,itdoesnotevencontaintrigonometricfunctions.Imagine

PAGE 74

683.RULESOFDIFFERENTIATION tryingtogetthisresult without implicitdierentiation,usingjustthe de“nitionofderivatives. Youwillbeaskedtocomputethederivativesoftheotherinverse trigonometricfunctionsintheexercises.17.3.Exercises.(1)Let C bethecirclegivenbytheequation x2+ y2=169.Use implicitdierentiationto“ndtheslopeofthetangentlineto C atthepoint(5 12). (2)Provethat(sinŠ 1x )=1 1 Š x2. (3)Provethat(cosŠ 1x )= Š1 1 Š x2. (4)Provethat(cotŠ 1x )= Š1 1+ x2. (5)Provethat(secŠ 1x )=1 x x2Š 1. (6)Provethat(cscŠ 1x )= Š1 x x2Š 1. (7)Compute(cosŠ 1(2 x +0 1)). (8)Compute sinŠ 1( x2) (9)Compute sinŠ 1(1 /x ) (10)Compute(tanŠ 1(2 x )). (11)Compute(cscŠ 1( x/ 3)). (12)Compute(secŠ 1( x Š 0 01)).18.DerivativesofLogarithmicFunctions18.1.TheFormulafor(logax ).Asanotherpowerfulapplicationofimplicitdierentiation,wecomputethederivativeofthefunction f ( x )= ln x Theorem 3.13 Wehave (ln x )= 1 x Proof. Set y =ln x .Then ey= x .Dierentiatingbothsideswith respectto x ,weget ey dy dx =1 dy dx = 1 ey. However, ey= x byde“nition,so dy dx = 1 x asclaimed.

PAGE 75

18.DERIVATIVESOFLOGARITHMICFUNCTIONS69 Itisnowabreezetodeterminethederivativeoflogarithmicfunctionsof any base. Corollary 3.1 Let a =1 bea“xedpositiverealnumber.Then (logax )= 1 x ln a Proof. Notethat x = eln alogax= e(ln a )(logax ). Soln x =(ln a )(logax )and f ( x )=logax = ln x ln a Asln a isaconstant,itfollowsthat f( x )= 1 ln a (ln x )= 1 x ln a asclaimed. 18.2.TheChainRuleandlnx.Aninterestingconsequenceof Theorem3.13isthefollowing. Corollary 3.2 Let f ( x ) beadierentiablefunctionthattakes positivevaluesonly.Then d dx ln f ( x )= f( x ) f ( x ) Proof. Bythechainrule, d dx ln f = d df df dx = f( x ) f ( x ) Example 3.6 Let f ( x )=cos x .Then d dx ln(cos x )= Š sin x cos x = Š tan x.18.3.LogarithmicDifferentiation.Sometimesweneedtocomputethe derivativeofacomplicatedproduct.Thisissometimeseasierbytaking thelogarithmoftheproduct,whichwillbeasum,andusingimplicit dierentiation.Thisprocedure,whichiscalled logarithmicdierentiation ,hastheinherentadvantagethatitdealswithsumsinsteadof products,andsumsaremucheasiertodierentiatethanproducts.

PAGE 76

703.RULESOFDIFFERENTIATION Example 3.7 Let y = x3 x +1 x Š 2 Compute dy/dx Solution: Takinglogarithms,weget ln y =3ln x + 1 2 ln( x +1) Š 1 2 ln( x Š 2) Nowtakingderivativeswithrespectto x andusingCorollary3.2,we have dy dx 1 y = 3 x + 1 2( x +1) Š 1 2( x Š 2) Finally,wecansolvethisequationfor dy/dx toget dy dx = y 3 x + 1 2( x +1) Š 1 2( x Š 2) = x3 x +1 x Š 2 3 x + 1 2( x +1) Š 1 2( x Š 2) 18.4.PowerFunctionsRevisited.Recallthatinanearliersection,we provedthatif n isa“xedpositiveinteger,then( xn)xn Š 1.Westated thatthiswasthecaseforallrealnumbers n ,notjustpositiveintegers, butwehavenotprovedthatclaim.Nowwehavethetools,namely logarithmicdierentiation,toproveit. Theorem 3.14 Let n beanyrealnumber.Thenwehave d dx xn= nxn Š 1. Proof. Set y = xn.Letusassumeforthecaseofsimplicitythat x ispositive.Takinglogarithms,wehave ln y = n ln x. Dierentiatingbothsideswithrespectto x ,weget dy dx 1 y = n x Solvingfor dy/dx yields dy dx = ny x = nxn x = nxn Š 1asclaimed.

PAGE 77

19.APPLICATIONSOFRATESOFCHANGE71 18.5.TheNumbereRevisited.Recallthatwehavede“nedthenumber e ,thebaseofthenaturallogarithm,asthenumberforwhich limh 0( ehŠ 1) /h =1.Ournewknowledgeletsusexpress e moredirectly,asalimit. Notethatif f ( x )=ln x ,then f( x )=1 /x ,so f(1)=1.Bythe de“nitionofderivatives,thismeansthat limh 0ln(1+ h ) Š ln1 h =1 Observingthatln1=0andusingthepowerruleoflogarithms,weget limh 0ln(1+ h )1 /h=1 or,applyingtheexponentialfunction eztobothsides,wehave limh 0(1+ h )1 /h= e. Equivalently,setting x =1 /h ,weget limx 1+ 1 x x= e. Eitherofthelasttwoformulascanhelptodeterminetheapproximate value2.712828of e .18.6.Exercises.(1)Computed dxln( x +1). (2)Compute(ln | x | ). (3)Compute( xx). (4)Compute f( x )if f ( x )= x43 x +4 x +1. (5)Computelimx 1+1 x2 x. (6)Let h ( x )=ln sinŠ 1x .Compute h( x ). (7)Let h ( x )=sinŠ 1(ln x ).Compute h( x ). (8)Let h ( x )=4Š x.Compute h( x ). (9)Let h ( x )=2 x.Compute h( x ). (10)Let h ( x )= esin x.Compute h( x ). (11)Uselogarithmicdierentiationto“nd y( x )= dy/dx if y ( x )= xln x. (12)Uselogarithmicdierentiationto“nd y( x )= dy/dx if y ( x )= x3 x.19.ApplicationsofRatesofChange Inthissection,weconsiderafewapplicationsofderivativesinvariousdisciplines.

PAGE 78

723.RULESOFDIFFERENTIATION 19.1.Physics.Recallthatifanobjectmovesalongalineandthe distanceitcoversintime t isdescribedbythefunction s ( t ),then (3.14) v ( t )= ds dt = s( t )=limh 0s ( t + h ) Š s ( t ) h isthe instantaneousvelocity oftheobjectattime t Wecantakethisconceptonestepfurther.Iftheobjectmovesat achangingvelocity,thenthe rateofchangeofthevelocity itselfcan beimportantinformation.Forinstance,whenconsideringavehicles performance,wemaybeinterestedinhowfastitcanreachitstop speed,notonlywhatitstopspeedis. Thecorrespondingnotioninphysicsiscalled acceleration ,andis denotedby a ( t ).Thatis,keepingthepreviousnotation,wehave (3.15) a ( t )= v( t )= dv dt = s( t ) Example 3.8 Thepositionofaparticleisdescribedbytheequation (3.16) s ( t )= 1 3 t3Š 3 t2+5 t. Here s ismeasuredinmetersand t inseconds. (I) Whatisthevelocityoftheparticleafter3seconds? (II) Findtheaccelerationoftheparticleafter10seconds. (III) Whendoestheparticlemovebackward? Additionalquestionsaboutthemovementofthisparticlewillbe givenintheexercises. Solution: (I)Thevelocityoftheparticleisdescribedbythefunction v ( t )= s( t )= t2Š 6 t +5.Thisyields v (3)=9 Š 18+5= Š 4.So thevelocityoftheparticleafter3secondsis Š 4m / s,meaning thattheparticleismovingbackwardataspeedof4meters persecondafter3seconds. (II)Theaccelerationoftheparticleisgivenbytheformula a ( t )= v( t )=2 t Š 6.So,after10seconds,theparticleisaccelerating at14m / s2. (III)Theparticleismovingbackwardwhenitsvelocity v ( t )isnegative.Thathappenswhen v ( t )= t2Š 6 t +5=( t Š 1)( t Š 5) < 0, thatis,when t (1 5).Inotherwords,theparticleismoving backwardbetweenthe“rstand“fthseconds.

PAGE 79

19.APPLICATIONSOFRATESOFCHANGE73 19.2.Economics.Letussaythatacompanyestimatesthatitcosts C ( x )dollarstoproduce x unitsofanewproduct.Itisoftenthecase that C ( x ),whichiscalledthe costfunction ,canbedescribedbya polynomialfunction,suchas C ( x )= a + bx + cx2+ dx3. Thereasonforthisisasfollows.Therewillbesomecosts,suchas designingtheproductandobtainingpermits,thatwillbepresentregardlessofthenumberofunitsproduced.Thesewillberepresentedby theconstantterm a .Thentherewillbecosts,suchasrentingalocation andbuyingsupplies,thatwillbemoreorlessindirectproportionto thenumberofunitsproduced.Thesewillberepresentedbythelinear term bx .Thentherewillbeotherfactors,suchashiringworkers,marketingtheproduct,andorganizingproduction,thatwillbeindirect proportiontoahigherpowerof x asthedierencesinsizeturninto dierencesinkind.Taxesmayfactorinatanevenhigherrate. Becausethecostfunction C ( x )isnotalinearpolynomial,producing the1001stunitdoesnotcostofthesameasproducingthe“rstunit orthe5001stunit.Thecostofincreasingproductionfrom n unitsto n +1units,inotherwords,thecostofproducingthe( n +1)thunit, canbecomputedbytheformula M ( n )= C ( n +1) Š C ( n ) The marginalcostfunction C( x )describeshowthecostfunction changes.Inthat, C( x )and M ( n )aresimilar.Thereisoneimportant dierence.Asweknow,thederivative C( x )isgivenby (3.17)lim x 0C ( x + x ) Š C ( x ) x However,itcouldwellbethatthesmallestmeaningfulpositivevalue of x is1,incasetheproductsaresuchthatfractionalunitsdonot makesense(e.g.,automobiles).Inthatcase, x 0isimpossible initsprecisemathematicalmeaning;theclosestthat x cangetto 0iswhen x =1.Inthatcase,however,theexpressionafterthe limitsymbolin(3.17)simpli“esto C ( x +1) Š C ( x ),justifyingthe approximation (3.18) M ( x )= C ( x +1) Š C ( x ) C( x ) Example 3.9 Thecostfunctionofabottleofanewmedicationis givenby C ( x )=106+20 x +0 001 x2+0 000001 x3.Findtheapproximate costofproducingthe101standthe1001stbottles.

PAGE 80

743.RULESOFDIFFERENTIATION Solution: Bytheprecedingdiscussion,weneedtocomputethefunction C( x ).Bytherulesofdierentiatingapolynomialfunction,we get C( x )=0 000003 x2+0 002 x +20.Sothe101stbottlecosts 0 0003 1002+0 002 100+20=20 23dollarstoproduce,whilethe 1001stbottlecosts0 000003 10002+0 002 1002+20=43dollarsto produce. Itisimportanttonotethattheresultofthepreviousexample,that is,thefactthatitcostsmoretoproducethe1001stbottlethanthe 101stbottledoes not meanthatthemorebottlesareproduced,the moreexpensiveitistoproducethe average bottle.Thisisbecause thecostofproducingthe “rst bottleisastronomical,since C (1) > 106. Comparedtothat,thecostofeachofthe“rstthousand,oreven,“rst tenthousandbottlesisverysmall,sotheproductionofeachofthem willbringthecostofproducingthe average bottledown.(Thecost ofproducingtheaveragebottleif n bottlesareproducedisofcourse C ( n ) /n .) Intheexercises,youareaskedtocomparetheseresultstotheresults obtainedbyusingtheformula C ( n +1) Š C ( n ).19.3.Exercises.(1)ConsidertheparticleofExample3.8.After6seconds,howfar fromitsstartingpointisthatparticle?Inwhatdirection? (2)Considertheparticleofthepreviousexercise.Arethereany momentswhentheparticleisnotmoving? (3)Thelocationofanobjectmovingverticallyisdescribedbythe function s ( t )= t Št2 5for t [0 5],wheretimeismeasured insecondsanddistanceismeasuredinmeters.Whenwillthe objecthaveaninstantaneousvelocityof0.2m/s? (4)Considertheobjectofthepreviousexercise.Whendoesit havethegreatestspeedgoingup?Whendoesithavethe greatestspeedgoingdown? (5)Considertheparticleofthepreviousexercise.Willitsaccelerationeverbe1m/s2? (6)Considertheparticleofthepreviousexercise.Whenwillits accelerationbenegative? (7)Usetheformula M ( n )= C ( n +1) Š C ( n )to“ndthecostof producingthe101stand1001stunitsinExample3.9.Compareyourresultswiththeestimatesthatwefoundusingthe function C( x ). (8)Tworacecarsspeedupfromastandingstartto60m/sso thateachcarhasconstantacceleration.The“rstcarreaches

PAGE 81

20.RELATEDRATES75 one-thirdofitstopvelocityin4seconds,whilethesecondcar reachesone-fourthofitstopvelocityin3seconds.Whichcar willhavecoveredmoredistancebythetimeitreachesitstop velocity? (9)Thecostfunctionforacompanytoproduce x laptopsis C ( x )=1500+2 x +0 4 x2+0 01 x3.Findthemarginalcost functionforthisproduct. (10)Considerthecostfunctionofthepreviousexercise.Explain themeaningof C(200). (11)Acertaininsectpopulationhasbeenexposedtoaninsecticide, whichresultedinthepopulationchangingaccordingtothe function f ( t )=10 000 Š 1000 t Š 500 t2,where t ismeasuredin hours.Findthegrowthrateoftheinsectpopulationafterone hourandafter“vehours.20.RelatedRates20.1.Preliminaries.Anintuitiveideaofthenotionof relatedrates comesfromasimplefactofeverydaylife:Iftherearetworelatedquantitiesthatarechangingwithtime,thentheirratesofchangeshould alsoberelated.Forexample,thevolume V ofwaterinapoolofarea 20m2isrelatedtothewaterlevel h (thepooldepthinmeters)as V =20h.Supposethewaterlevelislowandneedstobeincreased.A hoseisputintothepoolthatcanpumpwateratarateof0 2m3/ h. Atwhatratedoesthewaterlevelincrease?Thevolumeandthewater levelarebothfunctionsoftime, V = V ( t )and h = h ( t ).Forevery instanceoftime t ,theirvaluesarerelatedas V ( t )=20 h ( t )andso mustbetheirderivativesorratesofchange: (3.19) V ( t )=20 h ( t )= V( t )=20 h( t ) Nowthequestioniseasytoanswer.Since V( t )=0 2m3/ h, h( t )= V( t ) / 20=0 01m / h=1cm / h.Thewaterlevelrisesby1cmevery hour.Asomewhatpracticalestimate!Youwouldknow exactly when tocomebackandturnothewaterifyouneededaninchorsoofthe waterlevelincrease.Apparently,thesameideaofrelatedrateswould workforloweringthewaterlevelafterrain.20.2.Units.Itisimportanttobringallthequantitiestothesame systemofunits.Forexample,intheaboveproblemthepoolarea isoftengiveninsquarefeet,forexample,200ft2,whilethepumpingrateisgiveningallonsperhour,forexample, V=60gal / h. Onegallonis3 785 10Š 3m3andtherefore V=60 3 785 10Š 3= 0 2271m3/ h.Onesquarefootis9 29 10Š 2m2,sothepoolareais

PAGE 82

763.RULESOFDIFFERENTIATION 200 9 29 10Š 2=18 58m2.Hence, h=0 2271 / 18 58 1 2cm / h. In1999,NASAlosta$125millionMarsorbiterbecauseaLockheed MartinengineeringteamusedEnglishunitsofmeasurementwhilethe agencysteamusedthemoreconventionalmetricsystemforakey spacecraftoperation.20.3.FormalDe“nitionofRelatedRates.Theverynotionof related quantities canbestatedinpropermathematicalterms. Definition 3.2 Twoquantities y and x aresaidtoberelatedif thereisafunction f suchthat y = f ( x ) Inthepreviousexample, V = f ( h )=20 h .Supposenowthatthe quantities y and x arefunctionsofanothervariable t (e.g., t istime): x = x ( t )and y = y ( t ).Thentherateofchangeof x or y withrespect to t isnothingbutthederivative x( t )or y( t ).Theproblemofrelated ratesŽcannowbecastinthepropermathematicalterms:Whatisthe relationshipbetweenthederivatives x( t )and y( t )ifthevaluesof x ( t ) and y ( t )arerelatedby y = f ( x )?Thevaluesofthefunctions x ( t )and y ( t )arerelatedas y ( t )= f ( x ( t ))forany t .Takingthederivativeof bothsideswithrespectto t bymeansofthechainrule(Theorem3.11), weobtainageneralizationof(3.19): (3.20) y ( t )= f ( x ( t ))= y( t )= f( x ( t )) x( t ) Equation(3.20)establishesthesought-afterrelationbetweentherates yand x.However,itseemssomewhatdierentfrom(3.19):Therates arestillproportionaltooneanother,buttheproportionalitycoecient f( x )isnolongeraconstant,butafunction.Howdoweuseit?Take aparticularvalueof t = t0.Letthevaluesof x and y at t = t0be x0= x ( t0)and y0= y ( t0).Thenumber a = f( x0)canbecalculated.Then theequality y( t0)= ax( t0)determinestherelationbetweentherates yand xattheinstancewhen x hasthevalue x0(or y hasthevalue y0= f ( x0)). Example 3.10 Letalaserpointerbepositionedatadistance D =1 mfromawall.Thepointercanberotatedsothatthebright spotcreatedbythelaserbeamtravelshorizontallyonthewall. (I) Atwhats peeddoesthe brightspottravelalongthewallifthe pointerrevolvesataconstantrate rad/s? (II) Atwhatdirectionofthelaserbeamdoesthebrightspottravel atthespeed v =4 m/s if = rad/s ?

PAGE 83

20.RELATEDRATES77 D y v y Figure3.1. Alaserpointerispositionedatadistance D fromawallandrotatesclockwise.Itsbeammakesa brightspotthatmovestotherightalongthewall. Solution: (I)Theanalysisofanyproblemonrelatedratesmustbeginwith de“ningthequantitieswhoseratesarebeingstudied.Inother words,onehastoanswerthequestion:Howarethesequantitiesmeasured?Theorientationofthelaserbeamcanbe describedbytheangle betweentheperpendiculartothe wallandthelaserbeam.Thepositionofthebrightspotmay besetbythedistance y traveledbyitfromthepointonthe wallwhen =0,thatis,whenthelaserbeamisperpendiculartothewall.Ifthepointerrotates,theanglebecomesa functionoftime, = ( t ),andsodoesthepositionofthe brightspot, y = y ( t ).Thus,thequestionisabouttherelation betweentherates y( t )= v (thespeedatwhichthebrightspot travels)and ( t )= (therateatwhichthepointerrotates). (II)Thenextstepisto“ndafunctionthatdeterminestherelation betweenthequantitiesofinterest,thatis,betweenthedistance y andtheangle : y = f ( ).Itisclearthat D and y are relatedasthecathetioftherighttrianglewhosehypotenuse isthelaserbeam: y = D tan = f ( ).

PAGE 84

783.RULESOFDIFFERENTIATION (III)Oncetherelationbetweenthequantitiesofinteresthasbeen established,therelationbetweentheirratescanbefound. Since(tan )=1 / cos2 ,Equation(3.20)yields (3.21) y = D tan = y= D cos2 = v = D cos2 The“rstquestionisanswered. (IV)Notethattherate y= v isnotconstanteveniftherate = isconstant.Toanswerthesecondquestion,onehasto“nd thevalueof when v =4 m / s, D =1m,and = rad / s. ItfollowsfromEquation(3.21)that cos2 = D v = 1 4 = = 3 ; thatis,thebrightspotmovesatthespeed4 m / swhenthe laserbeammakes60withtheperpendiculartothewall. 20.4.CanAnythingTravelFasterThanLight?Thesolution(3.21)has aninterestingfeature.When approaches90,thatis,thelaserbeam isgettingclosertobeingparalleltothewall,thecosine,cos ,tendsto 0inEquation(3.21),andhencetherate y= v growsunboundedly.It seemslikejustwithmerelyalaserpointer,asuperluminalobjectcan becreatedinalecturehall!Letusinvestigatethis.Thespeedoflightis c 300 000km / s 186 000mi / sec.Thelightcanmakeatriparound theworldinmerely0 13seconds!Example3.10isnowsupplemented bytwoadditionalquestions: (I) Isitpossiblethat v canexceedthespeedof light?Ifso,at whichdirectionofthelaserbeamdoesithappen? (II) Atwhichpositionofthebrightspotdoesithappen? Theanswersread: (I)Setting D =1m=10Š 3km(watchtheunits:alldistances arenowinkilometers!)and v = c =3 105km / s,theangle atwhichthebrightspotexceedsthespeedoflightsatis“es theequationcos2 = D/c 1 05 10Š 8,andhence 89 99414.Sothebrightspotbecomessuperluminalif > 89 99414! (II)Since y = D tan v>c if y> 9772m.Well,alecture hallappearstobeabitŽsmallforthisexperiment!Take aDremelminiaturegrinder(soldinLowesstores)forwhich 103rad / s(itcanbeusedtorotatethepointer),andset D =0 1m,then v>c if y> 98m;notyetexactlyalecture hallexperiment,butitcanbemanagedonthecampus!

PAGE 85

20.RELATEDRATES79 Einsteinstheoryofrelativitystatesthatnomaterialobjectcan travelfasterthanlight.HasacounterexampletoEinsteinstheoryjust beenfound?Theanswerisno.ŽInthemotionofthebrightspot, nomaterialobjectactuallymovesalongthewall.Brightspotsat y and y + y arecreatedby dierent portionsofthelaserbeamthat areemittedbythelaserattwodistinctmomentsoftime.Alumpof lightthatarrivedat y wasre”ectedbythewall(thatiswhywesee thebrightspot!),andhenceitcouldnotappearatthenextposition y + y (atthispositionarriveda dierent lumpoflightemittedbythe laseratalatertime).Sotherate y/ t cannotpossiblybeassociated withthemotionofanymaterialobject.20.5.RelatedProblem.ThenexttimeyouwatchaFloridasunset,look atyourshadow.DoesthereexistapositionoftheSunabovethe horizonatwhichyourshadowextendsfasterthanthespeedoflight?20.6.MoreThanTwoRelatedRates.Therearesituationswhenseveral quantitiesarerelatedamongthemselves.Ifthesequantitiesbecome functionsofavariable t ,thentheirratesare linearly related.Aproofof thisstatementisgiveninamoreadvancedcourse,wherethefunctions ofseveralvariablesarestudied.However,thebasicideaof“nding relationsbetweentherateshasnotchanged:Theyareobtainedby dierentiatingtherelationsbetweenthequantitiesinquestionwith respectto t .Theprocedureisillustratedinthefollowingexample. Example 3.11 Considerarectanglewithsides x and y .Supposethat x and y changewithtime.Findtheirratesofchangewhen x =3 cm and y =1 cm if,atthatmoment,theareaoftherectangle decreasesatarateof 2 cm2/s whiletheperimeterdoesnotchange. Solution: (I)Therearefourquantitiesinvolved:therectangledimensions x and y ,thearea S ,andtheperimeter P (II)Therearetworelationsbetweenthem: S = xy,P =2( x + y ) (III)If x = x ( t )and y = y ( t ),then S ( t )= x ( t ) y ( t )and P ( t )= 2( x ( t )+ y ( t )).Usingthederivativeoftheproductandthesum oftwofunctions,thelinearrelationsbetweentheratesare S= xy + xy,P=2( x+ y) (IV)Since P=0(theperimeterdoesnotchange), x= Š yand S=( x Š y ) y.Nowlet S= Š 2cm2/ sbecause S decreases

PAGE 86

803.RULESOFDIFFERENTIATION ( Smustbenegative).With x =3cmand y =1cm,one has Š 2=(3 Š 1) yand y= Š 1cm / s.Itthenfollowsthat x= Š y=1cm / s. 20.7.Exercises.(1)Aladder24ftlongleansagainstaverticalwall.Ifthelower endisbeingmoved away fromthewallatarateof3ft/sec, howfastisthetopdescendingwhenthelowerendis8ftfrom fromthewall?Whenarethelowerandupperendsmovingat thesamerate? (2)Aman6fttallwalks away fromanarclight15fthighata rateof3milesperhour.Howfastisthefartherendofhis shadowmoving?Howfastishisshadowlengthening? (3)Thevolumeofasphereisincreasingatarateof16cm3/s. Howfastistheradiusincreasingwhenitis6cm?Howfastis thesurfaceareaincreasingwhenitis36cm2? (4)Sandisbeingpouredonthegroundfromanelevatedpipeand formsapilethathasalwaystheshapeofacircularconewhose heightisequaltotheradiusofthebase.Ifthesandfallsata rateof0.5m3/min,howfastistheheightofthepileincreasing whenitis2m? (5)Aparticlemovesalongthecurvede“nedbythealgebraicequation x2Š 2 y3=9sothatthecoordinate x increasessteadily atarateof3unitsoflengthpersecond.Findtherateof changeofthecoordinate y whentheparticleisatthepoint ( x,y )=(5 2). (6)Thevelocityofaparticlemovingalongastraightlinesatis“es thecondition v2= c +2 b/s ,where a and b areconstants and s isthedistancetraveledbytheparticle.Showthatthe acceleration(therateofchangeinvelocitywithrespectto time)is a = dv/dt = Š b/s2. (7)Considertwolines y + x =2 a and y Š x =0,where a isanumber.Supposethataparticlemovesalongthe“rstlinetoward thepointofintersectionofthelinesataconstantspeed v1, whileanotherparticlemovesalongthesecondlineinthedirection away fromthepointofintersectionataconstantspeed v2. Findtherateofchangeofthedistancebetweentheparticles whenthe“rstandsecondparticlesareatthedistances s1and s2fromthepointofintersection,respectively.Inparticular, whatisthevalueofthisrateif s1= s2and v1= v2? (8)Thebladesofapairofscissorshavewidth2 h .Findtherate atwhichthepointofintersectionoftheedgesoftheblades

PAGE 87

21.LINEARAPPROXIMATIONSANDDIFFERENTIALS81 ismovingiftheanglebetweenthebladesdecreasesataconstantrate .Assumethatthebladesareattachedbyascrew throughthemidpointofeachblade(i.e.,throughapointthat isatdistance h fromtheedgesoftheblade).If h =4mmand = Š 2rad/s,howlongshouldthebladesbetoseethepoint ofintersectiongoingsuperluminal? (9)If y2=2 x and x isdecreasingsteadilyatarateof0.25units persecond,“ndhowfasttheslopeofthegraphischangingat thepoint( x,y )=(8 Š 4). (10)Apoolhasasphericalbottomofradius R andthemaximal depth h
PAGE 88

823.RULESOFDIFFERENTIATION iscalledthe linearapproximation or tangentlineapproximation .The conceptofthetangentlineapproximationisillustratedinFigure3.2. f ( x0)x0x0x0x yy L ( x ) y f ( x ) Figure3.2. Tangentlineapproximation.Inaneighborhoodof x0(aninterval[ x0Š ,x0+ ]),thetangent line y = L ( x )staysclosetothegraph y = f ( x ).Byreducingthewidthoftheinterval ,onecanmaketheerror ofthetangentlineapproximationassmallasdesired, i.e., | f ( x ) Š L ( x ) | forall x [ x0Š ,x0+ ]. Example 3.12 Usethelinearapproximationtoestimatethe value 3 92 Solution: (I)Consider f ( x )= x .Theclosestvalueof x to3 92atwhich thesquarerootcanbeevaluatedwithoutacalculatoris x0=4: f ( x0)=2.Notethetwoimportantstepshere:thechoiceof f ( x )suitablefortheproblemandthechoiceof x0nearwhich thelinearapproximationistobeused. (II)Since f( x )=( x )=1 / (2 x )and f(4)=1 / 4,byEquation (3.22)thelinearizationof x near x =4is L ( x )=2+ 1 4 ( x Š 4) (III)Thelinearapproximationmeansthatthevalue f (3 92)= 3 92isapproximatedbythevalue L (3 92): 3 92 L (3 92)=2+ 1 4 (3 92 Š 4)=1 98 Acalculatorgives 3 92 1 9799.Sotheapproximationerroris | 3 92 Š L (3 92) | < 1 02 10Š 4.Itiseasytoseethat L (4 08)=2 02

PAGE 89

21.LINEARAPPROXIMATIONSANDDIFFERENTIALS83 and | 4 08 Š L (4 08) | < 1 02 10Š 4.Innotationsgiveninthecaption ofFigure3.2,thisobservationcanbesummarizedbythefollowing inequality: | x Š L ( x ) | < 1 02 10Š 4= if | x Š 4 | 0 08= Inotherwords,thevaluesof x anditslinearizationdierbynomore than1 02 10Š 4forall3 92 x 4 08.Naturally,adecrease(increase) intheupperboundfortheerrorwouldleadtoadecrease(increase)in thesizeofaneighborhoodof x =4wherethelinearapproximationis accurate.21.2.AccuracyoftheLinearApproximation.Thepreviousexampleleads toaproblemthatisextremelyimportantinapplications:Givenanupperboundfortheerror ofthelinearapproximationofafunction f ( x ) near x0,“nd suchthat | f ( x ) Š L ( x ) | if | x Š x0| or,alternatively,given ,thatis,theneighborhood x0Š x x0+ estimatetheerror ofthelinearapproximation.Thefollowingtheorem isusefultoanswerthesequestions. Theorem 3.15 Supposeafunction f ( x ) istwicedierentiablein ( a,b ) suchthat | f( x ) | M forall x ( a,b ) andsomenumber M Let L ( x ) bethelinearizationof f ( x ) at x0 ( a,b ) .Then | f ( x ) Š L ( x ) |1 2M ( x Š x0)2,x ( a,b ) ThistheoremisasimplerversionoftheTaylortheorem,which isprovedinadvancedcalculuscourses.Thefollowingexampleillustratestheuseofthistheoremtoassesstheaccuracyofthelinear approximation. Example 3.13 Considerthelinearizationof sin x at x =0 .Find aninterval | x | inwhichtheerrorofthelinearapproximationdoes notexceed =0 5 10Š 3. Solution: (I)Since f( x )=(sin x )=cos x f(0)=1,and f (0)=0,the linearizationis L ( x )= x (II)InTheorem3.15,let a = Š and b = .Next,onehasto“nd M .Thesimplestwaytodothisistotakethemaximalvalue of | f( x ) | intheinterval | x | .Notethatthereshouldbe
PAGE 90

843.RULESOFDIFFERENTIATION error .Sosin x ismonotonicin | x | ,andhence | (sin x )| = | sin x | sin = M forall | x | .ByTheorem3.15, (3.23) | sin x Š x |1 2Mx21 2M2= if | x | With M =sin ,thesolutionoftheequation 2sin =2 = 10Š 3determines .Ananalyticsolutionofthisequationisimpossible.Soavalueof hastobefoundnumerically(actually, 0 100057). (III)Otherwise,onecanchoosealarger M ,forexample,sin x 1forany x .So M =1isacceptable,too.Thissimpli“es Equation(3.23): 2=10Š 3andhence 0 0362.Thisvalue of appearstobesmallerthanthatinthecase M =sin .It followsfromEquation(3.23)thatalargervalueof M leadsto asmaller .Sothisoptionshouldnotbeabused.ŽAgood M isnottoolargeandyetissimpleenoughtosolveEquation (3.23).Thisrequiressomeskillstoachieve. (IV)Agoodcompromiseistousetheinequalitysin .So thechoice M = alsoful“llstheconditionsofTheorem3.15. Equation(3.23)becomes 3=10Š 3and =0 1,whichisto becomparedwith =0 0362when M =1and 0 100057 when M =sin Theconverseproblemissimpler:Findanupperboundforthe errorofthelinearapproximationofsin x at x =0intheinterval | x | 0 2.Bymonotonicityofsin x intheinterval( Š / 2 ,/ 2), | (sin x )| = | sin x | sin(0 2)= M for | x | 0 2and,hence, | sin x Š x | =1 2M2=0 5 sin(0 2) (0 2)2 3 9734 10Š 3.21.3.Differential.Forarealvariable x ,the dierential dx isde“nedas anincrementof x .Itcanbegiventhevalueofanyrealnumberindependentlyofthevalueof x ;thatis, dx isconsideredasanindependent variable.So,witheveryrealvariable,onecanassociateanotherreal variable,calledthe dierential .Iftworealvariablesarerelated,the followingrulepostulatestherelationbetweentheirdierentials. Definition 3.4 Lettwovariables y and x berelatedas y = f ( x ) where f isadierentiablefunction.Thedierential dy = df ( x ) is de“nedbythelineartransformationof dx : (3.24) dy = df ( x )= f( x ) dx. Notethatthevariables x and dx ontheright-handsideareindependent variables.Equation(3.24)statesthat,ifthevariables y and x are

PAGE 91

21.LINEARAPPROXIMATIONSANDDIFFERENTIALS85 related,thenthedierential dy isnolongeranindependentvariable andisdeterminedby x and dx ;speci“cally, dy depends linearly on dx .21.4.GeometricalSigni“canceoftheDifferential.Put dx = x ,where x isarealnumber.Fix x = x0andconsideranincrementofthe variable y = f ( x )between x0+ x and x0: y = f ( x0+ x ) Š f ( x0)= f ( x0). f ( x0 x ) y L ( x ) y f ( x ) y df f f ( x0) x0L ( x0 x ) x0 x x Figure3.3. Geometricalsigni“canceofthedierential. Thedierential df ( x0)= f( x0) dx istheincrementalong thetangentline: df ( x0)= L ( x0+ x ) Š L ( x0), dx = x .Thedierential df ( x0)doesnotcoincidewiththe incrementofthefunction f ( x0)= f ( x0+ x ) Š f ( x0) = df ( x0).Onlywhen x becomesin“nitesimallysmall, x 0,doesitcoincideuptotermsthatgoto0faster than x ,i.e.,[ f ( x0) Š df ( x0)] / x 0as x 0. Thedierential df ( x0)= f( x0) x doesnotgenerallycoincidewith theincrement f ( x0).Forexample,put f ( x )= x2, x0=1, x =0 2, then f (1)=(1+0 2)2Š 1=0 44,whereas df (1)= f(1) x =2 0 2= 0 4.Sincethederivative f( x0)determinestheslopeofthetangentline L ( x )= f ( x0)+ f( x0)( x Š x0)tothegraph y = f ( x ),thedierential df ( x0)istheincrementofthelinearization y = L ( x )ofthefunction at x = x0intheinterval[ x0,x0+ x ];thatis,foraparticularvalue x = x0andanarbitrarychosenincrement dx = x df ( x0)= L ( x0+ x ) Š L ( x0)= f( x0) x. Thus, df ( x0) = f ( x0)becausethetangentlinedoesnotgenerally coincidewiththegraph.ThisobservationissummarizedinFigure3.3. Inparticular,thetangentlineapproximationcannowbestatedas f ( x0+ x ) L ( x0+ x )= f ( x0)+ df ( x0) ,dx = x.

PAGE 92

863.RULESOFDIFFERENTIATION Anintuitiveunderstandingofthedierentialstemsfromitsgeometricalinterpretation.Let x tendto0.Theratio y Š dy x = f ( x ) Š f( x ) x x = f ( x ) x Š f( x ) 0as x 0 becausebytheexistenceof f( x ), f ( x ) / x f( x )as x 0.This meansthatthedierence y Š dy mustgoto0fasterthan x .An increment x issaidtobe in“nitesimallysmall if( x )n, n> 1,can alwaysbeneglected.Soonemightthinkofdierentialsasin“nitesimal variationsofvariables.Fromthispointofview,thede“nition(3.24) looksrathernatural:In“nitesimalvariationsoftworelatedvariables mustberelatedlinearlyastheirhigherpowerscanalwaysbeneglected. Theconceptofthedierentialbecomesratherpracticalwhenonehasto establishrelationsbetweenvariationsofrelatedquantitiesinsituations whenthesevariationsmaybeviewedasin“nitesimal.21.5.InverseFunctionandtheDifferential.Theconceptofthedierentialoersasimplewayto“ndthederivativeofaninversefunction. Supposethatafunction f hastheinverse g = fŠ 1and g isdierentiable(conditionsunderwhich g existsandisdierentiablearestated laterintheinversefunctiontheoremofSection23).If y = f ( x ), thenthedierentialsarerelatedas dy = f( x ) dx .Ontheotherhand, x = fŠ 1( y )= g ( y )andhence dx = g( y ) dy .Since theratioofthe dierentialsisthederivative ,itfollowsthat dx dy = 1 dy dx g( f ( x ))= 1 f( x ) g( y )= 1 f( g ( y )) Forexample, f ( x )=tan x and g ( y )=tanŠ 1y .Then f( x )= 1 cos2x =1+( f ( x ))2= g( y )= 1 f( g ( y )) = 1 1+ y2, wheretherelation f ( g ( y ))= y hasbeenused.21.6.RelatedErrors.Everyphysicalquantityisknownonlywitha certaindegreeofaccuracy.Errorsareinherentintheveryprocessof takingmeasurements.Asapointoffact,avalueofaphysicalquantitygivenwithoutitsmeasurementerrordoesnotmakemuchsense; neithershouldonedrawanyconclusionfromdatawithoutaproper analysisoftheerrors.Oneoftheimportantpracticalapplicationsof thedierentialliesintheerroranalysisofrelatedquantities. Supposethereisarelationbetweentwoquantities y and x y = f ( x ).Let x bemeasuredwithanerror.Thismeansthefollowing.

PAGE 93

21.LINEARAPPROXIMATIONSANDDIFFERENTIALS87 After n measurements,onegets n values x1,x2,...,xn.Theaverageis x =( x1+ x2+ + xn) /n isviewedastheactualvalue.Themeasured valuesdeviatefromtheaveragebyamounts x1= x Š x1,..., xn= x Š xn.If x =max {| x1| ,..., | xn|} (i.e., x isthemaximalofthe absolutevaluesofallthedeviations),thenallmeasuredvalueslieinthe interval[ x Š x,x + x ].Thequantity x isan absoluteerror ofthe measurement,andonewritesforthemeasuredvalue x x toindicate theaveragemeasuredvalueanditsabsoluteerror.Theerror x is knownanddeterminedbytheveryprocessoftakingmeasurements. Astandardquestioninerroranalysis:Whatistheaccuracyofthe value y = f ( x )?Apparently, x and x areindependentvariablesas theerror x dependsonthewayinwhichthevariableismeasured (therearemoreandlessaccuratemethodswhichwouldleadtosmaller andhighervaluesof x independentlyofthevalueof x ).Naturally, onemightassumethattheerrorsaresmall;thatis,theyarein“nitesimalvariationsofmeasuredquantities.Thentheerrorsoftherelated quantitiesmustberelatedastheirdierentials!Thisisastandard assumptionoftheerroranalysis.Theabsolutevalueofthedierential | dy | representsan absolute errorof y = f ( x ),thatis, y = f ( x ) df ( x ). Thequantity | dy/y | iscalleda relativeerror Example 3.14 Whataretheabsoluteandrelativeerrorsofthe volumeofacubeifitssideis 10 0 1 cm ? Remark .Whenmeasuringthelengthbyarulerwithagrid,the measurementerrorshouldnotexceedtherulergridspacing(e.g.,a rulerwithamillimetergrid). Solution: Thevolume V andside x arerelatedas V = x3.So dV = 3 x2dx .Setting dx =0 1cmand x =10cm, dV =30cm3and V =1000 30cm3.Therelativeerroris dV/V =0 03or3%(note that dx/x =0 01,i.e.,only1%). Theerroranalysisforseveralrelatedquantitiesisstudiedinmultivariablecalculuscourses.Itisbasedontheconceptofthedierential offunctionsofseveralvariables.21.7.Exercises.(1)Findthelinearizationsofeachofthefollowingfunctionsatthe speci“edpoint: (i)cos x,x = / 4 (ii)tan x,x =0 (iii) eŠ x2,x =0

PAGE 94

883.RULESOFDIFFERENTIATION (iv)ln x,x = e (v) 1+ x,x =3 (2)Estimatetheerrorofthetangentlineapproximationofeach ofthefollowingfunctionsoveraninterval | x Š x0| forthe speci“edpoint x0andthewidth : (i) 1+ x,x0=3 =0 1 (ii)ln x,x =1 =0 2 (iii)tan x,x =0 = / 4 (3)Findthedierentialsofeachofthefollowingfunctions: (i) x (1 Š x2)3(ii)( y Š 2) / ( y +1) (iii) 1+ x2/x (iv)sin2t +cos( t2) (v)ln( x +2)+ xex(4)Usedierentials(thetangentlineapproximation)toestimate thefollowingnumbersandassesstheaccuracyoftheestimates: (i) 24 6 (ii) e0 08(5)Find dy intermsof x y ,and dx if (i) x + y =4 (ii) y3+ x3=2 xy (iii)cos( x +3 y )=sin( xy ) (6)Findanapproximateformulafortheareaofacircularringof radius r andwidth dr .Whatistheexactformula? (7)Findanapproximateformulaforthevolumeofaspherical shellofradius r andthickness dr .Assesstheaccuracyofthe approximationbystatingtheconditionon r and dr sothat therelativeerrordoesnotexceed =0 01(i.e.,1%). (8)Usedierentialsto“ndthederivativesoftheinversefunctions sinŠ 1x andcosŠ 1x .

PAGE 95

CHAPTER4 ApplicationsofDifferentiation 22.MinimumandMaximumValues Someofthemostimportantapplicationsofcalculusareoptimizationproblems.Anexampleofanancientoptimizationproblem:Aman canthrowastoneataspeedof v0.Atwhatangleshouldthestonebe throwninordertogetthemaximalrange?Anexampleofamodern optimizationproblem:Howcanoneoptimizetheinformation”owin theWorldWideWebtoavoidcrashesofservers?Manyoftheseproblemscanbereducedto“ndingthemaximalandminimalvaluesofa givenfunction. Definition 4.1(AbsoluteMaximumandMinimum) Afunction f hasanabsolutemaximumat c if f ( x ) f ( c ) forall x inthedomain D of f .Similarly,thevalue f ( c ) iscalledthe maximumvalue of f .A function f hasanabsoluteminimumat c if f ( x ) f ( c ) forall x in thedomain D of f .Thevalue f ( c ) iscalledthe minimumvalue of f Themaximumandminimumvaluesof f arecalledthe extremevalues of f Forexample,thefunction f ( x )=cos x attainsitsmaximumvalue 1at x =2 n ,where n =0 1 2 ,... ,anditsminimumvalue Š 1at x = +2n .Afunctiondoesnotalwayshaveamaximumorminimum value.Forinstance,thefunction f ( x )=1 /x de“nedforallreal x =0 hasneithermaximumnorminimumvaluebecause,foranyreal M ,one canalways“nd x suchthat f ( x ) >M (0
PAGE 96

904.APPLICATIONSOFDIFFERENTIATION relative)minimumat c if f ( x ) f ( c ) forall x insomeopeninterval containing c Example 4.1 Doesthefunction f ( x )= x3Š x = x ( x2Š 1) have anabsolutemaximum(minimum)valueandrelativemaxima(minima) ontherealaxis? 3 3 1 1 1 y x Figure4.1. Graphofthefunction f ( x )= x3Š x = x ( x2Š 1).Itdoesnothaveanabsolutemaximumor minimumvalue.However,itdoeshavearelativemaximumat x = Š 1 / 3andarelativeminimumat x = 1 / 3. Solution: (I)Thefunctionhasneitheranabsolutemaximumnoranabsoluteminimumbecauseitgrowsunboundedlywithincreasing x anditdecreasesunboundedlyas x attainslargernegative values. (II)Thefunctionvanishesatthreepoints x =0 1.Itcanhave relativeminimaandmaximabetweenitszerosbecausethe valuesof f areboundedfromaboveandbelow: | f ( x ) || x |3+ | x | 2for | x | 1,thatis, Š 2 f ( x ) 2if Š 1 x 1. (III)Considertheopeninterval x (0 1).Thefunctionisstrictly negativeinitandboundedfrombelow: M
PAGE 97

22.MINIMUMANDMAXIMUMVALUES91 (IV)Similarly, f isstrictlypositivein( Š 1 0)andboundedfrom above0
PAGE 98

924.APPLICATIONSOFDIFFERENTIATION existenceofitsrelativemaximumandminimum!Thefollowingexampleillustratesthepoint.Considerthefunction f ( x )=2 x if x [0 1) and f ( x )=1if x [1 2].Sothefunctionisde“nedontheclosed interval[0 2]andboundedfromabove f ( x ) 2).An attempttoestablishtheexistenceofamaximumvalueof f bylowering M fails!Indeed,thelowestupperboundis M =2,butthereisno c suchthat f ( c )=2.Thevaluesof f approach2as x approaches1from theleft,but f (1)=1!Foranypositive > 0, f (1 Š )
PAGE 99

22.MINIMUMANDMAXIMUMVALUES93 interval a 0suchthat c
PAGE 100

944.APPLICATIONSOFDIFFERENTIATION (IV)Afunctionde“nedonaclosedinterval[ a,b ]canhaveitsabsolutemaximumorminimumattheendpoints.When“nding theabsolutemaximumandminimumvalues,thevaluesof f atthecriticalpointsmustbecomparedwith f ( a )and f ( b ). Thelargest(smallest)ofthemistheabsolutemaximum(minimum)value. Example 4.2 Ifastoneisthrownatas peed v0m/s andanangle withthehorizontalline,thenitstrajectoryisaparabola: (4.4) y = x tan Š x2g 2 v2 0cos2 where y isthestoneheight(verticalposition), x isthehorizontalposition(allthepositionsareinmeters),and g =9 8 m/s2isaconstant universalforallobjectsnearthesurfaceoftheEarth(thefree-fallacceleration).ThisisaconsequenceoftheNewtonssecondlaw.Atwhat angleshouldonethrowastonetoreachthemaximalrangeatagiven speed v0? Solution: (I)Therangeasafunctionoftheangle hastobefound“rst. Thestonelandswhenitsheight y vanishes.Theequation y =0hastwosolutions x =0(naturally,thisiswherethe stonewasthrown)and x = L ( ),where L ( )= 2 v2 0 g tan cos2 = 2 v2 0 g sin cos = v2 0 g sin(2 ) (II)Therange L ( )isadierentiablefunctionof sothevalues of atwhich L attainsitsextremevaluesmaybefoundfrom theequation L( )=0= v2 0 g 2cos(2 )=0= cos(2 )=0 Thisequationhascountablymanysolutions2 = / 2+ n where n isanyinteger.Butintheintervalofthephysical valuesof [0 ,/ 2],ithasonlyonesolution = / 4.Since sin(2 / 4)=1(theabsolutemaximumofthesine), L attains itsmaximumvalueat = / 4;thatis,therangeismaximal, Lmax= v2 0/g ,whenastoneisthrownat45. Remark .Theconclusionintheprecedingexampleisindependent ofthestonesmassanditsinitialspeed v0.Inreality,forlargervalues of v0,likeaprojectileshotbyagun,thetrajectorywoulddeviateabit

PAGE 101

22.MINIMUMANDMAXIMUMVALUES95 fromtheparabola(duetofrictionwiththeair).Sotheoptimalangle woulddeviateabitfrom / 4.Thedeviationwouldalsodependonthe massandtheinitialspeed.Therangeoptimizationproblembecomes moreinvolvedandwouldrequirethetheoryof dierentialequations Itshouldalsobenotedthattheangleatwhichthemaximalrangeis attaineddependsontheinitialheightatwhichthestoneisthrown.So theanglewouldbedierentfrom45when,forexample,thestoneis thrownfromacli.22.3.Exercises.(1)Examinethefollowingfunctionsformaximaandminima. Drawthegraphineachcase. (i) y = x2Š 5 x +3 (ii) y =2 x3Š 3 x2+6 x Š 3 (iii) y = x2+16 /x (iv) y = x2Š 1 /x2(v) y =4 x/ ( x2+1) (vi) y =sin x +cos x (vii) y = xex(2)Findallcriticalpointsofthefollowingfunctionsanddetermine whetherthereisalocalmaximum,alocalminimum,ornone oftheaboveateachcriticalpoint. (i) y = | x Š 3 | (ii) y = | x2Š 4 | +2 x (iii) y = | x3Š 1 | (iv) y = | sin(2 x ) | (v) y = | x3| ex(vi) y =( x Š 1)1 / 3(vii) y = x ( x +1)2 / 3(viii) y =(1 Š x2)3 / 2(3)Findextremevaluesofthefollowingfunctionsonthespeci“ed intervalorshowthatsuchvaluesdonotexist. (i) y = x4Š 4 x2, Š 3
PAGE 102

964.APPLICATIONSOFDIFFERENTIATION (6)Findthemaximumareaofanisoscelestrianglewith“xed perimeter p (7)Letthesumoftwonumbersbe s .Findthenumbersineach ofthefollowingcases: (i)Thesumoftheirsquaresisaminimum. (ii)Thesumoftheircubesisaminimum. (iii)Theirproductisamaximum. (iv)Thedierencebetweenoneandthereciprocaloftheother isamaximum.23.TheMeanValueTheorem Theorem 4.3(RollesTheorem) Let f beafunctionthatsatis“es thefollowingthreehypotheses: (I) f iscontinuousontheclosedinterval [ a,b ] (II) f isdierentiableontheopeninterval ( a,b ) (III) f ( a )= f ( b ) Thenthereisanumber c in ( a,b ) suchthat f( c )=0 y x a cb Figure4.3. Rollestheorem.Thecontinuityof f guaranteestheboundednessof f .Sothegraphof f liesbetweentwohorizontallines.Byloweringanupperbound orincreasingalowerbounduntiloneofthehorizontal lines(orboth)touchesthegraphandbecomesitstangentline,dierentiabilityof f ensurestheexistenceof thetangentlineateverypointin( a,b ).Theslopeofthe horizontaltangentlineis0andsoisthederivativeat thatpoint.

PAGE 103

23.THEMEANVALUETHEOREM97 Thistheoremprovidesausefulmethodtoprovetheexistenceofalocalmaximumorminimumofafunction f whenanalyticsolutionsofthe equation f( x )=0arehardto“nd.Infact,ithasalreadybeenusedin Example4.1:Thefunction f ( x )= x3Š x ontheintervals[ Š 1 0],[0 1], [ Š 1 1]satis“esthehypothesesofRollestheorembecause f ( 1)= f (0)=0.TheprooffollowscloselytheargumentsofExample4.1. ProofofTheorem4.3. (I)If f ( x )= f ( a )= k isaconstantfunction,then f( x )=0 everywhere. (II)Let f ( x ) >f ( a )forsome x ( a,b )(cf.Example4.1for x [ Š 1 0]).Since f iscontinuous,theextremevaluetheorem applies,andtherefore f hasamaximumin[ a,b ].Since f ( a )= f ( b ),themaximalvaluemustbeattainedat c ( a,b ).ByFermatstheorem, f( c )=0because f isdierentiablein( a,b ). (III)If f ( x ) 0.Bycontinuity, f hastotakeallintermediate valuesbetween Š 4and2(theintermediatevaluetheorem).So f has atleast onerootin( Š 1 1). (II)Supposeithastworoots a and b ,thatis, f ( a )= f ( b )=0. Then,byRollestheorem, f( x )hastovanishsomewherein ( a,b ).Butthisisnotpossiblebecause f( x )=5 x4+3 x2+1 > 0 forany x .Thus, f hastheonlyrealroot. Theorem 4.4(TheMeanValueTheorem) Let f beafunction thatsatis“esthefollowinghypotheses: (I) f iscontinuousontheclosedinterval [ a,b ] (II) f isdierentiableontheopeninterval ( a,b ) Thenthereisanumber c ( a,b ) suchthat (4.5) f( c )= f ( b ) Š f ( a ) b Š a or f ( b ) Š f ( a )= f( c )( b Š a ) .

PAGE 104

984.APPLICATIONSOFDIFFERENTIATION Thegeometricalinterpretationofthetheoremissimple(see Figure4.4).Considerthelinethroughthepoints( a,f ( a ))and( b,f ( b )). Itsslopeis( f ( b ) Š f ( a )) / ( b Š a ).Thetheoremassertstheexistence ofapointwherethegraph y = f ( x )hasatangentlinewiththesame slope(cf.Equation(4.5))(as f( c )istheslopeofthetangentlineat x = c ).Letusturntoaformalproof. x y c ( a, f ( a )) ( b, f ( b )) Figure4.4. Meanvaluetheorem.Thesecantlineof thegraphof f throughthepoints( a,f ( a ))and( b,f ( b )) hastheslopetan =[ f ( b ) Š f ( a )] / ( b Š a ),where istheanglebetweenthesecantlineandthehorizontal line.If f doesnotcoincidewiththesecantline,thennear x = a theslopeofthetangentlinedoesnotcoincidewith tan .Herethecasewhenthisslopeisgreaterthantan isshown.Thenthegraphof f liesabovethesecantline near x = a .Butthegraphhastoreturntothesecantline again.Nearthepointwherethegraphandthesecant linesmeetagain,thetangentlinehastohaveasmaller slopethantan .Soatsomepoint c thetangentlinehas tobeparalleltothesecantline,meaningthat f( c )= tan ProofofTheorem4.4. (I)Considerthelinethroughthepoints( a,f ( a ))and( b,f ( b )).Its equationis

PAGE 105

23.THEMEANVALUETHEOREM99 y = L ( x )= f ( a )+ f ( b ) Š f ( a ) b Š a ( x Š a ) (4.6) L ( a )= f ( a ) ,L ( b )= f ( b ) Next,considerthefunction (4.7) h ( x )= f ( x ) Š L ( x )= f ( x ) Š f ( a ) Š f ( b ) Š f ( a ) b Š a ( x Š a ) Itsvaluesdeterminethedeviationofthegraph y = f ( x )from thesecantline y = L ( x )ontheclosedinterval[ a,b ]. (II)Thefunction h ( x )satis“esthethreehypothesesofRollestheorem.First,itiscontinuouson[ a,b ]asthesumoftwocontinuousfunctions f ( x )and Š L ( x )(alinearfunctioniscontinuous).Second,itisdierentiableon( a,b )asthesumoftwo dierentiablefunctions: (4.8) h( x )= f( x ) Š f ( b ) Š f ( a ) b Š a ,x ( a,b ) Finally,by(4.6)and(4.7), h ( a )= f ( a ) Š L ( a )=0and h ( b )= f ( b ) Š L ( b )=0,thatis, h ( a )= h ( b ). (III)ByRollestheorem,thereisanumber c ( a,b )suchthat h( c )=0= f( c )= f ( b ) Š f ( a ) b Š a whereEquation(4.8)hasbeenused. Example 4.4 Aspeedingcarwaspulledoveronaninterstateroad andastatetroopergaveawarningtothedriver.Forty“veminutes laterandpassed65milesontheroad,thecarstoppedatarestarea. Anotherstatetrooperapproachedthedriverandissuedas peedingticket, claimingthatthedriverex ceeded 86 milesperhour.Wasthetroopers claimcorrect? Solution: Let s ( t )bethedistancetraveledbythecarafteritwas pulledoverthe“rsttime.Therateofchange s( t )= v ( t )isthespeed ofthecaratanymomentoftime.Thefunction s ( t )isde“nedbetween t =0and t =45min=0 75hrsothat s (0)=0and s (0 75)=65mi. Itisdierentiableas s( t )isthecarspeed!Bythemeanvaluetheorem, thereisatimemoment t = c (0 0 75)when s( c )= v ( c )= s (0 75) Š s (0) 0 75 Š 0 = 65 0 75 86 7mi / hr Thespeedingticketisjusti“ed.

PAGE 106

1004.APPLICATIONSOFDIFFERENTIATION Foranytwomomentsoftime a and b ,theratio( s ( b ) Š s ( a )) / ( b Š a ) istheaveragespeedonthetimeinterval[ a,b ].Themeanvaluetheorem simplystatesthatamovingobjectalwaysattainsitsaveragespeedat leastatonemomentoftimebetween a and b .So,ifattimemoment b theobjectappearstobetravelingslowerthanitsaveragespeed,prior tothatitmusthavebeentravelingfasterthanitsaveragespeed. Example 4.5 Supposethederivative fexistsandisboundedon ( a,b ) ,thatis, m f( x ) M .If f ( a ) isgiven,howsmallandhow largecan f ( b ) possiblybe? Solution: Bythemeanvaluetheorem,thereisa c ( a,b )suchthat f ( b )= f ( a )+ f( c )( b Š a ).Since m f( c ) M f ( a )+ m ( b Š a ) f ( b ) f ( a )+ M ( b Š a ) Thisequationiseasytounderstandwiththehelpofamechanical analogy:Howfarcanacartravelintime b Š a ifitsspeedisnotlower than m ,butcannotexceed M ? 23.1.PropertiesoftheFirstDerivative.Thederivativeofaconstant functionvanishes.Howabouttheconverse?Thefollowingtheorem answersthisquestion. Theorem 4.5 If f( x )=0 forall x inaninterval ( a,b ) ,then f is constanton ( a,b ) Proof. Takeanytwonumbers x1and x2between a and b .Bythe meanvaluetheorem,thereisanumber c between x1and x2suchthat f ( x1) Š f ( x2)= f( c )( x1Š x2).Byhypothesis, f( c )=0forany c Thus, f ( x1) Š f ( x2)=0or f ( x1)= f ( x2)forany x1and x2in( a,b ); thatis, f isconstant. Thehypothesisthat f( x )=0ina singleinterval iscrucial.For example,thesignfunction f ( x )=1if x> 0,and f ( x )= Š 1if x< 0,haszeroderivativeatanypointofitsdomain,butitisnot constant.Thekeypointtonoteisthatthedomainisnotasingle interval,butaunionoftwodisjointintervals( Š 0)and(0 )So themeanvaluetheoremisnotapplicabletoanyintervalcontaining x =0.Thisexampleiseasilyextendedtothecasewhenthedomainis anycollectionofdisjointintervalsand f takesdierentconstantvalues ondierentintervals. Corollary 4.1 If f( x )= g( x ) forall x inaninterval ( a,b ) then f Š g isconstant,thatis, f ( x )= g ( x )+ k ,where k isaconstant.

PAGE 107

23.THEMEANVALUETHEOREM101 Proof. Let h ( x )= f ( x ) Š g ( x ).Since h= fŠ g=0in( a,b ), h isconstant,andtheconclusionfollows. Thesingofthe“rstderivativede“nesintervalsofgrowthanddecreaseofafunction. Theorem 4.6(Increasing-DecreasingTest) (I) If f> 0 onaninterval,then f isincreasingonthatinterval. (II) If f< 0 onaninterval,then f isdecreasingonthatinterval. Proof. Take any twonumbers x1and x2intheintervalsothat x1f ( x2).Since f isdierentiable,themeanvaluetheoremstates thatthereisanumber c between x1and x2suchthat (4.9) f ( x2) Š f ( x1)= f( c )( x2Š x1) If f> 0,thenitfollowsfrom(4.9)that f ( x2) Š f ( x1) > 0because,by assumption, x2>x1;thatis,thefunctionisincreasing.Similarly,for f< 0, f ( x2) Š f ( x1) < 0,andthefunctionisdecreasing. Theincreasing-decreasingtestisfurtherillustratedontheinteractivewebsiteathttp://www.math.u”.edu/ mathguy/ufcalcbook/inc dec.html.23.2.TheInverseFunctionTheorem.ABabyVersion.Givenafunction f itsinversefunctionexistsif f isone-to-oneasexplainedinSection5. AsimpleruletocalculatethederivativeoftheinversefunctionwaspresentedinSections17and21.However,theveryquestionofwhether theinversefunctionisactuallydierentiablehasnotbeenaddressed.It appearsthatthequestionsabouttheexistenceoftheinversefunction anditsdierentiabilitycanbeansweredbylookingatthesignofthe derivative. Theorem 4.7 (ABabyVersionoftheInverseFunctionTheorem). Let f beafunctionon Š a 0 (or f( x ) < 0 )forall x ( a,b ) .Then f hastheinverse g = fŠ 1on ( c,d ) forsome Š c 0.Theothercaseissimilar.Bythe increasing-decreasingtest, f ( x1)
PAGE 108

1024.APPLICATIONSOFDIFFERENTIATION y f ( x ) y f 1( x ) f f f 1 x Figure4.5. Inversefunctiontheorem.Anincreasing function f f> 0,isone-to-oneandhencehastheinverse g = fŠ 1.Thegraphsof f and g areobtainedfrom oneanotherbythere”ectionabouttheline y = x .Asecantlineofthegraphof f withtheslopetan = f/ x ismappedonthesecantlineofthegraphof g withthe slopetan = fŠ 1/ f bythisre”ection.Theangles and arerelatedas + = / 2andhence tan =1 / tan .Inthelimit x 0,whichalso implies f 0,thesecantlinesbecomethetangent linessothattan f( x )andtan g( y ),where y = f ( x ).Hence, g( f ( x ))=1 /f( x ). 2 .Thefunction f iscontinuouson( a,b )becauseitisdierentiable on( a,b ).Bytheintermediatevaluetheorem, f takesallintermediate valuesbetween f ( x1)and f ( x2) >f ( x1)foranyinterval[ x1,x2]in ( a,b ).Thisshowsthattherangeof f isasingleinterval( c,d )for some Š c
PAGE 109

23.THEMEANVALUETHEOREM103 Iwiththecorrespondingendpoints f ( x )= y and f ( x + x )= y + y where y = f .Put g = g ( y + y ) Š g ( y ).Since g mapsthe interval Ionto I g = x .Thelimit y 0impliesthat x 0. Therefore, g/ y = x/ y =1 / ( y/ x ) 1 /f( x )as x 0, whichshowsthat g( y )exists,and g( y )=1 /f( x ),where y = f ( x ).A graphicillustrationisgiveninFigure4.5. Forexample,theexponentialfunction f ( x )= exhasthederivative f( x )= ex> 0forall Š 0,then f ( a )isarelativeminimumvalueof f Remark .Thederivativeat a isde“nedastherightlimit, f( a )=limh 0+f ( a + h ) Š f ( a ) h (4)Let f beafunctionwhosedomainofde“nitionistheclosed interval[ a,b ],whichiscontinuousat a and b ,whichisdierentiableintheinterval( a,b ),andforwhich f( a )=0ordoes notexist.Showthatifthereisapoint c suchthat f( x ) > 0 for a
PAGE 110

1044.APPLICATIONSOFDIFFERENTIATION (6)Investigatewhethereachofthefollowingequationshassolutions.Ifso,foreachsolution,“ndanintervalinwhichno othersolutionlies. (i)cos x = x2(ii) ex=4 Š x4Š x2(iii)ln x =4 Š x2(iv)2tan x =tanŠ 1x (v) x6+3 x4+3 x2+ x Š 7=0 (7)Provethegeneralizedmeanvaluetheorem: If f and g arefunctionsthataredierentiableinaninterval [ a,b ]andif g( x ) =0forany x in[ a,b ],thenthereisanumber c ,with a
PAGE 111

24.THEFIRSTANDSECONDDERIVATIVETESTS105 24.1.TheFirstDerivativeTest.Bytheincreasing-decreasingtest, f is increasingonintervalifitsderivativeispositive,and f isdecreasing onanintervalifitsderivativeisnegative.Suppose fiscontinuous suchthat f( a )= m and f( b )= M .Then,ontheinterval[ a,b ], fmusttakeallintermediatevaluesbetween m and M .Suppose m< 0 and M> 0or m> 0and M< 0,thatis,thederivativechanges itssignontheinterval[ a,b ],then fmustvanishbetween a and b Thismeansthat f hasacriticalpoint af ( c + h )forsomesmallpositive h .Wecanthenconclude that f attainsitslocalmaximumat c .Similarly,ifthederivative fchangesfromnegativetopositiveat c ,then f changesfromdecreasing toincreasingat c f ( c Š h ) >f ( c )and f ( c ) f ( c ) >f ( c + h );thatis,ineithercasethefunction f hasneitheralocalminimumnoralocalmaximum.The“ndingsare summarizedinthefollowingtheorem. Theorem 4.8(TheFirstDerivativeTest) Supposethat c isacriticalpointofacontinuousfunction f (I) If fchangesfrompositivetonegativeat c ,then f hasalocal maximumat c (II) If fchangesfromnegativetopositiveat c ,then f hasalocal minimumat c (III) If fdoesnotchangesitssignat c ,then f hasneitheralocal maximumnoralocalminimumat c Itisimportanttonotethattheveryexistenceof fat c is not requiredinthe“rstderivativetest.Recallthede“nitionofacritical point( f( c )=0or f( c )doesnotexist).Infact,intheprecedingproof ofthe“rstderivativetest,thecondition f( c )=0canbedropped becauseallthatisneededtoapplytheincreasing-decreasingtestis thesignofthederivative f( x )for xc .Forexample, f ( x )= | x | .Then f( x )= Š 1for x< 0(thefunctionisdecreasing) and f( x )=1for x> 0(thefunctionisincreasing).Hence, f ( x ) hasaminimumat x =0,eventhough fdoesnotexistat x =0.The continuityhypothesis isalsocrucial.Considerthefunction f ( x )=1 /x2for x =0and f (0)=0.Then f( x )= Š 2 /x3for x =0and f(0)does

PAGE 112

1064.APPLICATIONSOFDIFFERENTIATION notexist.So x =0isacriticalpoint.Thefunctionisincreasingfor x< 0because f> 0,anditisdecreasingfor x> 0because f< 0. However, f hasnomaximumat x =0because f isdiscontinuousat x =0.Infact,itattainsitsabsoluteminimumat x =0! Thereareplentyofmechanicalanalogiesofthe“rstderivativetest. Let H ( t )betheheight(relativetotheground)ofastonethrownupwardasafunctionoftime t .Atthebeginning,thestonemovesupward so H> 0(theheightisincreasing).Whenthestonecomesbackto theground,itmovesdownwardso H< 0(theheightisdecreasing). Naturally,atsomemomentoftime,thestonehastoreachthemaximalheight.Analyzethemotionofapendulum(orasee-saw)fromthis pointofview!Theheightwouldhavetwomaximaandoneminimum. Example 4.6(Example4.1Revisited) Findalllocalmaximaand minimaof f ( x )= x3Š x andtheintervalsonwhichthefunctionis increasingordecreasing(thefunctionisdepictedinFigure4.1). Solution: (I)Since f isdierentiable(itisapolynomial),allitscritical pointssatisfytheequation f( x )=3 x2Š 1=3 x Š 1 / 3 x +1 / 3 =0 Hence,thecriticalpointsare c1= Š 1 / 3and c2=1 / 3. (II)For x 0( f is increasingon( Š ,c1)).For c1c2,theproduct( x Š c1)( x Š c2)ispositive(astheproduct oftwopositivenumbers),andhence f> 0( f isincreasingon ( c2, )). (III)Thederivativechangesfrompositivetonegativeat c1.Therefore, f hasalocalmaximumat c1.Thederivativechanges fromnegativetopositiveat c2.Therefore, f hasalocalminimumat c2. 24.2.PropertiesoftheSecondDerivative:In”ectionPoints.Definition 4.4(Concavity) Thegraphofafunction f iscalled concaveupward onaninterval I ifitliesaboveallofitstangentlines on I .Thegraphiscalled concavedownward on I ifitliesbelowallof itstangentlineson I .

PAGE 113

24.THEFIRSTANDSECONDDERIVATIVETESTS107 Notethatthenotionofconcavityimpliesthat f isdierentiable (otherwise,thetangentlinesdonotexist).If f istwicedierentiable, thentheconcavityisdeterminedbythesignofthesecondderivative f.Supposethatthegraphof f isconcaveupwardon I .Considerthe tangentlinesattwopoints c and c + h in I : L1( x )= f ( c )+ f( c )( x Š c ) ,L2( x )= f ( c + h )+ f( c + h )( x Š c Š h ) Thegraphof f liesabovethelines L1and L2,thatis, f ( x ) Š L1( x ) > 0 and f ( x ) Š L2( x ) > 0forall x in I .Putting x = c inthelastinequality and x = c + h intheformerone,weobtain f ( c ) Š L2( c )= f ( c ) Š f ( c + h )+ f( c + h ) h> 0 f ( c + h ) Š L1( c + h )= f ( c + h ) Š f ( c ) Š f( c ) h> 0 Thesumoftheright-handsidesoftheseinequalitiesispositiveasthe sumoftwopositivenumbers: (4.10) h [ f( c + h ) Š f( c )] > 0= f( c + h ) Š f( c ) h > 0 wherethe“rstinequalityhasbeendividedbyapositivenumber h2. Inequality(4.10)istrueforany h .Therefore,bytakingthelimit h 0,wecanconcludethat f( c ) > 0ifthegraphisconcaveupward. Inequality(4.10)showsthat f( c + h ) >f( c )for h> 0and f( c ) > f( c + h )for h< 0.Inotherwords,thederivative f,ortheslopeofthe tangentlineofthegraphof f ,increasesfortheupwardconcavity,and hence( f)= fmustbepositivebytheincreasing-decreasingtest. Similarly,thedownwardconcavityimpliesthat fisnegative.Itturns outthattheconverseisalsotrue. Theorem 4.9(TheConcavityTest) Let f betwicedierentiable onaninterval I (I) If f( x ) > 0 forall x in I ,thenthegraphof f isconcave upwardon I (II) If f( x ) < 0 forall x in I ,thenthegraphof f isconcave downwardon I Howdoesthegraphof f looknearapoint c where f( c )=0?There arefourpossibilities.First, f( c h ) > 0forsomesmall h> 0.This meansthatthegraphisconcaveupwardtotheleftandrightof c .As anexample,consider f ( x )= x4.Second, f( c h ) < 0.Thisimplies thatthegraphisconcavedownwardtotheleftandrightof c .Asan example,take f ( x )= Š x4.ThesetwocasesaredepictedinFigure4.6. Third, f( c Š h ) > 0and f( c + h ) < 0,thatis,theconcavitychanges fromupwardtodownward(e.g., f ( x )= Š x3).Fourth, f( c Š h ) < 0

PAGE 114

1084.APPLICATIONSOFDIFFERENTIATION x0x0x x f" ( x ) > 0 f" ( x ) < 0 y y Figure4.6. Concavitynearapoint x0atwhich f( x0)=0.Thegraphisconcavedownwardif f( x ) > 0 for x>x0and xx0and x 0,thatis,theconcavitychangesfromdownwardto upward(e.g., f ( x )= x3). Definition 4.5(In”ectionPoint) Apoint P onthegraph y = f ( x ) iscalledan in”ectionpoint if f iscontinuousthereandthegraph changesfromconcaveupwardtoconcavedownwardorfromconcave downwardtoconcaveupward. Let c beacriticalpointof f .Suppose fiscontinuousnear c Whatcan f( c )tellusaboutthenatureofthecriticalnumber(local minimumormaximum)?Therearethreepossibilities.First, f( c ) > 0. Thismeansthat f( x ) > 0forall x insomeneighborhoodof c (by thecontinuityof f).Hence, f isconcaveupwardnear c ;thatis, itsgraphlies above thetangentlineat c ,whichisahorizontalline because f( c )=0.So f musthavealocalminimum.Similarly,the condition f( c ) < 0impliesthattheconcavityisdownwardnear c and f hasalocalmaximum.If f( c )=0,thentheconcavitymay ormaynotchangeat c asdiscussedearlier.Thefunctionmayhave alocalmaximum,alocalminimum,oranin”ectionpoint;thatis,no conclusionaboutthenatureofthecriticalpointcanbereached.

PAGE 115

24.THEFIRSTANDSECONDDERIVATIVETESTS109 y x x0f" > 0 f" < 0 Figure4.7. Concavitynearapoint x0atwhich f( x0)=0(continuedfromFigure4.6).In”ectionpoint. Thesecondderivativechangesitssignatthein”ection point x = x0.Theconcavityofthegraphof f also changesatthein”ectionpoint.Suchalocalbehavior canbeillustratedby f ( x )= c + a ( x Š x0)+( x Š x0)3, where c and a areconstants,sothat f( x )=6( x Š x0) changesitssignat x0.Notethat f( x0)= a ;i.e., a de“nestheslopeofthegraphat x = x0. Theorem 4.10(TheSecondDerivativeTest) Suppose fiscontinuousnear c (I) If f( c )=0 and f( c ) > 0 ,then f hasalocalminimumat c (II) If f( c )=0 and f( c ) < 0 ,then f hasalocalmaximumat c (III) If f( c )=0 and f( c )=0 ,then f mayhavealocalmaximum, alocalminimum,oranin”ectionpoint. InExample4.6,thefunction f ( x )= x3Š x isshowntohavetwo criticalpoints: x = 1 / 3asdepictedinFigure4.1.Since f( x )=6 x f( Š 1 / 3)= Š 2 3 < 0(alocalmaximum)and f(1 / 3)=2 3 > 0 (alocalminimum).Thefunctionalsohasanin”ectionpointat x =0: f( x )=6 x< 0if x< 0and f( x )=6 x> 0if x> 0.Notethatan in”ectionpointmay not berelatedacriticalpoint!Inotherwords,the tangentlineatanin”ectionpointcanhave any slope.Intheexample discussed, f(0)= Š 1(seealsoFigure4.7).

PAGE 116

1104.APPLICATIONSOFDIFFERENTIATION yf" ( c ) < 0 f" ( c ) > 0 f" ( c ) = 0 y xx x c c c y(inflection) Figure4.8. Secondderivativetest.Thegraphof f has ahorizontaltangentlineatacriticalpoint, f( c )=0.If f( c ) < 0,thegraphisconcaveupwardnear x = c and f hasalocalmaximumat c (leftpanel).If f( c ) > 0, thegraphisconcavedownwardnear x = c and f has alocalminimumat c (middlepanel).If f( c )=0,the graphmayhaveanin”ectionpointwhen fchangesits signat c (rightpanel),but fmaynotchangeitssign at c ,andhencethebehaviordepictedintheleftand middlepanelsisalsopossibleinthecase f( c )=0.The secondderivativetestisinconclusive.Thefunctionmay havealocalminimum,alocalmaximum,oranin”ection. ExamplesaregiveninthecaptionsofFigures4.6and4.7 (with a =0).24.3.Exercises.(1)Findallcriticalpointsofthegivenfunctioninitsdomain.Use eitherthe“rstorsecondderivativetesttodeterminewhether thereisalocalmaximum,alocalminimum,oranin”ection ateachcriticalpoint. (i) f ( x )=8 x3Š 9 x2+1 (ii) f ( x )= x4Š 6 x2+8 x +2 (iii) f ( x )= x2Š 2 /x (iv) f ( x )= x 1 Š x2(v) f ( x )= x x Š x2(vi) f ( x )=sin(2 x )+ x (vii) f ( x )=ln x Š x3(viii) f ( x )=x2Š 4 x +1(ix) f ( x )=tan x (x) f ( x )=( exŠ eŠ x) / ( ex+ eŠ x) (xi) f ( x )= x2 / 3(1 Š x )1 / 3

PAGE 117

25.TAYLORPOLYNOMIALSANDTHELOCALBEHAVIOR111 (2)Forwhichvaluesof a and b willthefunction h ( x )= ax2+ b/x3haveahorizontaltangent( x,y )=(1 5)?Does h havea relativemaximumorminimumat(1 5)? (3)Atapoint P inthe“rstquadrantonthecurve y =7 Š x2, atangentisdrawn,meetingthecoordinateaxesat A and B Findthepositionof P thatmakesthedistancebetween A and B aminimum. (4)Water”owsoutofahemisphericalbasinthroughaholeatthe bottomsothatthevolumeofthewaterremainingatanytime decreasesatarateproportionaltothesquarerootofthedepth ofthewaterremaining.Provethatthelevelofthewaterfalls mostslowlywhenthedepthistwo-thirdsoftheradiusofthe basin. Hint :Thevolumeofasphericalsegmentofonebaseis V = 3 (3 Rh2Š h3) where R istheradiusofthesphereand h istheheightofthe segment. (5)Afunction f issuchthat fiscontinuousontheinterval[ a,b ]. Theequation f ( x )=0hasthreedierentsolutionsintheopen interval( a,b ).Showthattheequation f( x )=0hasatleast onesolutionin( a,b ). (6)If f isafunctionthathasasecondderivativeateachpoint ofaninterval[ a,b ],showthatthereisanumber c insidethis intervalsuchthat f ( b )= f ( a )+ f( a )( b Š a )+ f( c ) 2 ( b Š a )2. Hint :Considerthefunction h de“nedby h ( x )= f ( x ) Š f ( a ) Š f( a )( x Š a ) Š k ( x Š a )2, wherethenumber k issochosenthat h ( b )=0.25.TaylorPolynomialsandtheLocalBehaviorofaFunction Thetangentlineapproximation L ( x )isthebestlinearapproximationof f ( x )near x = a because L ( x )and f ( x )havethesamerate ofchangeat a .Intheprevioussection,itwasshownthatthesecond derivativeat a providesimportantinformationaboutthebehaviorof f ( x )near a ,namelytheconcavity.Thetangentline L ( x )hasnoconcavityas L( x )=0.Thequestionariseswhetherthereisasystematic methodtoimprovetheaccuracyofthetangentlineapproximationto

PAGE 118

1124.APPLICATIONSOFDIFFERENTIATION capturemoreessentialfeaturesofthebehaviorof f ( x )near a (i.e.,the localbehavior of f ).25.1.TaylorPolynomials.Thefunction L ( x )isapolynomialofthe“rst degree.Considerthesecond-degreepolynomial T2( x )= f ( a )+ f( a )( x Š a )+ c2( x Š a )2= L ( x )+ c2( x Š a )2, where c2isanarbitrarycoecient.Thispolynomialhasthesame featuresas L ( x ),thatis, T2( a )= L ( a )= f ( a )and T 2( a )= L( a )= f( a )because T 2( x )= f( a )+2 c2( x Š a ).Soitmightprovideabetter approximationof f ( x )than L ( x )near a ifthecoecient c2ischosen sothat T2( x )hasthesameconcavityas f ( x )near a .Bytheconcavity test,itisthenreasonabletoassumethat T 2( a )= f( a ),whichyields 2 c2= f( a )or c2= f( a ) / 2.Theideacanbeextendedtoapolynomial ofdegree n : Tn( x )= c0+ c1( x Š a )+ c2( x Š a )2+ + cn( x Š a )n, wherethecoecientsare“xedbytheconditions Tn( a )= f ( a ) ,T n( a )= f( a ) ,T n( a )= f( a ) ,...,T( n ) n( a )= cn. Theresultingpolynomialiscalled the n th-degreeTaylorpolynomial : Tn( x )= f ( a )+ f( a )( x Š a )+ f( a ) 2! ( x Š a )2+ + f( n )( a ) n ( x Š a )n.25.2.AccuracyofTaylorPolynomials.Theaccuracyofthetangentline approximationisassessedinTheorem3.15.Letuscompareitwith theaccuracyofhigher-degreeTaylorpolynomials.ConsiderTaylor polynomialsoftheexponentialfunction exnear x =0.Since( ex)= exand e0=1,theTaylorpolynomialsare f ( x )= ex: Tn( x )=1+ x + 1 2 x2+ 1 6 x3+ + 1 n xn. Letustakeafewvaluesof x near x =0andcomparethevaluesofthe Taylorpolynomialswiththevalueofthefunction: x =1: f =2 718 T1=2 000 T2=2 500 T3=2 667 x = Š 0 5: f =0 607 T1=0 500 T2=0 625 T3=0 604 x =0 25: f =1 284 T1=1 250 T2=1 281 T3=1 284 Twoobservationscanbemadefromthistable.First,theaccuracy increaseswithincreasingthedegreeoftheTaylorpolynomial(reading therowsofthetablefromlefttoright).Second,lower-degreeTaylor polynomialsbecomemoreaccurateastheargumentgetsclosertothe pointatwhichtheTaylorpolynomialsareconstructed(readingthe

PAGE 119

25.TAYLORPOLYNOMIALSANDTHELOCALBEHAVIOR113 columnsofthetablefromtoptobottom).Forexample,theapproximation ex T3( x )isaccurateuptofoursigni“cantdigitsif | x | 1 / 4. Sotheaccuracyoftheapproximation ex T2( x )isdeterminedby thedierence T2Š T3= Š x3/ 6,thatis,bythenextmonomialtobe addedto T2togetthenextTaylorpolynomial.Thisobservationisa characteristicfeatureofTaylorpolynomials: Theorem 4.11 Let f becontinuouslydierentiable n +1 timeson anopeninterval I containing a .Let f( n +1)beboundedon I | f( n +1)( x ) | M .Then (4.11) | f ( x ) Š Tn( x ) M ( n +1)! | x Š a |n +1, where TnistheTaylorpolynomialat a Theorem3.15isaparticularcaseofthistheoremfor n =1.Inequality(4.11)isaconsequenceoftheTaylortheoremwhoseproofis givenamoreadvancedcalculuscourse.Forexample,whatistheaccuracyoftheTaylorpolynomial T5( x )near a =0fortheexponential exintheinterval[ Š 1 1]?Togettheupperboundonerrors,oneshould takethemaximalvalueoftheright-handsideof(4.11)for n =5inthe interval,thatis,( ex)( n )= ex M = e ,and | x | 1,sotheabsolute errorcannotexceed e/ 6! 0 0038.25.3.TaylorPolynomialsnearCriticalPoints.Let a beacriticalpoint of f .Provided f isenoughtimesdierentiable,Taylorpolynomialscan beconstructednear a .Thelineartermvanishesbecause f( a )=0. Thesecondderivativetestiseasytounderstandbylookingat T2( x )= f ( a )+ f( a )( x Š a )2/ 2.If f( a ) > 0,then f lookslikeadownward parabolanear a asdepictedintheleftpanelofFigure4.8(alocal maximum).If f( a ) < 0,then f lookslikeanupwardparabolanear a asdepictedinthemiddlepanelofFigure4.8(alocalminimum). Forexample,cos x hasalocalmaximumat a =0,anditbehaves near a =0ascos x T2( x )=1 Š x2/ 2.Thesecondderivativetest isinconclusiveif f( a )=0.Inthiscase, f ( x )behavesnear a as T3( x )= f ( a )+ f( a )( x Š a )3/ 6.So,if f( a ) =0, f hasanin”ection pointat a asdepictedintherightpanelofFigure4.8.If f( a )=0,one shouldlookat T4( x )= f ( a )+ f(4)( a )( x Š a )4/ 24.Afunctionhasalocal maximum(minimum)at a if f(4)( a ) > 0( f(4)( a ) < 0)astheconcavity doesnotchangeat x = a .Thisistobecomparedwithexamplesgiven inthecaptionofFigure4.6.Itisnowclearthatthelocalbehavior of f nearitscriticalpointisdeterminedbyaTaylorpolynomialthat

PAGE 120

1144.APPLICATIONSOFDIFFERENTIATION hasthe“rstnonvanishingcorrectionto f ( a ),providedthefunctionis dierentiablesucientlymanytimes. Example 4.7 Investigate f ( x )= x Š tan x near x =0 Solution: FindaTaylorpolynomialfortan x withtwonontrivial terms.Inthiscase,itis T3:tan x T3( x )= x Š x3/ 3.Therefore, f ( x ) x Š T3( x )= x3/ 3.Sothereisanin”ectionpointat x =0. 25.4.Asymptotes.Howcanthebehaviorofafunctionnear a beanalyzedifthefunctionisnotdierentiableat a ,ornotevende“nedat a ,orhowdoesitbehaveintheasymptoticregions x ? Definition 4.6(VerticalAsymptotes) Theline x = a isaverticalasymptoteofthegraph y = f ( x ) ifatleastoneofthelimits limx af ( x ) isin“nite( or Š ). Inotherwords,thefunction f ( x )increases(decreases)unboundedly as x approaches a fromeithertheleftortheright.Forexample,the function (4.12) f ( x )= x ( x2+3) x2Š 1 = x ( x2+3) ( x Š 1)( x +1) hastwoverticalasymptotesbecausethedenominatorvanishesat x =1 and x = Š 1.When x approaches Š 1fromtheleft, f ( x )tendsto Š whileittendsto if Š 1isapproachedfromtheright.Similarly, f ( x ) Š as x 1Šand f ( x ) as x 1+. Suppose f hasaverticalasymptoteat a .Howdoesitbehavenear a ?HowfastŽdoesitdivergewhen x getscloserto a ? Definition 4.7(AsymptoticBehavior) Thefunctions f ( x ) and g ( x ) onanopeninterval x>a (including x> Š )or x

PAGE 121

25.TAYLORPOLYNOMIALSANDTHELOCALBEHAVIOR115 Example 4.8 Findtheasymptoticbehaviorofthefunction(4.12) at x = 1 Solution: Thefunctionhastobeinvestigatednear x = 1andalso when x (I)Near x = Š 1,theunboundedgrowthof f ( x )isassociatedwith thedivergentfactor1 / ( x +1)sothat f ( x )= h ( x ) / (1+ x ), where h ( x )is“nitenear x = Š 1.Then f ( x ) h ( Š 1) / ( x + 1)= g ( x ): f ( x )= 1 x +1 x ( x2+3) x Š 1 2 x +1 = g ( x ) Apparently,limx Š 1( f ( x ) Š g ( x ))=0.Thegraphsof f ( x ) and g ( x )=2 / ( x +1)areclosenear x = Š 1, (II)Similarly,near x =1 f ( x )= 1 x Š 1 x ( x2+3) x +1 2 x Š 1 = g ( x ) 10 5510 20 10 10 20 Figure4.9. Graphof f ( x )givenin(4.12)(theblue solidcurve).Ithasaslantasymptote g ( x )= x as x (thedashedline).Intheseasymptoticregions, f ( x ) g ( x )= x .Thefunctionalsohastwovertical asymptotes x =1(theredverticalline)and x = Š 1(the blueverticalline).Theredsolidcurveisthegraphof g ( x )=2 / ( x Š 1),whichshowstheasymptoticbehaviorof f ( x )near x =1.Inaneighborhoodof x =1, f ( x ) g ( x )near x =1.Thefunction f exhibitsasimilarbehaviornear x = Š 1(notdepictedhere).

PAGE 122

1164.APPLICATIONSOFDIFFERENTIATION (III)When x isalargenegativeorpositivenumber, f ( x )= x x2(1+3 x2) x2(1 Š1 x2) x3 x2= x = g ( x ) where1 /x3and3 /x2aresmallascomparedto1forlarge x2andcanbeneglected,thatis,limx ( f ( x ) Š x )=0.Sothe graphof f asymptoticallyapproachestheline y = x .Since 1+3 /x2> 1 Š 1 /x2,theratio(1+3 /x2) / (1 Š 1 /x2) > 1forall x2> 1,andhence f ( x ) >x when x> 0,while f ( x ) 0.

PAGE 123

25.TAYLORPOLYNOMIALSANDTHELOCALBEHAVIOR117 (3)Let f and g betwicedierentiableat a and g ( a )=0.Find thesecond-degreeTaylorpolynomialforthefunction F ( x )= f ( g ( x ))near x = a Hint :Use f ( u ) T2( u )= f (0)+ f(0) u + f(0) u2/ 2,where u = g ( x )and g ( x )isalsoapproximatedbythecorresponding T2near a (4)Findthethird-degreeTaylorpolynomialforthefollowingfunctionsataspeci“edpointbyusingtheresultsfromtheprevious exercises(i.e.,byusingTaylorpolynomialsofasuitablychosenargument). (i) f ( x )=sin( x3) ,x =0 (ii) f ( x )=sin(sin x ) ,x =0 (iii) f ( x )=tan(1 Š cos x ) ,x =0 (5)UseTaylorpolynomialstoinvestigatethelocalbehaviorofa givenfunctionnearaspeci“edcriticalpoint(whetherithasa localmaximum,alocalminimum,oranin”ection). (i) f ( x )=sin( x4) ,x =0 (ii) f ( x )=1 Š x2/ 2 Š cos x,x =0 (iii) f ( x )=ln(1+ x ) Š x + x2,x =0 (6)UseTaylorpolynomialsofsuccessivedegreesfor f ( x )=ln(1+ x )near x =0toevaluateln2.Whatdegreeisrequiredto calculateln2with“vesigni“cantdigits? (7)Findverticalandslantasymptotes,ifany,ofagivenfunction. Investigatetheasymptoticbehaviorofthefunctionnearthe pointswhereithasverticalasymptotesandintheasymptotic regions x (i) f ( x )= x +4 /x (ii) f ( x )= x2/ ( x2Š 1) (iii) f ( x )=( x3Š 3 x2) / ( x2Š 2 x +1) (iv) f ( x )= x2 / 3( x2Š 1)Š 1 / 3(v) f ( x )=(cos x Š 1) /x2(vi) f ( x )=( x Š sin x ) /x4(8)Approximatethegivenfunctionnearaspeci“edpointbyusing Taylorpolynomials. (i) f ( x )= xŠ 2 / 3ln(1+ x ) ,x =0 (ii) f ( x )= xŠ 4 / 3sin2(2 x2 / 3) ,x =0 (iii) f ( x )= xŠ 5 / 3( x Š tan( x1 / 3)) ,x =0 (iv) f ( x )=[sin( x Š 1) Š x +1] / ( x Š 1)3,x =1 (9)Supposethatthefunctions f and g aresuchthat f ( a )= g ( a )=0, f( k )( a )=0for k =1 2 ,...,n ,and g( k )( a )=0 for k =1 2 ,...,m ,while f( n +1)( a ) =0and g( m +1)( a ) =0.

PAGE 124

1184.APPLICATIONSOFDIFFERENTIATION Investigatethelocalbehaviorofthefunction h ( x )= f ( x ) /g ( x ) near x = a if n = m ,if n>m ,andif m
PAGE 125

26.LHOSPITALSRULE119 Forthespecialcaseinwhich f ( a )= g ( a )=0,thederivatives fand garecontinuous,and g( a ) =0,itisnotdiculttoseewhy lHospitalsrule(4.16)holds: limx af ( x ) g ( x ) =limx a f ( x ) Š f ( a ) x Š a g ( x ) Š g ( a ) x Š a= limx a f ( x ) Š f ( a ) x Š a limx a g ( x ) Š g ( a ) x Š a= f( a ) g( a ) =limx af( x ) g( x ) The“rstequalityfollowsfrom f ( a )= g ( a )=0,thesecondandthird equalitiesaretheconsequenceofthelimitlawsandtheassumption that g( a ) =0,andthelastequalityfollowsfromthecontinuityofthe derivatives.Thissimpli“edversionoflHospitalsrulecanbeunderstoodgeometrically.Thefunctions f and g canbeapproximatedby theirtangentlinesat a f ( x ) f( a )( x Š a )and g ( x ) g( a )( x Š a ), sothat f ( x ) /g ( x ) f( a ) /g( a )near a ItisnotsoeasytoprovethegeneralversionoflHospitalsrule (theproofisomittedhere).LHospitalsruleisalsovalidforone-sided limits x aandforthelimitsat .TheconditionsoflHospitals rulemustbeveri“edforthecorrespondinglimits. Whathappensif f( a )= g( a )=0?Apparently,theconditionsof lHospitalsrulearesatis“edforthederivatives f( x )and g( x )inthis case.SolHospitalsrulemaybeappliedagaintotheratio f( x ) /g( x ). Forfunctionsdierentiablemanytimes,lHospitalsruleiseasytounderstandviatheTaylorpolynomials.Supposethatfunctions f and g arecontinuouslydierentiablesucientlymanytimesnear a .Thenby Theorem4.11thefollowingapproximationholds f ( x ) g ( x ) f ( a )+ f( a )( x Š a )+1 2f( a )( x Š a )2+ g ( a )+ g( a )( x Š a )+1 2g( a )( x Š a )2+ If f ( a )= g ( a )=0,thenthelimitoftheratioisdeterminedby f( a ) /g( a ).If f ( a )= g ( a )=0and f( a )= g( a )=0,thenthe limitisdeterminedby f( a ) /g( a )andsoon. Example 4.10 Investigatetheindeterminateforms(4.14)and (4.15). Solution: (I)Let f ( x )= exŠ 1and g ( x )= x .Then f (0)= g (0)=0(the conditionsoflHospitalsruleareful“lled).Hence, limx 0exŠ 1 x =limx 0( exŠ 1) ( x )=limx 0ex 1 =1 .

PAGE 126

1204.APPLICATIONSOFDIFFERENTIATION (II)Let f ( x )=1 Š cos x and g ( x )= x2sothat f (0)= g (0)=0. Then f( x )=sin x and g( x )=2 x .Since f(0)=0and g(0)=0,lHospitalsrulecanbeappliedagain: limx 01 Š cos x x2=limx 0sin x 2 x =limx 0(sin x ) (2 x )=limx 0cos x 2 = 1 2 (III)Let f ( x )=tan x Š x and g ( x )= x3sothat f (0)= g (0)=0. Thederivatives f( x )=sec2x Š 1and g( x )=3 x2vanish at x =0.LHospitalsrulecanbeusedagaintoresolvethe indeterminateform.Forcomplicatedfunctions,takinghigherorderderivativesmightbequiteanalgebraicexercise.Sometimes,simplealgebraictransformationsofanindeterminate formincombinationwithbasiclimitlawsmayleadtothe answerfasterthanasuccessiveuseoflHospitalsrule: limx 0tan x Š x x3=limx 0sec2x Š 1 3 x2=limx 01 Š cos2x 3 x2cos2x =limx 0sin2x 3 x2= 1 3 limx 0sin x x 2= 1 3 Thethirdequalityfollowsfromcos x 1as x 0,and thereforecos2x inthedenominatorcanbereplacedby1in accordwiththebasiclimitlaws. (IV)Let f ( x )=ln x and g ( x )= xŠ 1sothat f ( x ) Š and g ( x ) as x 0+.SotheconditionsoflHospitalsrule areful“lled.Therefore, limx 0+ln x xŠ 1=limx 0+(ln x ) ( xŠ 1)=limx 0+xŠ 1 Š xŠ 2= Š limx 0+x =0 26.2.IndeterminateProducts0.Supposethat f ( x ) and g ( x ) 0as x a .Howcantheindeterminateproduct f ( x ) g ( x ) beinvestigatedwhen x a ?Itturnsouttheindeterminateproductcanbetransformedintooneoftheindeterminateformstowhich lHospitalsruleisapplicable: (4.17) fg = f 1 /g 0 or fg = g 1 /f 0 0 0 Thefunction x ln x isanindeterminateproductofthetype0 as x 0+.Itcanbetransformedintoanindeterminateformofthe type asin(4.15),whichisthenresolvedbylHospitalsrule(see Example4.10).Notethat,althougheitherofthetransformationsin

PAGE 127

26.LHOSPITALSRULE121 (4.17)maybeappliedwiththesubsequentuseoflHospitalsrule,the technicalitiesinvolvedmightdiersubstantially.Forinstance,ifthe secondoptionin(4.17)isappliedto x ln x = x/ (1 / ln x ),then limx 0+x ln x =limx 0+x 1 ln x=limx 0+1 Š1 ln2x 1 x= Š limx 0+x ln2x. Althoughourgoalhasnotbeenachieved,oureorthasnotbeenin vain.Sincetheleft-handsidevanishesbyExample4.10,itfollowsthat x ln2x 0as x 0+.Byrepeatingthisprocedurerecursively,one caninferthat x lnnx 0as x 0+forany n =1 2 ,... .26.3.IndeterminatePowers00,0,and1.Severalindeterminate formsarisefromthelimitsof[ f ( x )]g ( x )as x a : 00( f ( x ) 0 ,g ( x ) 0); 0( f ( x ) ,g ( x ) 0); 1( f ( x ) 1 ,g ( x ) 0) Note c0=1if c =0and c = .Similarly, c=0if0 c< 1and c= if c> 1.Theindeterminatepowerscanbetransformedinto anindeterminateproductwiththehelpoftheidentity y = eln y: limx a[ f ( x )]g ( x )=limx aeln([ f ( x )]g ( x ))=limx aeg ( x )ln( f ( x ))= elimx ag ( x )ln( f ( x )). Thelimitof g ( x )ln( f ( x ))isoftype0 andcanbetreatedbythe rule(4.17).Theprocedureisillustratedwithanexampleofthetype 0indeterminatepower: limx x1 /x=limx eln( x1 /x)=limx eln( x ) /x= elimx ln( x ) /x= e0=1 .26.4.IndeterminateDifferencesŠ.Suppose f ( x ) and g ( x ) as x a .Thelimitof f ( x ) Š g ( x )as x a iscalled an indeterminatedierence .Thefollowingtransformationsmightbe helpfultoinvestigateit: f Š g = f 1 Š g f = 1 Š g/f 1 /f or f Š g = g f g Š 1 = f/g Š 1 1 /g If f ( x ) /g ( x ) 1,thentheindeterminatedierenceisequivalenttoan indeterminateformoftype0 / 0andcanbeinvestigatedbylHospitals rule.Thelimitof f/g isanindeterminateformoftype / andcan alsobeinvestigatedbylHospitalsrule.Supposethat f ( x ) /g ( x ) k as x a ,where k canbeeitheranon-negativenumberor k = If k< 1,then f Š g = g ( f/g Š 1) ( k Š 1)= Š ;that is, g increasesfasterthan f as x a .If k> 1or k = ,then

PAGE 128

1224.APPLICATIONSOFDIFFERENTIATION f Š g = g ( f/g Š 1) ( k Š 1)= ;thatis, f increasesfaster than g as x a .Forexample, limx 0+ ln x + 1 x =limx 0+1 x 1+ x ln x =limx 0+1 x (1+0)= If k =1,thenitisalsopossiblethat f Š g c ,where c isanumber.In thiscase, f and g increaseasymptoticallyatthesamerate: fŠ g 0. If c =0,thefunctions f and g havethesameasymptoticbehavior.For example, limx 0 1 sin x Š cot x =limx 01 sin x 1 Š cos x =limx 0sin x cos x =0 wherelHospitalsrulehasbeenusedinthesecondequality.Notethat Taylorpolynomialsallowusto“ndthelocalbehaviorofthisfunction near x =0.Use T2toapproximatecos x and T3forsin x : 1 sin x Š cot x = 1 Š cos x sin x x2/ 2 x Š x3/ 6 = x/ 2 1 Š x2/ 6 x 2 where x2/ 6issmallascomparedto1when x iscloseenoughto0and canthereforebeneglectedinthedenominator.26.5.Exercises.(1)Findthelimits. (i)limx 0ln(1+ x ) x (ii)limx 0x Š sin x x3(iii)limx / 4sin x Š cos x cos(2 x ) (iv)limx 0+sin x ln x (v)limx 0+[1 Š cos x ]ln x (vi)limx 0sin x Š x exŠ 1 Š x2/ 2 (vii)limx 0tan x Š x ex3Š 1 (viii)limx 0xŠ 2 x +1 Š 1 x Š 1 2

PAGE 129

27.ANALYZINGTHESHAPEOFAGRAPH123 (ix)limx 1+ 2 x x(x)limx (1 Š eŠ x)ex(xi)limx x Š (ln x )n ,n> 0 (xii)limx x Š ln x 1 /x(xiii)limx 0+(sin( ax ))sin( bx ),a> 0 ,b> 0 (2)Considerthefunction f ( x )= eŠ 1 /x2if x =0and f (0)=0. Show“rstthat limx 0xneŠ 1 /x2=0 ,n> 0 Usethisfactandthede“nitionofderivativestoshowthat f( k )(0)=0forall k .CanTaylorpolynomialsbeusedto investigatethelocalbehaviorofthefunctionnear x =0and establishthenatureofthecriticalpoint x =0? (3)Findallcriticalpointsofthefunction f ( x )= eŠ 1 /x,x> 0 0 ,x =0 Š e1 /x,x< 0 andinvestigatethebehaviorofthefunctionnearthem.27.AnalyzingtheShapeofaGraph Toanalyzetheshapeofagraph y = f ( x ),itisusefultohave aclearideaofhowthebasicfunctionsbehave.Forexample,sin x andcos x areregulareverywhere,bounded(e.g., | sin x | 1),and periodicwithaperiodof2 .Inaddition,sin x haszerosat x = n n =0 1 2 ,... ,whilecos x vanishesat / 2+ n .Thefunctionsin x isodd,whilecos x iseven.Theirratiotan x =sin x/ cos x isnotde“ned atrootsofcos x .Howdoestan x behave,say,near x = / 2?Since bothsin x andcos x aresmoothnear x = / 2,thebehavioroftan x near / 2canbeunderstoodwiththehelpofTaylorpolynomials.Let usapproximatesin x by T1( x )=1+( x Š / 2)andcos x by T3( x )= Š ( x Š / 2)+( x Š / 2)3/ 6.Tosimplifythenotation,write x = x Š / 2 (thedeviationof x from / 2).Then tan x 1+ x Š x +( x )3/ 6 = Š 1 x 1+ x 1 Š ( x )2/ 6 Š 1 x = Š 1 x Š / 2 ,

PAGE 130

1244.APPLICATIONSOFDIFFERENTIATION wherethesecondratiointheproducthasbeenapproximatedby1 because x issmall.Sincetan( x + )=tan x ,thisbehaviorrepeats itselfatneareveryrootofcos x .27.1.GrowthofthePower,Exponential,andLogarithmicFunctions.Let uscomparethegrowthofthepowerfunction xn,theexponentialfunction ex,andthelogarithmicfunctionln x as x Theexponential functiongrowsfasterthanthepowerfunction .Let f ( x )= exand g ( x )= xn.Letusanalyzetheratio f/g as x .Theconditions oflHospitalsrulearesatis“ed: ex and xn as x LHospitalsrulecansuccessivelybeapplieduntiltheindeterminate formisresolved: limx ex xn=limx ex nxn Š 1=limx ex n ( n Š 1) xn Š 2= =limx ex n = Theconclusionistrueforanyreal n .Foranyreal n ,thereexists apositiveinteger N suchthat n 1.But exgrowsfasterthan xN.Similarly,itisstraightforwardtoshowthatthe logarithmicfunctiongrowsslowerthananypowerfunction : limx ln x xn=limx (ln x ) ( xn)=limx 1 x nxn Š 1=limx 1 nxn=0 forany n> 0( n maybeanypositiverealnumberhere).27.2.Asymptotesatx .Theasymptoticbehaviorofrational functionsiseasilydeterminedbythehighestpowersofthenumerator anddenominator,asinExample4.8.Ingeneral,iflimx f ( x )is in“nite,thenthelimitof f/g canbestudiedfortrial g swithdierent growth, g = mx (forslantasymptotes), g = xn, g =ln x ,andsoon. Suppose g ( x )isfoundsuchthat f ( x ) /g ( x ) 1as x .Doesthis meanthat g and f havethesameasymptoticbehavior?Theansweris no.ŽIftheindeterminateform f ( x ) Š g ( x )oftype Š converges to0as x ,thentheindeterminateform f ( x ) /g ( x )oftype convergesto1.Indeed,itfollowsfrom1 /g ( x ) 0and f ( x ) Š g ( x ) 0 that(1 /g ( x ))( f ( x ) Š g ( x ))= f ( x ) /g ( x ) Š 1 0. Theconverseisnot true .Considerthefollowingsimpleexample: f ( x )= x +sin x and g ( x )= x .Evidently, f ( x ) /g ( x )=1+sin x/x 1as x .But thelimitlimx ( f ( x ) Š g ( x ))=limx sin x doesnotexist.So,even if g isfoundtohavetheproperty f ( x ) /g ( x ) 1as x ,the indeterminateform f Š g oftype Š muststillbeinvestigatedin ordertodeterminewhetherornot g hasthesameasymptoticbehavior as f .

PAGE 131

27.ANALYZINGTHESHAPEOFAGRAPH125 27.3.GuidelinesforAnalyzingtheShapeofaGraph.Thefollowing guidelinesareusefulforsketchingthegraphofafunction.Itshould benotedthatnotallthestepscanalwaysbecarriedout.Thisdepends verymuchonthecomplexityofthefunctioninquestion.Sotheseare reallyguidelines,notamust-doŽalgorithm.Givenafunction f ,“nd: (I) Domain. Thedomainconsistsofallvaluesof x atwhich f ( x )isde“ned.Typically,itisacollectionofintervals.If f isde“ned for x>a or x

PAGE 132

1264.APPLICATIONSOFDIFFERENTIATION (VI) Intervalsofpositiveandnegativevaluesof f Thesearetheintervalswherethegraph y = f ( x )liesaboveor belowthe x axis.Rootsof f generallyseparatetheintervalsof positiveandnegativevaluesof f .However,thisisnotalways thecase.Let c bearootof f .If f( c ) =0,thenthefunction f isincreasingordecreasingat c andhencemustchangeits sign.If f( c )=0or fdoesnotexistat c ,thatis,aroot of f coincideswithitscriticalpoint,then f isnegativenear c if f hasalocalmaximumat c and f ispositivenear c if ithasalocalminimumat c .Sothesignofthederivative fmustbeinvestigatednear c (the“rstderivativetest).Vertical asymptotescanalsoseparateintervalsofpositiveandnegative valuesof f .Forexample,thefunction(4.12)hasoneroot x =0 andtwoverticalasymptotesat x = Š 1and x =1.So f is negativeon( Š Š 1),positiveon( Š 1 0),negativeon(0 1), andpositiveon(1 ).ThegraphisshowninFigure4.9. (VII) Intervalsofincrease( f> 0)anddecrease( f< 0 ). If f> 0( f< 0)onaninterval,then f increases(decreases) onit(theincreasing-decreasingtest).Theseintervalsaregenerallyseparatedbycriticalpointsandverticalasymptotes.As aconsequenceofthisstudy,thenatureofeachcriticalpoint isestablishedbythe“rstderivativetest. (VIII) Intervalsofupwardanddownwardconcavity. Theseintervalsareseparatedbyin”ectionpointsandvertical asymptotes.Thesignof f( x )mustbestudied.Yet,thesecond derivativetestandTaylorpolynomialscanbeusedtoestablish thenatureofacriticalpointof f (IX) Valuesof f atcriticalpointsandin”ectionpoints. Thesevaluessetrelativescalesofthegraph(e.g.,theyshow howmuchthefunctionincreasesbetweentwocriticalpoints). Example 4.11 Sketchthegraphof f ( x )= x1 / 3( x Š 6)2 / 3. Solution: Followingtheprecedingguidelines: (I)Thedomainisthewholerealline. (II)Therootsof f are x =0and x =6(theinterceptswiththe x axis).Theinterceptwiththe y axisis f (0)=0. (III)Thefunctionisnotperiodic,anditisneitheroddnoreven. (IV)Thereisnoverticalasymptote.Tostudytheasymptoticbehavioras x ,itisconvenienttofactoroutthelargest powerof x : f ( x )= x (1 Š 6 /x )2 / 3sothat f ( x ) x .Thisshows thatthegraphhasaslantasymptoteoftheform y = x + m ,

PAGE 133

27.ANALYZINGTHESHAPEOFAGRAPH127 Figure4.10. Graphof f ( x )= x1 / 3( x Š 6)2 / 3.Theroots of f are x =0and x =6,andtheyde“netheintercepts withthe x axis.Ithastheslantasymptote f ( x ) x Š 4as x .Thederivativevanishesat x =2(a localmaximum).Itdivergesat x =0and x =6;the graphhasverticaltangentlinesatthesepoints.The secondderivativeisnegativeif x< 0sothatthegraph isdownwardconcave.Itispositiveon(0 6)and(6 ). Thegraphisconcavedownward.Thepoint x =0isan in”ectionpointastheconcavitychangesatit. where m isyettobe“xedbycalculatingthelimitsof f ( x ) Š x as x .Thereaderisadvisedtodosoandshowthat m =4.HereTaylorpolynomialsareusedinstead,whichis provedtobemoreadvantageous.Put u = Š 6 /x sothat u 0 as x .Thenthefactor(1 Š 6 /x )2 / 3=(1+ u )2 / 3isapproximatedby T2( u )=1+2 3u Š1 9u2=1 Š 4 /x Š 1 / (324 x2). Therefore, f ( x ) xT2( u )= x Š 4 Š 1 / (324 x ).Thisequation shows,“rst,thatthegraphhastheslantasymptote y = x Š 4 and,second,that f ( x ) Š ( x Š 4) Š 1 / (324 x ) < 0if x islarge andpositive(i.e.,thegraphapproachestheslantasymptote

PAGE 134

1284.APPLICATIONSOFDIFFERENTIATION frombelow)and,similarly, f ( x ) Š ( x Š 4) > 0if x islargeand negative(i.e.,thegraphapproachestheslantasymptotefrom above). (V)Thederivativereads f( x )= x Š 2 x2 / 3( x Š 6)1 / 3. Itvanishesat x =2anddoesnotexistat x =0and x =6. Thecriticalpointsare0,2,and6.Inparticular, f( x ) as x 0andittendsto as x 6,respectively.Therefore, thegraphhasverticaltangentlinesat x =0and x =6.Near x =0,thegraphlookslike y = f ( x ) 62 / 3x1 / 3,whilenear x =6,ithasadownwardcusp y = f ( x ) 61 / 3( x Š 6)2 / 3. (VI)Thegraphliesbelowthe x axison( Š 0)as f ( x ) < 0and aboveiton(0 )as f ( x ) 0.Thefunctiondoesnotchange itssignattheroot x =6( f musthavealocalminimumat6, whichisalsoveri“edbythe“rstderivativetestbelow). (VII)Thederivativeisaproductofthreefactors x Š 2, xŠ 2 / 3,and ( x Š 6)Š 1 / 3.Byinvestigatingthesignsofthesefactorsonthe intervalsseparatedbythecriticalpoints,wecanconcludethat f> 0( f isincreasing)on( Š 0), f> 0( f isincreasing) on(0 2), f< 0( f isdecreasing)on(2 6),and f> 0( f is increasing)on(6 ).Also, f hasalocalmaximumat x =2 andalocalminimumat x =6bythe“rstderivativetest. (VIII)Thesecondderivativereads f( x )= Š 8 x5 / 3( x Š 6)4 / 3. Thefactor( x Š 6)4 / 3cannotbenegative.Thesignof fis determinedonlybythatof x5 / 3.Thus, f> 0on( Š 0)(the graphisupwardconcave)and f< 0on(0 6)and(6 )(the graphisdownwardconcave).So x =0isthein”ectionpoint. Also,near x =2,thegraphlookslikethedownwardparabola y = T2( x )= f (2)+ f(2)( x Š 2)2/ 2=(4 Š1 4( x Š 2)2) /3 2. Intheageofgraphingcalculators,theprecedingguidelinesmight lookratherobsoletebecause“ndingtheshapeofagraphcanbedone justbyhittingtheappropriatecalculatorbuttons.Butwhatacalculatorcannotdoistoprovidedetailsofthelocalbehaviorofafunctionnearpointsofinterest(e.g.,criticalpoints,asymptotes,etc.).In scienceandengineering,thisisoftenmuchmoreimportantthanthe

PAGE 135

28.OPTIMIZATIONPROBLEMS129 overallshapeofagraph.Inthepreviousexample,acalculatorwould showthatthereisaslantasymptote,acuspat x =6,andalocal maximumat x =2,butitwouldnotbeabletodeterminethelocal behaviorofthefunctionnearthecusp,oratthelocalmaximum,or intheasymptoticregion.Hereagoodworkingknowledgeofcalculus becomesindispensable,whileagraphingcalculatorisjustausefultool thatgreatlyfacilitatesthestudyofafunction.27.4.Exercises.(1)Sketchthegraphofeachofthefollowingfunctions: (i) f ( x )= x2( x Š 1)(2 Š x ) (ii) f ( x )= x2/ (4+ x2) (iii) f ( x )= x3/ ( x2Š 3 x +2) (iv) f ( x )=( x5Š x2) / ( x +1) (v) f ( x )= x sin x (vi) f ( x )=sin( nx ) / sin x,n =2 3 4 (vii) f ( x )=( ex+ eŠ x)cos x (viii) f ( x )= xŠ 1 / 3( x Š 6)Š 2 / 3(ix) f ( x )=sin2x/x2(x) f ( x )= ( x Š a )2+ b2(xi) f ( x )= | x2Š 1 |Š x (xii) f ( x )=[( x Š 1) / ( x +1)]1 / 3(2)Sketchthegraphofthepolynomialwith k realroots: f ( x )= A ( x Š x1)n1( x Š x2)n2 ( x Š xk)nk, where A> 0and n1,n2,...,nkarepositiveintegers.Investigate“rstthecasewhen n1= n2= = nk=1,thenthe casewhenoneofthepowers n1,n2,...,nkisgreaterthan1 (howdoesthegraphlookwhenthispowerisoddorwhenit iseven?).Thenproceedtothegeneralcase. (3)Let f and g besecond-degreepolynomialssuchthat f> 0 and g> 0.Sketchallpossibleshapesofthegraph y = f ( x ) /g ( x ).28.OptimizationProblems Supposethataquantity Q dependsonsomevariables.Theproblemofoptimizing Q implies“ndingthevaluesofthevariablesatwhich thequantity Q attainsitmaximalorminimalvalue.Thesimplestoptimizationproblemariseswhen Q dependsonasinglevariable x such that Q isafunction f ( x ).Thentheoptimizationproblemisreduced totheproblemof“ndingextremevaluesof f ( x ).Thelatterproblem

PAGE 136

1304.APPLICATIONSOFDIFFERENTIATION hasbeenanalyzedinSection22.Todetermineextremevaluesof f ,one hasto: (I)Findallcriticalpointsof f (II)Investigatethenatureofthecriticalpoints(localminimaand localmaxima).The“rstorsecondderivativetestscanbeused forthispurpose. (III)Calculatethevaluesof f attheendpointoftheinterval[ a,b ] (ifextremevaluesaresoughtonlyin[ a,b ])andcomparethem withvaluesof f atitslocalmaximaandminimatodetermine absoluteextremevaluesof f Thefollowingtestcanalsobeusedto“ndabsoluteextremevaluesof afunction. Theorem 4.13(FirstDerivativeTestforAbsoluteExtreme Values) Suppose c isacriticalpointofacontinuousfunction f de“nedonaninterval. (I) If f( x ) > 0 forall xc ,then f ( c ) istheabsolutemaximumvalueof f (II) If f( x ) < 0 forall x 0 forall x>c ,then f ( c ) istheabsoluteminimumvalueof f Theconclusionofthetheoremiseasytounderstand.Consider case(I).Since f( x ) > 0forall xc ,thefunctiondecreasesforall x>c .Bycontinuityof f ,thenumber f ( c )mustbethelargestvalue of f .Case(II)isprovedsimilarly. RecallExample4.2.Thisisatypicaloptimizationproblem.Its solutionisratherstraightforward,providedEquation(4.4)isgiven. Withoutit,theproblemof“ndinganoptimalangleforaprojectile becomesfarmoredicult.Itsmajorpartnowinvolvesaderivation ofEquation(4.4)!Thisisquitetypicalforoptimizationproblems.As arule,theyariseinvariousdisciplines,andtheirformulationasthe mathematical problemofextremevaluesrequiresaspeci“cknowledge outsidemathematics,forexample,thelawsofphysicsasinExample 4.2,chemistry,biology,economics,andsoon.Atypicaloptimization problemmaybesplitintothreebasicsteps: (I)Identifyavariablewithrespecttowhichaquantity Q istobe optimized. (II)Usethelawsofaspeci“cdisciplinetoexpress Q asafunction f ofthatvariable, Q = f ( x ). (III)Solvethemathematicalproblemofextremevaluesof f .

PAGE 137

28.OPTIMIZATIONPROBLEMS131 Example 4.12 Analuminumcanhastheshapeofacylinderof radius r andheight h .Designanaluminumcanofvolume V =300 cm3tominimizethecost(ortheamount)ofmaterialneededtomakethe can. Solution: Followingtheprecedingguidelines: (I)Apparently,theleastamountofmaterialisusedwhenthe surfaceareaofthecanisminimal.Soonehastominimizethe surfacearea S ,whichdependson r and h .Butthevariables r and h arenotindependentbecausethevolumeis“xed. (II)Thesurfaceareaisthesumoftheareasoftheside,top,and bottomofthecan: S =2 rh + r2+ r2=2 rh +2 r2.The volumeis V = r2h .Sincethevolumeis“xed,thevariables r and h arerelatedas h = V/ ( r2).Hence, S canbewrittenas afunctionoftheradius r only: S ( r )=2 r V r2+2 r2= 2 V r +2 r2. Onehasto“ndthevalueof r> 0atwhich S ( r )attainsits absoluteminimum.Thecorrespondingvalueof h isthenfound fromtherelation h = V/ ( r2). (III)Thefunction S ( r )isdierentiableforall r> 0.Therefore,all itscriticalpointsarerootsofthederivative: S( r )= Š 2 V r2+4 r = 4 r2 r3Š V 2 =0 Sothecriticalpointis rc= V 2 1 / 3. Since S( r ) < 0forall0 0forall r>rc,thefunction S ( r )attainsitsabsoluteminimumat rcbythe“rstderivativetestforabsoluteextremevalues.The dimensionsofthecanwithminimalcostsofmaterialfora givenvolume V are r = V 2 1 / 3 3 6cm ,h = V r2 c= 4 V 1 / 3=2 rc 7 2cm Theanalysishasshownthattheheightanddiameterofacan ofagivenvolumemustbeequalinordertominimizethecost ofmaterial(orthesurfaceareaofthecan).Checkoutalocal supermarkettoseeifmanufacturersusethisfact!

PAGE 138

1324.APPLICATIONSOFDIFFERENTIATION Thisexampleisfurtherillustratedontheinteractivewebsite athttp://www.math.u”.edu/ mathguy/ufcalcbook/optimize cylinder .html Remark .Inthepreviousexample, S hasbeenexpressedasafunctionof r .Thesameconclusioncouldbereachedif S isexpressedasa functionoftheheight h only,thatis,whentherelation r = V/ ( h ) issubstitutedintotheexpressionforthesurfaceareatoobtain S ( h ). Thecriticalpointof S ( h )canbeshowntobe hc=2 rc.Verifythis! ACuriousFact .Theprecedingproblemisessentialtoreduce wastefromplastic,glass,andaluminumcontainers.Itcanbestated moregenerally.Whatistheshapeofacontainerthathasthesmallest surfaceareaatagivenvolume?Itcanbeprovedbythecalculusof variationsthatsuchacontainermustbeasphere.Evenintheexample ofanaluminumcan,theoptimaldimensionsappeartobeascloseto thoseofasphereasthecylindricalgeometrywouldallow:Theheight anddiameterarethesame.Shouldonlysphericalcontainersbeusedto gogreenŽ?Toanswerthisquestion,afarmorecomplicatedoptimizationproblemmustbestudied.Forexample,spheresarenotoptimal forstorageandhencefortransportation;rectangularcontainersarefar better.Storagemaintenanceandtransportationrequireenergy(hence carbonemissions).Theproductionwasteforcontainersofdierent shapesisdierent.Finally,whataboutconsumersreactiontosphericalCokecansinavendingmachineorsphericalaluminumcansinthe supermarket?28.1.ApplicationstoEconomics.InSection19,weintroducedthecost function C ( x ),whichisthecostofproducing x unitsofacertainproduct.Thederivative C( x )isthe marginalcost .Itdeterminesthecostof increasingproductionfrom x unitsto x +1units.Let p ( x )betheprice perunitthatacompanycanchargeifitsells x units.Thefunction p ( x ) isalsocalledthe pricefunction .Naturally,itisgenerallyexpectedto beadecreasingfunctionbecausethepriceperunitusuallygoesdown whenalargernumberofunitsissold.Thetotalrevenue R ( x )= xp ( x ) iscalledthe revenuefunction .Thederivative R( x )iscalledthe marginalrevenuefunction .Itdeterminesthechangeintherevenuewhen thenumberofunitssoldincreasesfrom x to x +1.Finally,the pro“t function P ( x )= R ( x ) Š C ( x )= xp ( x ) Š C ( x ) determinesthetotalpro“tif x unitsaresold.Itsderivative P( x ) determinesthechangeinthetotalpro“twhenthenumberofunits

PAGE 139

28.OPTIMIZATIONPROBLEMS133 soldincreasesfrom x to x +1.Thestandardoptimizationproblem hereistominimizecostsandmaximizerevenuesandpro“t. Example 4.13 Asmallstoresellsjeansatapriceof $ 80per pair.Everyweek60unitsaresold.Thecosttothestorefor60unitsis $ 2500,includingthecostoftransportation.Amarketsurveyindicates that,foreach $ 10rebateoeredtobuyers,thenumberofunitssoldwill increaseby20aweek.Also,thepurchaseandtransportationcostswill godownby $ 10pereachweeklyorderincreaseof5units.Howlargea rebateshouldthestoreoertomaximizeitspro“t? Solution: (I)Whatisknownaboutthepricefunction p ( x )?First,itsvalue ataparticularnumberofsoldunits x = x0=60is p0= p (60)=80.Also,if x increasesbyanamountof x =20, thepricefunctiondecreasesby p =10(therebate).Thus, theratio m = Š p/ x = Š 1 / 2istherateofchangeof p ( x ) (theminussignindicatesthedecreasein p ( x )).Sotheprice functionis p ( x )= p0+ m ( x Š x0)=80 Š1 2( x Š 60)=110 Š1 2x. (II)Whatisknownaboutthecostfunction C ( x )?First,itsvalue ataparticularnumberofsuppliedunits x = x0=60is C0= C (60)=2500.Also,thecostfunctiondecreasesby C =20 if x increasesby x =5.Sotheratio M = Š C/ x = Š 4is therateofchangeof C orthemarginalcost.Therefore, C ( x )= C0+ M ( x Š x0)=2500 Š 4( x Š 60)=2740 Š 4 x. (III)Onehastomaximizethepro“tfunction: P ( x )= xp ( x ) Š C ( x )=114 x Š1 2x2Š 2740 Since P( x )=114 Š x, thefunctionhasonecriticalpoint x =114atwhich P ( x )attainsitsabsolutemaximalvalueby the“rstderivativetestforabsoluteextremevalues. (IV)If x =114unitscanbesold,thepriceperunitis p (114)= 110 Š 57=53;thatis,therebateshouldbe p (60) Š p (114)= 80 Š 53=27.Thus,thestoreshouldoerarebateof$27to maximizeitspro“t.Notealsotheincreaseintheweeklypro“t: P (60)=$2300whereas P (114)=$3758. Remark .Infact,thelinear(tangentline)approximationhasbeen usedtogettheunknownpriceandcostfunctionsinthepreviousexample.Thisisabene“tofmarketsurveys:Theyestimatethederivatives

PAGE 140

1344.APPLICATIONSOFDIFFERENTIATION (ortrends)ofthepricefunctions.Naturally,anincreaseinsalesleads toadecreaseinthedemandforthatparticularitem.So,afterasuccessfulrebatecampaign,thestorewouldneedanewmarketsurvey toestimate p(114)andgetthelinearapproximationat x =114.The pricemaygoupthen.Similarly,thecostfunctionisgenerallyhighly nonlinear.Itslinearizationnearaparticular x = x0cannotbevalidfor all x>x0.Indeed,inthepreviousexample,itvanishesat x =685and becomesnegativeafterthat,whichcannotpossiblybetrue.28.2.Exercises.(1)Apieceofwire1mlongiscutintotwopieces.Oneisbent intoasquareandtheotherintoacircle.Whereshouldcuts bemadeifthesumofareasofthesquareandcircleistobean extreme?Whichoftheseextremesarerelativemaximaand whichrelativeminima? (2)Showthatofalltrianglesinscribedinacircletheequilateral trianglehasthegreatestarea. (3)Atankhastheformofacylinderwithhemisphericalends. Ifthevolumeistobe V m3,whatarethedimensionsfora minimumamountofmaterial? (4)Thedemandforacertainarticlevariesinverselyasthecubeof thesellingprice.Ifthearticlecosts20centstomanufacture, “ndthesellingpricethatyieldsthemaximumpro“t. (5)Amanisinaboat1milefromthenearestpoint, A ,ofa straightshore.Hewishestoarriveassoonaspossibleata point, B ,3milesalongtheshorefrom A .Hecanrow2miles perhourandwalk4milesperhour.Whereshouldheland? (6)Thestinessofarectangularbeamvariesastheproductof thebreadthandthecubeofthedepth.Findthedimensionsof thestiestbeamthatcanbecutfromacylindricallogwhose radiusis R (7)Ifthecostperhourforfuelrequiredtooperateagivensteamer variesasthecubeofitsspeedandis$40perhourforaspeed of10milesperhour,andifotherexpensesamountto$200per hour,“ndthemosteconomicalratetooperatethesteamera distanceof500miles. (8)Arailroadcompanyagreedtorunaspecialtrainfor50passengersatauniformfareof$10each.Inordertosecuremore passengers,thecompanyagreedtodeduct10centsfromthis uniformfareforeachpassengerinexcessofthe50(i.e.,if therewere60passengers,thefarewouldbe$9each).What

PAGE 141

29.NEWTONSMETHOD135 numberofpassengerswouldgivethecompanythemaximum grossreceipt? (9)Asheetofpaperforaposteristocontain16squarefeet.The marginsatthetopandthebottomaretobe6inches,and thoseonthesides4inches.Whatarethedimensionsifthe printedareaistobemaximal? (10)Ataxicompanycharges15centsamileandlogs600passengermilesaday.Twenty-“vefewerpassenger-milesadaywouldbe loggedforeachcentincreaseintheratepermile.Whatrate yieldsthegreatestgrossincome? (11)Tworoadsintersectatrightangles,andaspringislocatedin anadjoining“eld10mfromoneroadan d5mfromtheother. Howshouldastraightpathjustpassingthespringbelaidout fromoneroadtotheothersoastocutotheleastamountof land?Howmuchlandiscuto? (12)Arectangularboxwithasquarebaseandanopentopistobe made.Findthevolumeofthelargestboxthatcanbemade from A cm2ofmaterial. (13)Arectangular“eldcontaining S m2istobefencedoalong thebankofastraightriver.Ifnofenceisneededalongthe river,whatmustbethedimensionsrequiringtheleastamount offencing? (14)Ifastoneisthrownfromacliofheight h ataspeed v0m/s andanangle withthehorizontalline,thenitstrajectoryis aparabola: y = h + x tan Š x2g 2 v2 0cos2 where y isthestoneheight(verticalposition), x isthehorizontalposition(allthepositionsareinmeters),and g =9 8 m/s2isaconstantuniversalforallobjectsnearthesurface oftheEarth(thefree-fallacceleration).ComparewithEquation(4.4).Atwhatangleisastonetobethrowntoreachthe maximalrangeatagivenspeed v0?29.NewtonsMethod Findingrootsofafunction f ( x )isanimportantprobleminvarious applications.Unfortunately,ananalyticsolutionoftheequation f ( x )= 0isimpossibleinmanypracticalcases.Forexample,consider f ( x )= x Š eŠ x.Theequation f ( x )=0isequivalentto x = eŠ x.Thegraphs y = x and y = eŠ xintersectatsome x between0and1.So f ( x )hasa root.Buthowcanitbecalculated?Herewepresentoneofthesimplest

PAGE 142

1364.APPLICATIONSOFDIFFERENTIATION methods,knownasNewtonsmethod.Itprovidesarecurrencerelation thatallowsustocomputearootofadierentiablefunctionwithany desiredaccuracy.29.1.NewtonsRecurrenceRelationforFindingaRoot.Suppose f ( x ) hasarootnear x0.Considerthetangentlineapproximationof f near x0: L ( x )= f ( x0)+ f( x0)( x Š x0).Itiseasyto“ndtherootof L ( x ), whichisdenotedby x1: L ( x )=0= x = x1= x0Š f ( x0) f( x0) Notethattherootof L ( x )existsif f( x0) =0(otherwise,thetangent lineishorizontalandcannothaveanyroot). x2x1x0 Figure4.11. DiagramforNewtonsmethod.Pick x0neartherootof f .Findthetangentlineofthegraph of f at x0.Determinetheintersectionpoint x1ofthe tangentlinewiththe x axis.Findthetangentlineto thegraphof f at x1anditsintersection x2withthe x axis.Byrepeatingthisprocedureasequenceofnumbers x0, x1,x2,... isobtainedthatconvergestotherootof f provided x0waschosencloseenoughtotheroot. Since L ( x )isonlyanapproximationto f ( x ),thenumber x1is closertotherootof f than x0,butdoesnotcoincidewithit.Inother words,thevalue f ( x1)iscloserto0than f ( x0):0 < | f ( x1) | < | f ( x0) | (theabsolutevalueisnecessaryifthefunctiontakesnegativevalues). Therefore,thetangentlineconstructedat x = x1, L ( x )= f ( x1) Š

PAGE 143

29.NEWTONSMETHOD137 f( x1)( x Š x1),canbeexpectedtoapproximate f ( x )evenbetternear itsrootbecause x1isclosertotherootthan x0.Therootofthenew tangentlineisgivenbythesameexpressionasbeforewhere x0should bereplacedby x1: x2= x1Š f ( x1) /f( x1).Theproceduremaybe recursivelyrepeatedtogenerateasequenceofvalues xn: (4.18) xn +1= xnŠ f ( xn) f( xn) ,n =0 1 2 ,..., provided f( xn) =0. Theorem 4.14 If f hasasingleroot r inanopenintervaland f( x ) =0 ontheinterval,thenthereexists x0sucientlycloseto r suchthatthesequence(4.18)convergestotheroot limn xn= r. Inpracticalterms,thesequenceelementsarecalculatedwithaparticularnumberofsigni“cantdigits(decimalplaces).Newtonsrecurrenceisapplieduntil xn +1and xnagreetoalltherelevantdecimal places.Then r = xn +1iscorrecttotherelevantdecimalplaces. Example 4.14 Findtherootof f ( x )= x Š eŠ xthatiscorrectto sixdecimalplaces. Solution: (I)Determinethepositionoftheroot“rst.Thegraphs y = x and y = eŠ xintersectonlyonceatapointbetween0and1. So,inanyopenintervalcontainingtheinterval(0 1), f has onlyoneroot. (II)Verifythecondition f( x ) =0: f( x )=1+ eŠ x> 0forall x (III)PickaninitialvalueofNewtonssequenceasclosetotheroot aspossible,e.g. x0=0.ThenNewtonssequenceforsixdecimalplacesis: x0=0 ,x1=0 5 ,x2=0 566311 ,x3=0 567143 ,x4=0 567143 Sotheroot r =0 567143iscorrecttosixdecimalplaces(in fact, f (0 567143)= Š 4 5 10Š 7). 29.2.PitfallsinNewtonsMethod.Unfortunately,thereisnounique recipeforchoosinganinitialpointinNewtonssequence.Thechoice dependsverymuchonthefunctioninquestion.Inpractice,itisdeterminedbytryingdierentvalues.Afewpossiblebadbehaviorsof Newtonssequenceareusefultokeepinmind.

PAGE 144

1384.APPLICATIONSOFDIFFERENTIATION (I)Abadchoiceoftheinitialpoint x0canproducethevalueof x1thatisaworseapproximationtotherootthan x0.Consider, forexample,thefunction f ( x )= x3Š 3 x2+2intheinterval [0 2]and f ( x )=2when x< 0and f ( x )= Š 2when x> 2. ThisisdepictedinFigure4.12.Thefunctioniscontinuously dierentiablebecause f( x )=3 x2Š 6 x approaches0as x 0+and x 2Š.Thefunctionhastheroot x =1and f( x ) < 0in theopeninterval(0 2).If0
PAGE 145

29.NEWTONSMETHOD139 xn xn 2 xn 1 xn Figure4.13. AcycleinaNewtonssequencefor f ( x )= | x | .Thefunctionhastheroot r =0, f (0)=0,and italsohastheverticaltangentlineattherootbecause thederivativediverges f( x ) as x r =0.A Newtonssequenceexhibitsacycle;itoscillatesaround theroot r neverconvergingtoit.Ingeneral,suchacycle canoccurinaNewtonssequenceforafunctionwhose behaviorneararoot r is f ( x ) a ( x Š r )2 ,where a is aconstantand =1 / 4.Furthermore,if0 << 1 / 4, thenaNewtonssequencedoesnotconverge. oscillatearound r ,neverconvergingtoit,oritmaydivergefor anyinitialpoint.Tounderstandthisphenomenon,suppose f ( x )behavesnearitsroot r as f ( x ) a ( x Š r )2 > 0, and a isconstant.Thederivative f( x ) 2 a ( x Š r )2 Š 1divergesas x r when < 1 / 2.Since f ( x ) /f( x )=1 2 ( x Š r ), Newtonssequence(4.18) xn +1= xnŠ1 2 ( xnŠ r )canalsobe writtenas xn +1Š r = q ( xnŠ r ),where q =1 Š1 2 .Apparently, xn r isequivalentto yn= xnŠ r 0.Butthesequence yn +1= qyn= q2yn Š 1= = qn +1y0convergesonlyif | q | = | 1 Š1 2 | < 1or > 1 / 4unless y0=0(i.e.,iftherootis accuratelyguessed!).Forexample,for f ( x )= x1 / 3( =1 / 6), Newtonssequencediverges: xn +1=(1 Š 3) xn= Š 2 xnfor any choiceoftheinitialpoint x0 =0.For f ( x )= | x |1 / 2( =1 / 4), Newtonssequenceoscillates xn +1=(1 Š 2) xn= Š xn(see Figure4.13).29.3.UnderstandingMoneyLoans.Supposethatonetakesaloanof P dollars(the principal )for n monthswithanannualinterestrateof

PAGE 146

1404.APPLICATIONSOFDIFFERENTIATION I %.Whatisthemonthlypayment?Itiscalculatedasfollows.The interestratepermonthis x = I/ 12.Forexample,anannualinterest rateof6%meansthat I =0 06and x =0 06 / 12=0 005.Each paymentincludesthepaymenttowardtheprincipalandtheinterest. Let Fkbetheamountyettobepaidafter k monthlypayments.It iscalledthe futurevalue oftheloan.Thesequence Fksatis“esthe conditions: F0= P and Fn=0(theloanandinterestarepaidoafter n payments).Let A bethemonthlypayment.Then F1= P + Px Š A,F2= F1+ F1x Š A,...,Fk= Fk Š 1+ Fk Š 1x Š A. Here F1isthefuturevalueoftheloanafteronepayment,whichisthe loan P plusthemonthlyinterest Px minusthepayment A .Afterone payment,theloanvalueis F1.So,afteronemorepayment,itsvalueis thevalue F1plusinterest F1x minusthepayment A ,andsoon.After n payments, Fn= Fn Š 1(1+ x ) Š A = Fn Š 2(1+ x )2Š A [(1+ x )+1]= = F0(1+ x )nŠ A [(1+ x )n Š 1+(1+ x )n Š 2+ +(1+ x )+1] = P (1+ x )nŠ A (1+ x )nŠ 1 x where,inthelastequality,thegeometricsumformula sn=1+ q + q2+ + qn Š 1=( qnŠ 1) / ( q Š 1)hasbeenused.Thelatterisproved bynotingthatthesums qsn= q + q2+ + qn Š 1+ qnand snhavethe sametermsexcept1in snand qnin qsnsothat qsnŠ sn= qnŠ 1 Š 1 or sn=( qnŠ 1) / ( q Š 1).Since Fn=0,themonthlypaymentis (4.19) A = Px 1 Š (1+ x )Š n. Forexample,aloanof$200,000for10yearsata“xedannualinterestrateof6%implies120monthlypaymentsof$2220 41.Indeed,in Equation(4.19),substitute x =0 06 / 12=0 005, n =120,and P = 200 000,then A 2220 41004.Thetotalamountpaidafter10yearsis 120 A =$266,449.20.Theinterestpaidis nA Š P =$66,449.20. Whensellingacar,adealermightoeramonthlypaymentfora fewyearsifacustomercannotaordtopaythepriceinfull.Inthis case,theloanamount P isthepriceofthecar;themonthlypayment A anditsnumber n areknown.Toassessthedeal,onehasto“gureout theinterestratebeforesigningup.Itmightbethecasethattheloan forahigher-qualitycar,meaningahigherpriceandhighermonthly payments,mighthavealowerinterestrate,thantheloanforacheaper

PAGE 147

29.NEWTONSMETHOD141 car(smallermonthlypayments).Knowingtheinterestrate,onecan alsoshopforaloanatalowerrateelsewhere(e.g.,banks)tobuya car.If A P ,and n aregiven,then x canbefoundbysolvingEquation (4.19),whichcanbewritteninamoreconvenientformas (4.20) f ( x )= Px (1+ x )nŠ A (1+ x )n+ A =0 Inotherwords,thisistheroot-“ndingproblem!Itcanbesolvedby Newtonsmethod.Thenumber x shouldbefoundupto“vedecimal places,whichissucientourpurposes. Example 4.15 Adealeroersacaratapriceof $ 10,000.Itcan alsobesoldforpaymentsof $ 217 42 permonthfor5years.Thereis anothercarbeingoeredatapriceof $ 15,000,whichcanalsobesold forpaymentsof $ 311 38 permonthfor5years.Whichloanhasalower interestrate? Solution: (I)Forthe“rstcar,onehasto“ndtherootofEquation(4.20)if A =217 42, P =10 000,and n =5 12=60.Itisconvenient toinitiateNewtonssequenceat x1=0 01,whichcorresponds toanannualinterestrateof12%(i.e., I =0 12and x = 0 12 / 12=0 01).Upto“vedecimalplaces,Newtonsmethod yields x =0 00917,whichcorrespondsto I =12 x =0 11004, oranannualinterestrateof11%. (II)Forthesecondcar,onehasto“ndtherootofEquation(4.20) if A =311 38, P =15 000,and n =5 12=60.Newtons method,initiatedagainat x1=0 01,yieldstheroot x = 0 00750(upto“vedecimalplaces).Thiscorrespondstoan annualinterestrateof9%.Sothesecondloanhasalower interestrate. Itisinterestingtonotethatthecarpricesdierby50%.Similarly, themonthlypaymentsappearinasimilarproportion311 38 / 217 42 1 43.Theoersmightlooklikeasnearlythesamedeal.Infact,they arenot!29.4.Exercises.(1)Findtherootsofagivenequationupto“vedecimalplaces. (i)cos x =2 x (ii) exŠ eŠ x=1 Š x2(iii)tanŠ 1x = x3(iv)tanŠ 1x =ln x

PAGE 148

1424.APPLICATIONSOFDIFFERENTIATION (v)ln(1+ x2)=4 Š x (vi) x5+ x Š 4=0 (vii)(4 Š x2)2Š x +4=0 (2)Newtonsmethodisbasedonthelinearapproximationofthe functionatasamplepoint xntogeneratethenextpoint xn +1ofNewtonssequence.Thisapproximationdoesnottakeinto accounttheconcavityofthefunctionat xn.GeneralizeNewtonsmethodbyusingtheTaylorpolynomial T2( x )at xnto generate xn +1asarootof T2.TakeanyoftheaboveexercisesandcomparetheconvergenceofNewtonsmethodwith itsgeneralization(i.e.,thenumbersofstepsneededtoobtain therootcorrecttothesamenumberofdecimalplaces,e.g.,6, 7,or10,startingwiththesameinitialpoint x0). (3)Consideraloanof$250,000atanannuallowinterestrateof 4%for15years.Findthemonthlypayments.Theinterest ratewasnot“xedandsubjecttochangesothatthemonthly paymentsmayincreaseupto20%.Howmuchmaytheannual interestrateincrease(percentagewise)? (4)Acardealeroersacaratapriceof$15,000for36monthly paymentsof$477.Whatistheinterestrate? (5)The“xedannualinterestrateonamortgageis7%.Forhow longshouldonetakealoanifonewantstopayintotalinterest nomorethanahalfoftheprincipal?Doesthemaximum loanperiodincreaseordecreasewithincreasingordecreasing interestrate?Doestheanswerchangeifthepaymentswillbe doneeverytwoweeks(i.e.,30paymentsperyearinsteadof 12)? (6)FindtherootoftheequationtanŠ 1x =1 Š x correctupto fourdecimalplacesbyinitiatingNewtonssequenceat x0=1. Investigatethedependenceofthenumberofneedediterations toachievethisaccuracyontheinitialpointbytaking x0= n where n =1 2 ,..., 10.30.Antiderivatives Inmanypracticalproblems,afunctionistoberecoveredfromits derivative.Forexample,ifthevelocityisgivenasafunctionoftime, v = v ( t ),onemightwantto“ndthepositionasafunctionoftime, s = s ( t ),where s( t )= v ( t ).Whatis s ( t )? Definition 4.8 Afunction F iscalledanantiderivativeof f on aninterval I if F( x )= f ( x ) forall x in I .

PAGE 149

30.ANTIDERIVATIVES143 Formanybasicfunctions,itisnotdicultto“ndthecorresponding antiderivative.Forexample,fromtherule( xn +1)=( n +1) xn,itfollows thatif f ( x )= xn, n = Š 1,theantiderivativeis F ( x )= xn +1/ ( n +1).It hasalsobeenprovedthat(ln | x | )=1 /x .Sothefunction F ( x )=ln | x | istheantiderivativeof f ( x )=1 /x forall x =0.30.1.UniquenessoftheAntiderivative.Suppose F( x )= f ( x )forall x inaninterval( a,b ).Issuchan F ( x )unique?Thisquestionisanswered byCorollary4.1.Indeed,let F ( x )and G ( x )beantiderivativesof f ( x ), thatis, F( x )= G( x )= f ( x )on( a,b ).ByCorollary4.1, F and G mayonlydierbyaconstant: G ( x )= F ( x )+ C .RecallthatCorollary 4.1doesnotholdfortheunionof disjoint intervals.Thus,anytwo antiderivativesofthesamefunctionmaydieratmostbyaconstant onaninterval. Theorem 4.15 If F isanantiderivativeof f onaninterval I thenthemostgeneralantiderivativeof f on I is F ( x )+ C, where C isanarbitraryconstant. Forexample,thegeneralantiderivativeofthepowerfunction f ( x )= xnwithapositiveinteger n is F ( x )= xn +1/ ( n +1)+ C because xnis de“nedonthesingleinterval( Š ).Thefunction f ( x )=1 /x isde“nedontheunionofdisjointintervals( Š 0)and(0 ).Sothegeneralantiderivativeis F ( x )=ln x + C1if x> 0and F ( x )=ln( Š x )+ C2if x< 0,where C1and C2arearbitraryconstants. Thenonuniquenessoftheantiderivativeisnotadrawbackofthe conceptbutratheragreatadvantage.Thisisexplainedbythefollowing example.Thevelocityofapieceofchalkthrownverticallyupwardwith avelocityof v0is v ( t )= v0Š gt ,where g =9 8m / s2istheacceleration ofafreefall.At t =0,thechalkhasavelocityof v (0)= v0.Thenit beginstoslowdown( v ( t )decreasesbecauseofgravity).Eventually,at t = v0/g ,thechalkstopsandbeginstofallback.If h ( t )istheheightof thechalkrelativetothe”oor,then h( t )= v ( t );thatis,theheightis anantiderivativeof v ( t ).Itiseasyto“ndaparticularantiderivativeof v ( t )usingtheantiderivativeofthepowerfunction: h ( t )= v0t Š gt2/ 2 (indeed, h( t )= v0Š gt ).Whatisthephysicalsigni“canceofthegeneral antiderivative h ( t )= C + v0t Š gt2/ 2?Itappearsasifthepositionof thechalkrelativetothe”oorisnotuniquelydetermined.Inparticular, h (0)= C istheheightattheverymomentwhenthechalkwasthrown upward.Butthechalkcouldbethrownupwarda t1mabovethe”oor or2maboveit withtheverysameinitialvelocity.So,inbothcases,

PAGE 150

1444.APPLICATIONSOFDIFFERENTIATION v ( t )isthesame,whilethe h ( t )arenot.Inthe“rstcase, h (0)=1, whereasinthesecondcase, h (0)=2. Thus,theconstant C canbe“xed byspecifyingthevalueoftheantiderivativeataparticularpoint Thisfeatureofthegeneralantiderivativecanalsobevisualizedby plottingthegraphs y = F ( x )+ C fordierentvaluesof C .Allsuch graphsareobtainedfromthegraph y = F ( x )byrigidtranslations alongthe y axis.Ifonedemandsthatthegraph y = F ( x )+ C should passthroughaparticularpoint( x0,y0),then C is“xed: y0= F ( x0)+ C or C = y0Š F ( x0).Forexample,“nd f ( x )if f( x )=3 x2and f (2)=1. Thegeneralantiderivativeof3 x2is f ( x )= x3+ C .From f (2)=1,it followsthat f (2)=8+ C =1or C = Š 7.Therefore, f ( x )= x3Š 7.30.2.LinearityoftheAntiderivative.Let F and G beantiderivativesof f and g ,respectively.Thenanantiderivativeof f + g is F + G .An antiderivativeof kf ,where k isanarbitraryconstant,is kF .These propertiesareeasilyveri“ed.Indeed,( F + G )= F+ G= f + g and( kF )= kF= kf ,wherethelinearityofthederivativehasbeen used.Inotherwords,antidierentiationisalinearoperationjustlike dierentiationitself.30.3.AntiderivativesofBasicFunctions.Anantiderivativeofthepower functionhasbeenfoundbystudyingthederivativeofthepowerand logarithmicfunctions.Theideaisusefulforotherbasisfunctions.Their antiderivativescanbefoundbyreadingthetableofderivativesofbasic functionsbackward,thatis,fromtherighttoleft.Forexample, (sin x )=cos x, ( Š cos x )=sin x, ( ex)= ex, (tan x )=(sec x )2, (sinŠ 1x )= 1 1 Š x2, (tanŠ 1x )= 1 1+ x2. Inparticular,thistablesaysthatthegeneralantiderivativeof f ( x )= 1 / (1+ x2)is F ( x )=tanŠ 1x + C .Thetableofderivativesofbasic functionscombinedwiththelinearityofantidierentiationisagood sourceofantiderivativesofmorecomplicatedfunctions. Example 4.16 Findthegeneralantiderivativeof f ( x )= eŠ 2 x+ cos(4 x )+ x2/ (1+ x2) Solution: (I)Bythelinearityoftheantiderivative,itissucientto“nd antiderivativesof eŠ 2 x,cos(4 x ),and x2/ (1+ x2).Thegeneralantiderivativeisobtainedbyaddingageneralconstantto thesumoftheparticularantiderivativesofthepreviousthree functions.

PAGE 151

30.ANTIDERIVATIVES145 (II)From( eŠ 2 x)= Š 2 eŠ 2 x,itfollowsthat( Š eŠ 2 x/ 2)= eŠ 2 x. Hence,anantiderivativeof eŠ 2 xis Š eŠ 2 x/ 2. (III)Similarly,from(sin(4 x ))=4cos(4 x ),itfollowsthatanantiderivativeofcos(4 x )issin(4 x ) / 4. (IV)Thetableofderivativesdoesnotappearhelpfulinthecaseof x2/ (1+ x2).However,asimplealgebraicmanipulationleads tothegoal: x2 1+ x2= 1+ x2Š 1 1+ x2=1 Š 1 1+ x2. Soitsantiderivativeis x Š tanŠ 1x .Thus,thegeneralantiderivativereads: F ( x )= Š 1 2 eŠ 2 x+ 1 4 sin(4 x )+ x Š tanŠ 1x + C. because f ( x )isde“nedonthesingleinterval( Š ). 30.4.AntiderivativesofHigherOrder.Whatis F ( x )if F( x )= f ( x ) foragiven f ( x )?Or,moregenerally,whatis F ( x )if F( n )( x )= f ( x )? Afunction F thatsatis“esthelatterconditioniscalledanantiderivativeof f ofthe n thorder.To“ndit,onehastoantidierentiate f n times.Forexample, F( x )=6 x .Takingthe“rstantiderivativeof f ( x )=6 x ,onegets F( x )=3 x2.Takingtheantiderivativeonemore timeyields F ( x )= x3.Whatabouttheuniquenessofhigher-order antiderivatives?To“ndthegeneralantiderivativeofahigherorder, eachtimeantidierentiationiscarriedout,thecorrespondinggeneral antiderivativemustbeused.Intheprecedingexample,thegeneral antiderivativeof f ( x )=6 x is3 x2+ C1,where C1isanarbitraryconstant.Hence, F( x )=3 x2+ C1.Itsgeneralantiderivativereads F ( x )= x3+ C1x + C2,where C2isanotherarbitraryconstant.Thus,thegeneral secondantiderivativecanbeobtainedfromaparticularonebyadding ageneralfunctionwhosesecondderivativeis0,whichisagenerallinear function:( C1x + C2)=0.Similarly,if F ( x )isaparticularfunction thatsatis“esthecondition F( n )( x )= f ( x ),thenthegeneralantiderivativeofthe n thorderis F ( x )+ C1xn Š 1+ C2xn Š 2+ + Cn Š 1x + Cn, where C1,...,Cnarearbitraryconstants.Indeed,the n thderivativeof apolynomialofdegree n Š 1is0.Notethatthisanalysisisjusti“ed onlywhen f wasde“nedinaninterval.Why?Whathappensifthe domainofthefunction f consistsofdisjointintervals? Thefollowingexampleillustratesthesigni“canceofarbitraryconstantsingeneralhigher-orderantiderivatives.

PAGE 152

1464.APPLICATIONSOFDIFFERENTIATION Example 4.17 Anyfree-fallingobjectnearthesurfaceoftheEarth hasthefree-fallaccelerationof 9 8 m/s2.Apieceofchalkisthrown verticallyupwardatas peedof 7 m/s andat 1 5 m abovethe”oor.When doesthechalkhitthe”oor? Solution: (I)Let h ( t )betheheightofthechalkrelativetothe”oor.Then itsvelocityis v ( t )= h( t ),anditsaccelerationis a ( t )= v( t )= h( t ).Sinceallfree-fallingobjectshaveanacceleration of9 8m / s2,onehas h( t )= Š 9 8.Theminussignindicates thattheaccelerationisdirecteddownward. (II)Thegeneralsecondantiderivativeoftheconstantfunction Š 9 8is h ( t )= Š 9 8 t2/ 2+ C1t + C2,where C1and C2are arbitraryconstants. (III)To“x C1and C2,theinitialconditionsofthemotionmust beused.Theinitialvelocityis v (0)=7.Since v ( t )= h( t )= Š 9 8 t + C1,onecaninferthat v (0)= C1=7.Theinitialheight is h (0)=1 5.Hence, h (0)= C1=1 5. (IV)Theheightis h ( t )= Š 9 8 t2/ 2+7 t +1 5.Thechalkhitsthe ”oorwhenitsheightvanishes,thatis,atthetimemoment t> 0when h ( t )=0.Apositiverootofthequadraticequation Š 9 8 t2/ 2+7 t +1 5=0is t 1 62s.Themaximumheight reachedbythechalkis4m.Why? 30.5.Exercises.(1)Findanantiderivativeofeachofthefollowingfunctions: (i) f ( x )=sin(4 x )+ x Š 1 /x (ii) f ( x )=1 / ( x2+4) (iii) f ( x )= e3 x+ eŠ 3 x+ x (iv) f ( x )=cos2x ,( Hint :2cos2x =1+cos(2 x )) (v) f ( x )=sin( ax )cos( bx ) (2)Findthegeneralantiderivativeofeachofthefollowingfunctions (i) f ( x )= x1 / 3(ii) f ( x )= xŠ 1 / 3(iii) f ( x )=1 / ( x +2) (iv) f ( x )= | x | (v) f ( x )=1 / ( x2Š 1) Hint :1 / [( x Š a )( x Š b )]= A [1 / ( x Š a ) Š 1 / ( x Š b )];“nd A .

PAGE 153

30.ANTIDERIVATIVES147 (3)Findthegeneralsecondantiderivativeofeachofthefollowing functions. (i) f ( x )= f0=constif x [ a,b ]and f ( x )=0otherwise (ii) f ( x )= | x | +1 (iii) f ( x )=sin(2 x ) (iv) f ( x )= e3 x+ eŠ 3 x(4)Acarthatwasatrestacceleratesatarateof1m/s2forone minute.Thenitdeceleratesatarateof0.5m/s2untilitstops. Findthedistancetraveledbythecar.

PAGE 155

CHAPTER5 Integration 31.AreasandDistances Considerthelinearfunction f ( x )= x .Whatistheareabelowthe graph y = f ( x )andabovetheinterval0 x 1?Thisquestionis easytoanswerbecausetheareainquestionistheareaoftheright trianglewithcathetiofunitlength: A =1 / 2.Let f ( x )= x2.What istheareanow?Tocalculateit,considera partition oftheinterval [0 1]by n segmentsonlength1 /n .Thepartitionisde“nedbythe setofpoints x0=0 ,x1=1 /n,x2=2 /n,...,xn Š 1=( n Š 1) /n ,and xn= n/n =1,thatis, xk= k/n ,where k =0 1 2 ,...,n .Thearea undertheparabola y = x2overtheinterval[0 1]isthesumofthe areas Skundertheparabolaoverthepartitioninterval[ xk Š 1,xk]where k =1 2 ,...,n A = S1+ S2+ + Sn. Thearea Skcannotexceedtheareaofarectanglewithbase1 /n and height f ( xk)=( k/n )2.Letusdenotethisupperboundby SU k= k2/n3. Thearea Skisgreaterthantheareaofarectanglewithbase1 /n andheight f ( xk Š 1)=( k Š 1)2/n2.Thelowerboundisdenotedby SL k=( k Š 1)2/n3.Thus, SL k= ( k Š 1)2 n3
PAGE 156

1505.INTEGRATION 1.0 0.8 0.6 0.4 0.2 01 3 4 1 2 1 4 1.0 0.8 0.6 0.4 0.2 01 3 4 1 2 1 4 Figure5.1. Theupperandlowerboundsforthearea underthegraph y = f ( x )= x2for n =4partitionintervalsin[0 1].Theupperboundisobtainedbytaking themaximumvalueof f oneachpartitioninterval(left panel).Thelowerboundisobtainedbytakingtheminimumvalueof f oneachpartitionsegment(rightpanel). When n increases,theupperbounddecreases,whilethe lowerbounddecreases,bothapproachingtheareaunder thegraphas n Thus,ifthelimitlimn AU nexists,thenlim AU n=lim AL nbecause 0
PAGE 157

31.AREASANDDISTANCES151 1 /n andheight f ( x k)=( x k)2,thatis, S k= f ( x k) /n .Thenthetotal areaunderthegraphisapproximatedbythesum A nofall Sk.Since SL k S k SU k(owingtothemonotonicityofthefunction x2ineach interval[ xk Š 1,xk]),thefollowinginequalityholdsforany n : AL n S 1+ S 2+ + S n= A n AU n. Takingthelimit n inthisinequalityleadstoaremarkableresult limn A n= A ; thatis,thelimitof A ndoesnotdependonthechoiceof samplepoints x k.Theareacouldhavebeenapproximatedby,forexample, A nwith thesamplepointsasthemidpoints x k=( xk+ xk Š 1) / 2,oranyother convenientchoice.Thisanalysiscanbeextendedtoanycontinuous function. Thecalculationoftheareaunderthegraphisfurtherillustratedin thevideowebsiteathttp://www.math.u”.edu/ mathguy/ufcalcbook/ riemann.html.31.1.TheAreaUndertheGraphofaContinuousFunction.Let f ( x )be continuouson[ a,b ].Considerapartitionof[ a,b ]by n segmentsof length x =( b Š a ) /n .Theendpointsofthepartitionssegmentsare xk= a + k x with k =0 1 2 ,...,n ,suchthat x0= a and xn= b .Let x kbeasamplepointintheinterval[ xk Š 1,xk]. Definition 5.1 Thearea A oftheregionthatliesunderthegraph ofacontinuousfunction f ( x ) 0 onaninterval [ a,b ] is (5.2) A =limn A n=limn f ( x 1) x + f ( x 2) x + + f ( x n) x foranychoiceofsamplepoints x k. Letusassessthisde“nition.Anycontinuousfunctionattainsits maximumandminimumvaluesonaclosedinterval.Let Mkand mkbe,respectively,themaximumandminimumvaluesof f ( x )onthe interval[ xk Š 1,xk].If Skistheareaunderthegraph y = f ( x )onthe interval[ xk Š 1,xk],then SL k= mk x Sk SU k= Mk x .Thearea S k= f ( x k) x oftherectanglewithbase x andheight f ( x k)isa continuousfunctionof x kontheinterval[ xk Š 1,xk].Therefore, S kmust takeallthevaluesbetweenitsminimumandmaximumvalues, SL kand SU k.Inparticular, S k= Skforsome x k [ xk Š 1,xk].Thus, forany “xed n ,thereisachoiceofsamplepointssuchthat A n= A Continuingtheanalogywiththeexampleof f ( x )= x2,letusshow thatthelimit(5.2)isindependentofthechoiceofsamplepoints,

PAGE 158

1525.INTEGRATION providedthe lower sums AL n= SL 1+ + SL nandthe upper sums AU n= SU 1+ + SU nconvergetothe same numberas n .Indeed, foranychoiceofsamplepoints SL k S k SU kand,hence,bytakingthesumoverthepartitioninthelatterinequality,oneinfersthat AL n A n AU n.Thereforeboththenumbers A nand A liebetween AL nand AU n: AL n A n AU nAL n A AU n= | A nŠ A | AU nŠ AL n(5.3) Thus,if AU nŠ AL n 0as n ,then A n A foranychoiceof partition.Thefollowingtheoremholds. Theorem 5.1 Let f beacontinuousfunctionon [ a,b ] .Suppose thatforanypartition x0= a n +1.Thentheupperandlowersumsconvergetothesame limitas n Thistheoremjusti“esthede“nition(5.2).Notealsothatthepartitionisnotgenerallyrequiredtobeequispaced.Theabovetheoremonly requiresthatthelengthnofthelargestpartitionintervaldecreases withincreasingthenumberofpartitionintervals(n=( b Š a ) /n = x foranequispacedpartition).31.2.ApproximatingtheAreaUnderaGraph.Inpractice,Equation(5.2) canbeusedto“ndtheareaunderthegraphthatiscorrecttoanydesirednumberofdecimalplaces.Takeapartitionoftheinterval[ a,b ], e.g.,“xsome n sothat x =( b Š a ) /n .Choosesamplepoints xkŠ 1 x k xk.Convenientchoicesmightbetheleftpoints x k= xk Š 1,the rightpoints x k= xk,orthemidpoints x k=( xk Š 1+ xk) / 2.Calculate thesum A n,keepingthedesirednumberofdecimalplaces.Re“nethe partitionby,forexample,doublingthenumberofsegments,andcalculate A 2 n.If A nand A 2 ncoincideinthedesirednumberofdecimal places,then A = A 2 niscorrecttothatnumberofdecimalplaces.If not,re“nethepartitionfurtherandcompute A 4 nandcompareitwith A 2 nandsoon,untiltheneededaccuracyisreached.Forany n ,the absoluteerroroftheapproximationmayestimatedbytheinequality ontherightin(5.3).31.3.SigmaNotationforSums.Toavoidwritinglengthyexpressions forsumsofanarbitrarynumberofterms,itisconvenienttoadoptthe

PAGE 159

31.AREASANDDISTANCES153 followingnotation: A n= S 1+ S 2+ + S n=nk =1S k, wheretheindex k iscalledthe summationindex .Thesymbol means addingall S k,startingwith k =1upto k = n .Forexample,the geometricsumformulacannowbewrittenas (5.4)1+ q + q2+ + qn=nk =0qk= qn +1Š 1 q Š 1 .31.4.TheDistanceProblem.Ifanobjectmoveswithaconstantvelocity v duringatimeinterval a t b ,thenthedistancetraveledby theobjectis D = v ( b Š a ).Howdoesonecalculatethedistanceifthe speedisanonconstantcontinuousfunctionoftime v = v ( t ) 0? Let D ( t )bethedistanceasafunctionoftime a t b .It satis“esthecondition D ( a )=0.Since v ( t ) 0,theobjecttravelsin thesamedirectionallthetime,and D ( t )increasesbecause D( t )= v ( t ) 0.Thus, D = D ( b ).Tocalculate D ( b ),considerapartition of[ a,b ]byinterval[ tk Š 1,tk]where tk= a + tk t =( b Š a ) /n k =0 1 ,...,n .Thedistance Dk= D ( tk) Š D ( tk Š 1)traveledbythe objectinthetimeinterval[ tk Š 1,tk]canbefoundbythemeanvalue theorem: D ( tk) Š D ( tk Š 1)= v ( t k) t forsome t kin[ tk Š 1,tk].Recall that v ( t k)istheaveragevelocityoverthetimeinterval[ tk Š 1,tk].The totaldistanceis D = D1+ + Dn.Ontheotherhand,points t krepresentaparticularchoiceofsamplepointsinthede“nition(5.2) appliedtoacontinuousfunction v ( t ).Therefore, D istheareaunder thegraphof v ( t )and,hence,canbecalculatedwith any choiceof samplepoints t k,notnecessarilywiththoseatwhich v coincideswith theaveragevelocityineachpartitioninterval: D =limn nk =1v ( t k) t, Furthermore,bythecondition D( t )= v ( t )thefunction D ( t )is theantiderivativeof v ( t )satisfyingtheinitialcondition D ( a )=0.If D ( t )is any antiderivativeof v ( t ),then D ( t )and D ( t )candieronly byaconstant, D ( t )= D ( t )+ C .Theconstant C is“xedbythe condition D ( a )=0and,hence, C = Š D ( a ).Thedistancetraveled is D = D ( b )= D ( b ) Š D ( a )Thisestablishesthefollowingrelation

PAGE 160

1545.INTEGRATION betweentheareaunderthegraphof v ( t )anditsantiderivative: (5.5) D =limn nk =1v ( t k) t = D ( b ) Š D ( a ) Example 5.1 Amovingobjectslowsdownsothatitsvelocityis v ( t )= eŠ 2 t.Whatisthedistancetraveledbytheobjectduringthetime interval 0 t 1 ? Solution: Let t =1 /n sothat tk= k/n k =0 1 ,...,n .Take t k=( k Š 1) /n k =1 2 ,...,n (theleftpointsofpartitionintervals). Then v ( t k) t = qk Š 1/n ,where q = eŠ 2 /n.Thedistancetraveledis D =limn 1 nn Š 1k =0qk=limn 1 n qnŠ 1 q Š 1 = 1 Š eŠ 2 limn n (1 Š eŠ 2 /n) wherethesumformula(5.4)hasbeenused.Tocomputethelimitin thedenominator,let x =1 /n ,thatis, x 0.Thelimitbecomesthe indeterminateform(1 Š eŠ 2 x) /x oftype0 / 0,whichcanberesolvedby lHospitalsrule:(1 Š eŠ 2 x)/ ( x )=2 eŠ 2 x/ 1 2as x 0.Thus,the distancetraveledis D =(1 Š eŠ 2) / 2. Alternativesolution .Anantiderivativeof v ( t )= eŠ 2 tis D ( t )= Š eŠ 2 t/ 2.ByEquation(5.5), D = D (1) Š D (0)=(1 Š eŠ 2) / 2. Whencomparedtotheprevioussolution,thisonelookslikecheating!Moretothepoint,take v ( t )= t2(theexamplediscussedatthe beginningofthissection).Itsantiderivativeis D ( t )= t3/ 3.Sothedistancetraveled,ortheareaunderthegraphof t2,is D (1) Š D (0)=1 / 3. Itturnsoutthattherelation(5.5)betweenanantiderivativeofafunctionandtheareaunderthegraphofthefunctionisnotspeci“cforthe distanceproblem.Itsgeneralizationwillbeestablishedwiththehelp oftheconceptofthe de“niteintegral .31.5.Exercises.(1)Findexplicitformulasfortheupperandlowersums, AU nand AL n,for f ( x )=2 x +1on[0 2]usinganequispacedpartition.Findthelimitsof AU nand AL nas n .Whatisthe geometricalsigni“canceofthislimit? (2)Findexplicitformulasfortheupperandlowersums, AU nand AL n,for f ( x )= x3on[0 1]usinganequispacedpartition.Show

PAGE 161

31.AREASANDDISTANCES155 that AU,L n 1 / 4as n .Whatisthegeometricalsigni“canceofthislimit? Hint :nk =1k3= 1 4 n2( n Š 1)2. (3)Findtheareaunderthegraphof f ( x )= eŠ x2on[ Š 1 1]correct upto“vedecimalplaces. (4)Findtheareaunderthegraph f ( x )= 1 Š x2,where Š 1 x 1,correctuptothreedecimalplaces.Usethegeometrical interpretationofthisareato“nditsexactvalue. (5)Findtheareaunderthegraphofeachofthefollowingfunctions onthegivenintervalusingtherelation(5.2): (i) f ( x )=3 Š 3 x, 0 x 1 (ii) f ( x )=1+ x + x2, 0 x 2 (iii) f ( x )= e3 x, Š 1 x 1 (6)Usetherelationn Š 1k =0sin(2 kx )= sin( nx )sin[( n Š 1) x ] sin x to“ndthefollowing: (i)Theupperandlowersumsfor f ( x )=sin x ontheinterval [0 ].Calculate AU nŠ AL nandinvestigateitsbehavioras n increases.Whatisthesigni“canceofthisnumberfora“xed n ? (ii)Theareaunderthegraphof f ( x )=sin x ,0 x using(5.2). (iii)Theareaunderthegraphof f ( x )=cos x ,0 x / 2, using(5.2). (7)Anobjecttravelswithvelocity v ( t )=cos2t .Findthedistance passedbytheobjectoverthetimeinterval0 t 2 (8)Useantiderivativesto“ndtheareaunderthegraphofeachof thefollowingfunctions.Explainwhythismethodcanbeused to“ndtheareaineachcase. (i) f ( x )=1 / ( x2+1) 0 x 1 (ii) f ( x )=sin( ax ) 0 x /a (9)Let f ( x )=( x5Š 1) / ( x Š 1)if0 x< 1and f (1)=5.Show that f ( x )iscontinuouson[0 1].Useantiderivativesto“nd theareaunderthegraphof f Hint :SeeEquation(5.4). (10)Findtheareaofaplanarregionboundedbythecurves y = 2 Š x2and y =1.

PAGE 162

1565.INTEGRATION (11)Findtheareaofaplanarregionboundedbythecurves y = x2and y = x .32.TheDe“niteIntegral Ageneralizationoftheconceptoftheareaunderagraphleadsto oneofthemostfundamentalconceptsincalculus,thede“niteintegral.32.1.SupremumandIn“mum.Theareaunderagraphisalsowellde“nedifthefunctionhassomenumberofboundedjumpdiscontinuities. Thedierencewiththecaseofacontinuousfunction f isthatnow f mayormaynotattainitsmaximumorminimumvaluesoneachpartitioninterval.Whatshouldbechangedinthede“nitionofthearea toaccommodatepossiblejumpdiscontinuitiesofthegraph?Suppose afunction f isboundedonaninterval[ a,b ];thatis,therearenumbers m and M suchthat m f ( x ) M forall x [ a,b ].If m isalower bound,thenanynumber m1m mayormaynotbealowerbound.Soonecan“ndthegreatest lowerboundthatisuniquefor f on[ a,b ].Similarly,onecan“ndthe lowestupperboundof f on[ a,b ].Theseboundshavespecialnames. Definition 5.2 (In“mumandSupremum). Thenumber m is calledthe in“mum ofaboundedfunction f onaninterval I =[ a,b ] if m isalowerboundof f but m + a isnotalowerboundforany a> 0 .Thisnumberisdenotedas m =infIf .Thenumber M is calledthe supremum of f on [ a,b ] if M isanupperboundof f but M Š a isnotanupperboundforany a> 0 .Thisnumberisdenoted as M =supIf Naturally,ifthefunctioniscontinuous,thensup f isnothingbut themaximumvalueof f andinf f isitsminimumvalue.However,if afunctionhasjumpdiscontinuities,thensup f andinf f alwaysexist,whilethemaximumandminimumvaluesmaynotexist.Thisis illustratedinFigure5.2.32.2.De“nitionoftheDe“niteIntegral.Let f beaboundedfunction onaninterval[ a,b ].Considerapartitionof[ a,b ]by n intervals Ik= [ xk Š 1,xk], k =1 2 ,...,n ,where a = x0
PAGE 163

32.THEDEFINITEINTEGRAL157 1 acb 1/2 1 acb 1/2 1 acb 1/2 1 acb 1/2 sup f 1> f ( x ) f ( c ) 1/2 sup f max f 1 inf f mix f 1/2 inf f 1/2< f ( x ) Figure5.2. Relationsbetweenthesupremumandin“mumof f andthemaximalandminimalvaluesof f Upperleftpanel :Thevaluesofthefunctionapproach 1as x approaches c fromtheleft,but f ( c )=1 / 2 < 1. Themaximumvalueof f doesnotexist,butthelowest upperbounddoesexist,sup f =1. Lowerleftpanel :Thevaluesof f approach1 / 2as x approaches c fromtheright,but f ( c )=1.Thefunction hasnominimumvalue,butthegreatestlowerboundis inf f =1 / 2. Upperrightpanel :Thevaluesof f approach1as x approaches c fromtheleftand f ( c )=1.Inthiscase, themaximalvalue f ( c )=1coincideswithsup f =1. Lowerrightpanel :Thevaluesof f approach1 / 2as x approaches c fromtherightand f ( c )=1 / 2.Theminimumvalue f ( c )=1 / 2coincideswiththegreatestlower boundinf f =1 / 2.

PAGE 164

1585.INTEGRATION foreverypartitionof[ a,b ].Putn=maxk xkwhichisthelength ofthelargestpartitionintervalfora“xed n .Thesequencesoflower anduppersumarede“nedsothatn> n +1;thatis,thelengthof thelargestpartitionintervaldecreaseswithincreasingthenumberof partitionintervals. Definition 5.3(TheDe“niteIntegral) Aboundedfunction f is saidtobe integrableonaninterval[ a,b ] ifthesequencesofitslower anduppersumsconvergetothesamenumber.Thisnumberiscalled the de“niteintegral of f from a to b andisdenotedby b af ( x ) dx =limn AL n=limn AU n; thenumbers a and b arecalledthe lowerandupperintegrationlimits respectively,andthefunction f iscalledthe integrand Apparently,fora continuousandnonnegative f on[ a,b ],thede“nite integralcoincideswiththeareaunderthegraphof f .Similarlyto theareaunderthegraphofacontinuousnonnegativefunction,an integrablefunctionhastheproperty AL n b afdx AU nforany n (seeExercise32.9.4).32.3.RiemannSums.ThereisananalogofEquation(5.2)forthe de“niteintegral. Definition 5.4 Let Ikbepartitionintervalsof [ a,b ] xkbethe lengthof Ik,and x k Ik.Thesum Rn( f )=nk =1f ( x k) xkiscalleda Riemannsum ofafunction f on [ a,b ] Thesum Rn( f )isnamedaftertheGermanmathematicianBernhardRiemann(1826…1866).Evidently,thevalueoftheRiemannsum generallydependsonthechoiceofpartitionintervalsandsamplepoints x k.However,forintegrablefunctionsRiemannsumshavearemarkable property. Theorem 5.2 If f isintegrableon [ a,b ] ,then,foranynumber > 0 ,thereexistsaninteger N suchthat b af ( x ) dx Š Rn( f ) < foreveryinteger n>N andforeverychoiceof x kin Ik.

PAGE 165

32.THEDEFINITEINTEGRAL159 Aproofofthistheoremisgivenasanexercise(seeExercise32.9.5; seealsoExercises32.9.12and32.9.13).Thetheoremassertsthata Riemannsumforasucientlylarge n canapproximatethede“nite integralwithanydesiredaccuracy;thatis,forany(small)designated absoluteerror Rn( f )diersfrom b afdx nomorethan forasucientlylarge n .Inotherwords, (5.6)limn Rn( f )= b af ( x ) dx, foranychoiceofsamplepoints x k.Equation(5.6)istheanalogofEquation(5.2).Itcanbeunderstoodfromtheinequality AL n Rn( f ) AU n, whichfollowsfrom mk f ( x k) Mkforany x k(seeFigure5.3). ab x1 *x2 *x3 *x4 *x5 Figure5.3. Riemannsumfor n =5partitionintervals.Itsvaluealwaysliesbetweenthelowerandupper sums, AL 5 R5 AU 5,foranychoiceofsamplepoints x kbecause mk f ( x k) Mk. Foranintegrablefunction, AL nand AU nconvergetothesamenumber,whichisthevalueofthede“niteintegral,and,bythesqueeze principle,soshould Rn( f )independentlyofthechoiceofsamplepoints.32.4.ContinuityandIntegrability.Therelation(5.6)canbeusedto calculatethede“niteintegral, provided thefunction f is integrable Thequestionofintegrabilityrequiresinvestigatingtheconvergenceof

PAGE 166

1605.INTEGRATION thesequencesoftheupperandlowersums,whichmightbeatedious taskevenforsuchsimplefunctionsas,forexample, f ( x )= x2,as discussedintheprevioussection.Thefollowingtheoremishelpful whenstudyingthequestionofintegrability. Theorem 5.3 If f iscontinuouson [ a,b ] ,orif f hasonlya“nite numberofboundedjumpdiscontinuities,then f isintegrableon [ a,b ] ; thatis,thede“niteintegral b af ( x ) dx exists. Aboundedfunction f within“nitelymanyjumpdiscontinuities mayormaynotbeintegrable.So,ingeneral,theareaunderthe graphofsuchafunctioncannotbeunambiguouslyde“ned.Asan example,consideraboundednonnegativefunction f on[0 1]suchthat f ( x )=1if x isarationalnumber,and f ( x )=0otherwise(i.e.,if x isirrational).Thefunctionisnotcontinuousanywherein[0 1]and hasin“nitelymanyjumpdiscontinuities.Forexample, f (1 / 2)=1, butwhen x approaches1 / 2,thevalue f ( x )keepsjumpingfrom0to 1andback,nomatterhowclose x isto1 / 2because,forany > 0, theinterval(1 2Š ,1 2+ ) always containsbothrationalandirrational numbers.Thisfunctionisnotintegrable.Indeed,takeapartition xk= k/n k =0 1 ,...,n .Anypartitioninterval[( k Š 1) /n,k/n ] containsbothrationalandirrationalnumbers.Therefore, mk=0and Mk=1.Hence,thelowersumvanishesforanypartition, AL n=0, whereastheuppersumis AU n= n k =1 x =1,thatis,limn AL n=0 whilelimn AU n=1.Thefunctionisnotintegrable.Theintegral doesnotexist.NotethattheRiemannsumcanstillbede“ned,but itslimitwould depend onthechoiceofsamplepoints(e.g.,take x kto berationalnumbersortake x ktobeirrationalnumbers;bothoptions arepossiblesinceanypartitionintervalalwayscontainsrationaland irrationalnumbers).32.5.PropertiesoftheDe“niteIntegral.Suppose f ( x )= c ,where c is aconstant.Inthiscase,foranypartitioninterval Ik, Mk= mk= c and AU n= AL n= c n k =1 x = cn x = c ( b Š a ).Inotherwords,a constantfunctionisintegrableanditsintegralis c ( b Š a ): (5.7) b acdx = c ( b Š a ) Foranytwointegrablefunctions f ( x )and g ( x )andconstants c1and c2,itfollowsfromtheconvergenceoftheRiemannsums(5.6)for f and g that

PAGE 167

32.THEDEFINITEINTEGRAL161 b a[ c1f ( x )+ c2g ( x )] dx =limn nk =1[ c1f ( x k)+ c2g ( x k)] xk= c1limn nk =1f ( x k) xk+ c2limn nk =1g ( x k) xk= c1b af ( x ) dx + c2b ag ( x ) dx. (5.8) Sotheintegrationisa linearoperation .Inparticular,theintegralof thesumoftwofunctionsisthesumoftheirintegrals.Theintegralof afunctionmultipliedbyaconstantistheproductoftheconstantand theintegralofthefunction.Iftheintegrationlimitsarereversed,then all xkchangetheirsignsas xkbecomeslessthan xk Š 1.Therefore, (5.9) b af ( x ) dx = Š a bf ( x ) dx and,inparticular, (5.10) a af ( x ) dx =0 Itcanbeprovedthat (5.11) b af ( x ) dx = c af ( x ) dx + b cf ( x ) dx for f integrableon[ a,b ]andany a c b .Theproofisrather technicalandisomitted.If f iscontinuousandpositiveon[ a,b ],then theproperty(5.11)istrivial:Theareaunderthegraphof f on[ a,b ] isthesumoftheareasunderthegraphof f on[ a,c ]and[ c,b ].32.6.GeometricalSigni“canceoftheDe“niteIntegral.Asalreadynoted, thede“niteintegralof f from a to b coincideswiththeareaunderthe graphof f foracontinuousandpositive f .Suppose f iscontinuousand negative on[ a,b ].Considerthefunction g ( x )= Š f ( x ).Theintegralof g isthearea A underthegraphof g and,hence, A alsocoincideswith thearea above thegraphof f andbelowthe x axis.Bythelinearity oftheintegral, b af ( x ) dx = Š b ag ( x ) dx = Š A .So,foranegative f theintegralof f coincideswiththe negative areaoftheregionbounded belowbythegraphof f andabovebythe x axis.Nowlet f becontinuouson[ a,b ].Letitbepositiveon[ a,c ]andnegativeon[ c,b ],that is, f ( c )=0.Thenitfollowsfromtheproperty(5.11)that b af ( x ) dx = c af ( x ) dx + b cf ( x ) dx = A1Š A2,

PAGE 168

1625.INTEGRATION where A1isthearea under thegraphof f on[ a,c ]and A2isthearea above thegraphof f on[ c,b ].ThispropertyisillustratedinFigure5.4. a b A1A2I A2 A1 Figure5.4. Geometricalinterpretationofthede“nite integral.Ifanintegrablefunction f isnonnegativeonan interval[ c,b ],thenitsintegralover[ c,b ]isthearea A2underthegraphof f abovetheinterval[ c,b ].If f isnonpositiveonaninterval[ a,c ],thenitsintegralover[ a,c ]is Š A1,where A1istheareaabovethegraphof f andbelow theinterval[ a,c ]onthe x axis.Bytheadditivityofthe integral,theintegralof f overtheinterval[ a,b ]beingthe unionofintervals[ a,c ]and[ c,b ]isthedierence A2Š A1. Thisshowsthatthevalueontheintegralcanbeanyreal number.32.7.ComparisonPropertiesoftheIntegral.Thefollowingadditional propertiesofthede“niteintegralcanbeestablished: b af ( x ) dx 0 if f ( x ) 0in[ a,b ] (5.12) b af ( x ) dx b ag ( x ) dx, if f ( x ) g ( x )in[ a,b ] (5.13) m ( b Š a ) b af ( x ) dx M ( b Š a ) if m f ( x ) M in[ a,b ] (5.14) Theproperty(5.12)followsdirectlyfromthede“nition:0 mk Mkforanypartitionif f ( x ) 0;thatis,theupperandlowersums arenonnegativeandsomustbetheintegral.If f iscontinuous,the

PAGE 169

32.THEDEFINITEINTEGRAL163 M m ab Figure5.5. Geometricalinterpretationoftheproperty (5.14).Thegraphofafunction f liesbetweentwohorizontallines y = m and y = M because m f ( x ) M forall x [ a,b ].Sothearea A underthegraphof f lies betweentheareasofrectangleswiththebase b Š a and heights m and M ,i.e., m ( b Š a ) A M ( b Š a ). property(5.12)statestheobviousthattheareaunderthegraphof f is nonnegative.Theproperty(5.13)followsfrom(5.12)forthefunction f ( x ) Š g ( x ) 0andthelinearityoftheintegral(5.8).Theproperty (5.14)isalsoaconsequenceofthede“nition.Indeed,foranypartition, m mk Mk M .Hence, m ( b Š a ) AL n AU n M ( b Š a )for any n .Inthelimit n ,thisinequalityturnsinto(5.14).32.8.EvaluationoftheIntegralbytheRiemannSum.Iftheintegralexists( f isintegrable),thenitcanbeevaluatedasthelimitoftheRiemannsum(5.6).Thelimitisindependentofthechoiceofsample points.Thefollowingchoicesareoftenusedinpractice: x k= xk Š 1(theleft-pointrule) x k= xk(theright-pointrule) x k=( xk Š 1+ xk) / 2(themidpointrule) incombinationwiththebasicpropertiesoftheintegral.Theevaluation oftheRiemannsumisrathertechnical.Formulaslike(5.1),(5.4),and (5.15)nk =1k = n ( n Š 1) 2 ,nk =1k3= n ( n Š 1) 2 2

PAGE 170

1645.INTEGRATION canbehelpful.However,theRiemannsumismostlyusedtocalculate theintegral approximately withsomedesignatedaccuracybymeansof computersimulations,similarlytoapproximatecalculationsofthearea discussedintheprevioussection. Example 5.2 Findthede“niteintegralof f ( x )= eŠ 2 xŠ 2 x2+4 x3from 0 to 1 Solution: (I)Thefunctioniscontinuouson[0 1]andhenceintegrable;that is,Equation(5.6)appliesforanychoiceof x k.Theleft-point rulewillbeused. (II)Bythelinearityoftheintegral, 1 0f ( x ) dx = 1 0eŠ 2 xdx Š 2 1 0x2dx +4 1 0x3dx. The“rstintegralis(1 Š eŠ 2) / 2byExample5.1(wherethe areaunderthegraphof eŠ 2 xin[0 1]wascalculated).The areaunderthegraph x2in[0 1]canbefoundatthebeginning oftheprevioussectionandisequalto1 / 3.Theareaunderthe graphof x3canbefoundwiththehelpofthesecondrelation in(5.15).Let x =1 /n and xk=( k Š 1) /n (theleft-point rule),thentheRiemannsum(5.6)becomes 1 0x3dx =limn 1 n4 nk =1k3=limn 1 n4n2( n Š 1)2 4 = 1 4 (III)Thus, 1 0f ( x ) dx = 1 Š eŠ 2 2 Š 2 3 + 1 4 = 1 Š 6 eŠ 2 12 32.9.Exercises.(1)Let f ( x )=sin(1 /x )if x =0and f (0)= f0.Givenanynumber > 0,“ndthesupremumandin“mumof f on[ Š ]. (2)Findtheupperandlowersumsforthefunction f ( x )=1if x 0and f ( x )= Š 2if x< 0ontheinterval[ Š 1 1].Use themtoshowthat f isintegrable. (3)Findtheupperandlowersumsforthefunction f ( x )=1if x =0and f ( x )= f0 =1if x =0onaninterval[ a,b ].Use themtoshowthat f isintegrableonany[ a,b ].

PAGE 171

32.THEDEFINITEINTEGRAL165 (4)Supposethatthelengthofthelargestpartitioninterval,n= maxk xk,decreasesasthenumber n ofpartitionintervals increases(i.e.,n +1< n).Showthat AL n AL n +1and AU n AU n +1.Deducefromthispropertythat AL n b afdx AU nforany n andforanyintegrable f on[ a,b ]. (5)Usetheinequalityfromthepreviousexercisetoprove Theorem5.2. (6)Let a
PAGE 172

1665.INTEGRATION (11)Let f becontinuouson[ a,b ].Then g ( x )= | f ( x ) | isintegrable on[ a,b ].Why?Showthat b afdx b a| f ( x ) | dx. (12)Let f haveaboundedderivativeon[ a,b ],thatis, | f( x ) | M1forall x [ a,b ].Consideranequispacedpartitionof[ a,b ]with x =( b Š a ) /n .Foreverypartitioninterval Ik=[ xk Š 1,xk], xk= a + kx x k =0 1 ,...,n ,showthatthereis xk Iksuchthat Ak= f ( xk) x istheareaunderthegraphof f on Ik.Let x k IkbesamplepointsintheRiemannsumfor f Usethemeanvaluetheoremtoprovethat | f ( x k) x Š Ak| M1 x2forany k .Deducefromthisinequalitythat b afdx Š Rn( f ) M1( b Š a )2 n (13)(Thetrapezoidalrule).Let f haveaboundedsecondderivativeon[ a,b ],thatis, | f( x ) | M2forall x [ a,b ].Bythe meanvaluetheorem,thereis c such f ( b ) Š f ( a )= f( c )( b Š a ). De“nethefunction g ( x )= f ( a )+ f( c )( x Š a ).Thegraph y = g ( x )isthesecantlinethroughthepoints( a,f ( a ))and ( b,f ( b )).Then T = b ag ( x ) dx =( f ( a )+ f ( b ))( b Š a ) / 2isthe areaofthetrapezoidboundedbytheline y = g ( x )on[ a,b ]. Usethemeanvaluetheoremforthederivative ftoprovethat | f ( x ) Š g ( x ) | M2( b Š a )2. Usethisinequalitytoshowthat b afdx Š b agdx M2( b Š a )3. Thetrapezoidalruletocalculate b afdx usesthepiecewise linearapproximationof f by g oneachpartitioninterval Ik= [ xk Š 1,xk]oflength xk: b afdx Tn( f )=nk =1xkxk Š 1gdx =nk =1 1 2( f ( xk Š 1)+ f ( xk)) xk.

PAGE 173

33.THEFUNDAMENTALTHEOREMOFCALCULUS167 Provethatforanequispacedpartition b afdx Š Tn( f ) M2( b Š a )3 n2. Bycomparingthisresultwiththatinexercise11,onecansee thattheerrorofthetrapezoidalruledecreasesfasterthanthat intheRiemannsumapproximationasthenumberofpartition intervalsincreases.Soitisabetterwaytoapproximatethe integral.Oneshouldkeepinmind,however,thattheintegrand hastohaveabounded second derivativeforsuchasuperiority. (14)Evaluate 2 0sin( x2) dx correctuptothreedecimalplacesusing theRiemannsumandtrapezoidalapproximations.Howmany partitionintervalsarerequiredtoachievethisaccuracyineach oftheapproximations?33.TheFundamentalTheoremofCalculus Inthissection,therelationbetweenthede“niteintegralofafunctionanditsantiderivativewillbeestablished.Thisrelationprovides apowerfulmethodforcalculatingthede“niteintegralthatavoidsthe useofRiemannsums.33.1.IntegrationandDifferentiation.Considerthede“niteintegralof f ( t )= t from0to x forsome x> 0.Thisintegralrepresentsthearea underthegraphof f ( t )= t intheinterval[0 ,x ],whichistheareaofa righttriangle: A ( x )= x 0tdt = x2 2 Thearea A ( x )canbeviewedasafunctionofthevariable x ,which isthelengthofthetrianglecatheti.Thisfunctionhasaninteresting property: A( x )= x = f ( x ) Inotherwords,thederivativeofthede“niteintegralwithrespecttoits upperlimitequalsthevalueoftheintegrandattheupperlimit.Recall thatif v ( t ) 0isthespeedofamovingobject,thenthedistance traveledbytheobjectintime T isgivenbytheareaunderthegraph of v ( t ): D ( T )= T 0v ( t ) dt. Ontheotherhand,thespeedistherateofchangeof D ( T ),andthereforethereshouldbe D( T )= v ( T );thatis,thederivativeoftheintegral withrespecttoitsupperlimitisagainthevalueoftheintegrandat

PAGE 174

1685.INTEGRATION theupperlimit.Howgeneralisthisproperty?Doesitholdforall integrablefunctions?Thefollowingtheoremanswersthesequestions. Theorem 5.4 If f iscontinuouson [ a,b ] ,thenthefunctionde“ned by g ( x )= x af ( t ) dt,a x b, iscontinuouson [ a,b ] anddierentiableon ( a,b ) ,and g( x )= f ( x ) Proof. Bythede“nitionofthederivative,onehastoprovethat (5.16)limh 0g ( x + h ) Š g ( x ) h = f ( x ) for a 0.Then m f ( t ) M for x t x + h and,bytheproperty(5.14), (5.17) mh = f ( u ) h x + h xf ( t ) dt f ( v ) h = Mh. Since h> 0,bydividingthisinequalityby h ,onecaninferthat (5.18) f ( u ) 1 h x + h xf ( t ) dt f ( v ) forsome u and v in[ x,x + h ].Inequality(5.18)canbeestablished for h< 0inasimilarmanner.Indeed,inequality(5.17)holdsforthe integral x x + hf ( t ) dt .Afterdividingitby Š h> 0,inequality(5.18)

PAGE 175

33.THEFUNDAMENTALTHEOREMOFCALCULUS169 isobtainedbutwiththeminussignattheintegral.Bytheproperty (5.9),thesignisreversed,yielding(5.18).Thus, f ( u ) g ( x + h ) Š g ( x ) h f ( v ) Since u and v lieintheinterval[ x,x + h ], limh 0f ( u )= f ( x ) limh 0f ( v )= f ( x ) Thentherelation(5.16)followsfromthesqueezeprinciple: f ( x )=limh 0f ( u ) limh 0g ( x + h ) Š g ( x ) h limh 0f ( v )= f ( x ) Thistheorembasicallystatesthatifacontinuousfunctionis“rst integratedandthendierentiated,thenitremainsunchanged: (5.19) d dx x af ( t ) dt = f ( x ) ,a 1.But g(1)does notexist. Example 5.3 Let g ( x )= b xeŠ t2dt .Find g( x ) Solution: Thefunction eŠ t2isacontinuousfunctioneverywhereasa compositionoftwocontinuousfunctions,theexponentialandpower functions.Bytheproperty(5.9), g ( x )= Š x beŠ t2dt .Therefore, g( x )= Š eŠ x2by(5.19). Thisexampleillustratesthegeneralproperty: d dx b xf ( t ) dt = Š d dx x bf ( t ) dt = Š f ( x ) foracontinuous f .

PAGE 176

1705.INTEGRATION 33.2.TheDe“niteIntegralandAntiderivative.Thefollowingtheorem establishestherelationbetweenthede“niteintegralofafunctionand itsantiderivative. Theorem 5.5 (TheFundamentalTheoremofCalculus) .If f is continuouson [ a,b ] ,then b af ( x ) dx = F ( b ) Š F ( a ) where F isanyantiderivativeof f ,thatis,afunctionsuchthat F= f Proof. Let g ( x )= x af ( t ) dt .By(5.19),thefunction g ( x )isan antiderivativeof f ( x )inanopeninterval( a,b ).If F isanyother antiderivativeof f ,then F and g maydieronlybyaconstant, F ( x )= g ( x )+ C,a
PAGE 177

33.THEFUNDAMENTALTHEOREMOFCALCULUS171 Anantiderivativeof xnis xn +1/ ( n +1)foranyreal n =1.Bytaking n = Š 1 / 2and n =1 / 2,anantiderivativeisobtained: F ( x )=2 x1 / 2+ 2 x3 / 2/ 3.Hence, 1 01+ x x dx = F (4) Š F (1)= 4+ 16 3 Š 2+ 2 3 = 20 3 Ifanobjectmovesalongastraightline,itspositionrelativetoa “xedpointontheline(theorigin)maybede“nedbyasinglecoordinate x whichisafunctionoftime.Thevelocity v ( t )= x( t )ispositiveif theparticlemovesinthedirectioninwhich x increasesandisnegative ifitmovesintheoppositedirection.Theaccelerationis a ( t )= v( t )= x( t ).Alawaccordingtowhichtheaccelerationchangeswithtimeis usuallyestablishedbythelawsofphysics.Thenapracticalquestion isto“ndtheposition x ( t ).Since x ( t )isasecondantiderivativeof theacceleration,itisnotuniqueandtwo(initial)conditionsmustbe imposedtogetauniquesolution. Example 5.6 Aparticlemovesalongthe x axiswiththeacceleration a ( t )=2 Š 6 t .Findthepositionoftheparticleatthetime t =3 ifitspositionandvelocityat t =1 were x (1)=1 and v (1)=2 Solution: Since v( t )= a ( t ),thevelocityistheantiderivativeofthe accelerationsubjecttothecondition v (1)=2.Hence,bytheproperty (5.10), v ( t )= v (1)+ t 1a ( s ) ds =2+(2 s Š 3 s2) t 1= t Š 3 t2+3 Since x( t )= v ( t ),theposition x ( t )istheantiderivativeofthevelocitysubjecttothecondition x (1)=1.Byproperty(5.10)suchan antiderivativereads x ( t )= x (1)+ t 1v ( s ) ds =1+( s2Š s3+3 s ) t 1= t2Š t3+3 t Š 2 Therefore x (3)= Š 11. 33.3.Exercises.(1)Findthederivativeofeachofthefollowingfunctions: (i) f ( x )= x 1(1+ t6)Š 1dt (ii) f ( x )= x20sin( t2) dt

PAGE 178

1725.INTEGRATION (iii) f ( x )= x Š xcos( et) dt (iv) f ( x )= x sin xet2dt (2)Let f ( x )beapiecewiseconstantfunction: f ( x )=0if x< 0, f ( x )=1if x [0 2), f ( x )=2if x [2 4), f ( x )= Š 3 if x [4 6],and f ( x )=0if x> 6.Usethegeometrical interpretationofthede“niteintegraltodrawthegraphsof g ( x )= x 1f ( t ) dt and h ( x )= x 4f ( t ) dt (3)Foraparticlemovingdownaroughinclinedplane,thevelocity is v ( t )= t .Whereistheparticleattheendof2seconds? (4)Findthelocationofaparticlemovingalongalineattheendof 2secondsiftheaccelerationoftheparticleis a ( t )=6 t Š 12 t2andifitspositionandvelocityattheendof1secondwere s (1)=5and v (1)=10. (5)Aspacecrafthadaconstantvelocityof v0.Thenitsengines were“redforatime T1,thenstoppedforatime T ,andthen “redagainforatime T2.Ifinthe“rstandsecondtimes,the enginescreatedconstantaccelerations a1and a2,whatisthe “nalvelocityofthespacecraft? (6)Findtheareaoftheplanarregionboundedabovebythe parabola y =3 Š x2andbelowbytheparabola y =1+ x2. (7)Findtheareaoftheplanarregionboundedabovebythe parabola y =2 Š x2andbelowbytheline y = x (8)Evaluatetheintegrals: (i) 2 1( x +1 /x ) dx (ii) 2 Š 1eŠ 2 xdx (iii) 0sin xdx (iv) 2 0( x +3 x ) dx (v) 3 1x2/ (4+ x2) dx (vi) 1 0(1+ x + x2+ x3+ + xn) dx (vii) 2 0(1+cos x +cos(2 x )+ +cos( nx )) dx34.Inde“niteIntegralsandtheNetChange Ashasbeenshownintheprevioussection,thederivativeofthe de“niteintegralofacontinuousfunction f withrespecttotheupper limitequalsthevalueof f attheupperlimit.Sointegrationanddierentiationappearasoperationsinversetooneanother.Tofurtherstress thisrelationbetweentheintegrationanddierentiation,thenotionof aninde“niteintegralisintroduced.

PAGE 179

34.INDEFINITEINTEGRALSANDTHENETCHANGE173 Definition 5.5(Inde“niteIntegral) Thefunction F iscalledan inde“niteintegral of f andisdenotedby F ( x )= f ( x ) dx if F( x )= f ( x ) Itfollowsfromthisde“nitionthataninde“niteintegralisnothingbutthegeneralantiderivativeof f .Thereasonforintroducing theintegralsymbolintotheantiderivativenotationisthefundamental theoremofcalculus: b af ( x ) dx = F ( b ) Š F ( a ) where F isanyantiderivativeof f .Sinceallantiderivativesdieronly byaconstant,whichisalwayscancelledoutinthedierence F ( b ) Š F ( a ),thede“niteintegralisthedierenceinvaluesoftheinde“nite integralattheupperandlowerlimitsofthede“niteintegral.The inde“niteintegralhasthesamepropertiesastheantiderivative.Itis linear: (5.20) c1f ( x )+ c2g ( x ) dx = c1 f ( x ) dx + c2 g ( x ) dx foranyconstants c1and c2andanyfunctions f and g Usingthetableofantiderivativesofbasicfunctions,onecanmake atableofinde“niteintegralsofbasicfunctions.Let C beanarbitrary constant.Thenitiseasytoverifythefollowingrelationships: xndx = xn +1 n +1 + C,n =1 1 x dx =ln x + C, sin( ax ) dx = Š cos( ax ) a + C, cos( ax ) dx = sin( ax ) a + C, exdx = ex+ C, axdx = ax ln a + C,a> 0 1 1+ x2dx =tanŠ 1x + C, 1 1 Š x2dx =sinŠ 1x + C, sec2( ax ) dx = tan( ax ) a + C, csc2( ax ) dx = Š cot( ax ) a + C, sec x tan xdx =sec x + C, csc x cot xdx = Š csc x + C. Recallthatthegeneralantiderivativeona giveninterval isobtained fromaparticularantiderivativebyaddinganarbitraryconstant.This

PAGE 180

1745.INTEGRATION doesnotholdforadomainbeingadisjointunionoftwoormoreintervals(reviewthepropertiesofantiderivatives).So,inthepreceding table,theconventionisusedthatthegivenexpressionsforinde“nite integralsarevalidonlyinan interval Example 5.7 Findageneralinde“niteintegralfor xŠ 3. Solution: Thefunction xŠ 3isnotde“nedat x =0.Soitsdomain istheunionoftwo disjoint intervals( Š 0)and(0 ).Bythe“rst equalityintheprecedingtable( n = Š 3), xŠ 3dx = Š xŠ 2 2 + C1,x> 0; xŠ 3dx = Š xŠ 2 2 + C2,x< 0 where C1and C2are arbitrary constants. Thefollowingnotationisusedinthefundamentaltheoremof calculus: (5.21) b af ( x ) dx = F ( x ) b a= F ( b ) Š F ( a ) where F ( x )= f ( x ) dx Example 5.8 Evaluate 1 0[3 x2Š x +4(1+ x2)Š 1] dx Solution: Bythelinearityoftheinde“niteintegral(5.20),aninde“nite integraloftheintegrandis x3Š x2/ 2+4tanŠ 1x .Anarbitraryconstant intheinde“niteintegralmaybeomittedherebecause,asalreadynoted, itisalwayscancelledoutinthede“niteintegral.Therefore, 1 0 3 x2Š x + 4 1+ x2 dx = x3Š x2 2 +4tanŠ 1x 1 0= 1 2 Š wheretanŠ 1(1)= / 4hasbeenused. 34.1.TheNetChangeTheorem.Put f ( x )= F( x )inthefundamental theoremofcalculus(5.21).Theresultobtainedisknownasthenet changetheorem. Theorem 5.6 Theintegralofacontinuousrateofchangeisthe netchange: b aF( x ) dx = F ( b ) Š F ( a ) .

PAGE 181

34.INDEFINITEINTEGRALSANDTHENETCHANGE175 Notethat F( x )maybepositiveandnegativeintheinterval[ a,b ]so thatthequantity y = F ( x )may increaseanddecrease .Thedierence F ( b ) Š F ( a )representsthe net changeof y when x changesfrom a to b .Thenetchangevanishesif F ( b ) Š F ( a )=0.Thisdoesnotmean thatthequantity y doesnotchangeatall,butratherthismightmean, forexample,thatthequantity y increasesfromthevalue F ( a ),then, atsome c in[ a,b ],itbeginstodecrease,returningtoitsinitialvalue when x = b sothatitsnetchangevanishes. Ananalogywithanobjectmovingalongastraightlinecanbe madetoillustratethenetchange.Let x ( t )beapositionfunctionof theobjectrelativetosomepointontheline.Then x( t )= v ( t )isits velocity(notethatthevelocitycanbenegativesothattheobjectcan movebackandforth).Thenetchangeofthepositionoverthetime interval[ t1,t2]is t2t1v ( t ) dt = x ( t2) Š x ( t1) Example 5.9 Supposeanobjecttravelsalongastraightlinewith avelocityof v ( t )=1 Š 2 t .Findthenetchangeofitspositionoverthe timeinterval [0 1] andthetotaldistancetraveledbytheobjectoverthe sametimeinterval. Solution: (I)Theinde“niteintegralof v ( t )is x ( t )= t Š t2+ C .Sothenet changeoftheobjectpositionis 1 0v ( t ) dt = 1 0x( t ) dt = x (1) Š x (0)=0 (II)Notethatthevelocitychangesitssignat t =1 / 2.So,inthe interval[0 1 / 2],itispositive(i.e.,theobjectmovestotheright fromitsinitialposition),thenthevelocitybecomesnegativein [1 / 2 1](i.e.,theobjectgoesbacktotheinitialpoint).To“nd thedistancetraveledbytheobject,theabsolutevalue | v ( t ) | mustbeintegratedovertheinterval[0 1].Thinkof | v ( t ) | as thespeedshownonthespeedometerofyourcar;itisalways nonnegativeregardlessofthedirectioninwhichthecaris moving. 1 0| 1 Š 2 t | dt = 1 / 2 0(1 Š 2 t ) Š 1 1 / 2(1 Š 2 t ) dt =[ x (1 / 2) Š x (0)] Š [ x (1) Š x (1 / 2)]=1 / 2 ,

PAGE 182

1765.INTEGRATION wherethede“nition | v | = v if v> 0and | v | = Š v if v< 0has beenused. Otherexamplesofthenetchangeincludesthevolume V ( t )ofwater inareservoirbetweentwomomentsoftime t2t1V( t ) dt = V ( t2) Š V ( t1) where V( t )istherateofchangeofthevolume;thenetchangeofthe populationgrowth t2t1n( t ) dt = n ( t2) Š n ( t1) where n( t )isthegrowthrate;therelationbetweenthecostandmarginalcostfunctions: t2t1C( t ) dt = C ( t2) Š C ( t1); andsimilarlyformanyotherquantities.34.2.Exercises.(1)Findtheinde“niteintegrals: (i) 6 x5dx (ii) xŠ 3dx (iii) ( x2 / 3Š xŠ 2 / 3) dx (iv) ( x2Š 4)2dx (v) (1+ x )3dx (vi) ( x + a )2/ xdx (vii) x/ x2+1 dx (viii) x sin( x2) dx (ix) 1 Š xdx (x) ( x +cos(2 x )) dx (xi) (1+2 x ) / 1 Š x2dx (xii) (1 Š x + x3) /x2dx (xiii) x2/ ( a2+ x2) dx (2)Aparticletravelswithvelocity v ( t )=sin( t/ 2).Findthenet displacementoftheparticleoverthetimeinterval[0 2 ]and thedistancetraveledbytheparticle. (3)Abacteriapopulationgrowsatanexponentialrate n( t )= n0et,where n0istheinitialpopulationand isaconstant.If inthetime T thepopulationhasdoubled,“ndtheconstant .

PAGE 183

35.THESUBSTITUTIONRULE177 Whatisthepopulationat t =10 T ascomparedtotheinitial population?35.TheSubstitutionRule Aninde“niteintegralofthederivative F( x )isthefunction F ( x ) itself.Let u = F ( x ),where u isanewvariablede“nedasadierentiablefunctionof x .Considerthedierential du = F( x ) dx .Thenthe followingequalitieshold: F( x ) dx = F ( x )+ C = u + C = du, where C isanarbitraryconstantandthelastequalityfollowsfromthe factthataninde“niteintegralof f ( u )=1is u .Sowecanconclude that F( x ) dx = du ,providedthevariables u and x arerelatedas u = F ( x ).Thisalsoshowsthat itispermissibletooperatewith dx and du aftertheintegralsignasiftheyweredierentials .Thisobservationleadstoaneattechnicaltricktocalculateinde“niteintegrals. Forexample, 1 x +1 dx = d 2 x +1 =2 x +1+ C, wherethesubstitution u =2 x +1hasbeenused.Thistrickcanbe generalized. Let F ( u )beaninde“niteintegralofacontinuousfunction f ( u )on aninterval I .Let u = g ( x ),where g isdierentiableanditsrangeis theinterval I .Bythechainrule, F ( g ( x )) = F( g ( x )) g( x )= f ( g ( x )) g( x ) Inotherwords, F ( g ( x ))+ C isaninde“niteintegralof f ( g ( x )) g( x ).On aninterval,themostgeneralinde“niteintegralof f ( u )is f ( u ) du = F ( u )+ C .Therefore, F ( g ( x ))and f ( u ) du candieratmostbyan additiveconstant.Thisprovesthefollowingtheorem. Theorem 5.7 (TheSubstitutionRule) .If u = g ( x ) isadifferentiablefunctionwhoserangeisaninterval I and f iscontinuous on I ,then (5.22) f ( g ( x )) g( x ) dx = f ( g ( x )) dg ( x )= f ( u ) du. Thesubstitutionruleisoftenreferredtoasa changeoftheintegrationvariable .Itisapowerfulmethodtocalculateinde“niteintegrals. Example 5.10 Find x sin( x2+1) dx .

PAGE 184

1785.INTEGRATION Solution: x sin( x2+1) dx = sin( x2+1) 1 2 d ( x2+1)= 1 2 sin udu = Š 1 2 cos u + C = Š 1 2 cos( x2+1)+ C, wherethesubstitution u = x2+1hasbeenused. Example 5.11 Find tan xdx Solution: tan xdx = sin x cos x dx = Š d (cos x ) cos x = Š du u = Š ln | u | + C = Š ln | cos x | + C =ln | sec x | + C, wherethesubstitution u =cos x andthelogarithmpropertyln(1 /a )= Š ln a havebeenused. Thesubstitutionrulecanbeusedtoevaluatede“niteintegralsby meansofthefundamentaltheoremofcalculus. Example 5.12 Evaluate 2 0xex2dx Solution: First,“ndaninde“niteintegral: F ( x )= xex2dx = 1 2 ex2dx2= 1 2 eudu = 1 2 eu+ C = 1 2 ex2+ C. where u = x2.Bythefundamentaltheoremofcalculus, 1 0xex2dx = F (2) Š F (0)= 1 2 ( e4Š 1) Notethat,whenevaluatingtheintegral,theoriginalvariable x has beenrestoredintheinde“niteintegralinordertoapplythefundamentaltheoremofcalculus.Thefundamentaltheoremofcalculuscanalso beapplieddirectlyinthenewvariable u ,providedtherangeof u is properlychanged.Indeed,inthepreviousexample,theanswercould havebeenrecoveredfromtheinde“niteintegral1 2eu+ C if u = x2rangesfrom0=02to4=22as x rangesfrom0to2.Thisisespecially usefulwhenacalculationofade“niteintegralrequiresseveralchanges oftheintegrationvariable. Theorem 5.8 (TheSubstitutionRuleforDe“niteIntegrals) .If giscontinuouson [ a,b ] and f iscontinuousontherangeof u = g ( x ) then

PAGE 185

35.THESUBSTITUTIONRULE179 (5.23) b af ( g ( x )) g( x ) dx = g ( b ) g ( a )f ( u ) du. Proof. Let F beanantiderivativeof f .Then F ( g ( x ))isanantiderivativeof( F ( g ( x )))= F( g ( x )) g( x )= f ( g ( x )) g( x ).Bythefundamentaltheoremofcalculus, b af ( g ( x )) g( x ) dx = F ( g ( x )) b a= F ( g ( b )) Š F ( g ( a )) Ontheotherhand,since F ( u )isanantiderivativeof f ( u ),thefundamentaltheoremofcalculusyields g ( b ) g ( a )f ( u ) du = F ( u ) g ( b ) g ( a )= F ( g ( b )) Š F ( g ( a )) Sincetheright-handsidesoftheseequalitiescoincide,somusttheir left-handsides,whichimplies(5.23). Example 5.13 Evaluate e 1ln x/xdx Solution: Theintegrandcanbetransformedas ln x x dx =ln xd ln x. Sothesubstitution u =ln x canbemade.Therangeofthenew integrationvariable u isdeterminedbytherangeoftheoldone: u =0 when x =1and u =1when x = e .Thus, e 1ln x x dx = 1 0udu = u2 2 1 0= 1 2 35.1.Symmetry.Thecalculationofade“niteintegraloverasymmetricintervalcanbesimpli“ediftheintegrandpossessessymmetryproperties. Theorem 5.9 Suppose f iscontinuousonasymmetricinterval [ Š a,a ] .Then a Š af ( x ) dx =2 a 0f ( x ) dx if f ( Š x )= f ( x )( f iseven) (5.24) a Š af ( x ) dx =0if f ( Š x )= Š f ( x )( f isodd) (5.25)

PAGE 186

1805.INTEGRATION A a a A I 0 Figure5.6. Illustrationoftheproperty(5.25).Afunctionisskewsymmetricif f ( Š x )= Š f ( x ).Itsintegral overa symmetric interval[ Š a,a ]vanishes.Thearea A underthegraphof f andabovetheinterval[0 ,a ]isthe sameastheareaabovethegraphof f andbelowtheinterval[ Š a, 0]becauseoftheskewsymmetryofthefunction and thesymmetryoftheinterval[ Š a,a ]relativeto there”ection x Š x .BythepropertydepictedinFigure5.4,theintegralof f over[ Š a,a ]is A +( Š A )=0. Proof. Theintegralcanbesplitintotwointegrals: a Š af ( x ) dx = 0 Š a+ a 0 f ( x ) dx = Š Š a 0f ( x ) dx + a 0f ( x ) dx. Inthe“rstintegralontheveryright-handside,thesubstitution u = Š x ismadesothat u =0when x =0and u = a when x = Š a and dx = Š du .Hence, ŠŠ a0f ( x ) dx =a0f ( Š u ) du and a Š af ( x ) dx = a 0f ( Š u ) du + a 0f ( x ) dx. Now,if f iseven,then f ( Š u )= f ( u )and(5.24)follows.If f isodd, then f ( Š u )= Š f ( u )and(5.25)follows. Thegeometricalinterpretationofthistheoremistransparent(see Fig5.6).Suppose f ( x ) 0for0 x a .Theintegral a 0f ( x ) dx =

PAGE 187

35.THESUBSTITUTIONRULE181 A istheareaunderthegraphof f on[0 ,a ].If f iseven,then,by symmetry,thegraphof f on[ Š a, 0]isobtainedfromthaton[0 ,a ] byare”ectionaboutthe y axis.Therefore,thearea 0 Š af ( x ) dx must coincidewith A .If f isodd,thenitsgraphon[ Š a, 0]isobtainedbythe mirrorre”ectionabouttheoriginsothatthearea A appearsbeneath the x axis.Hence, 0 Š af ( x ) dx = Š A Example 5.14 Evaluate Š sin( x3) dx Solution: Unfortunately,anantiderivativeofsin( x3)cannotbeexpressedinelementaryfunctions,andthefundamentaltheoremofcalculuscannotbeused.Onecanalwaysevaluatetheintegralbytaking thelimitofthesequenceofRiemannsums.Analternativesolution isduetoasimplesymmetryargument.Notethatsin( x3)isanodd function,sin(( Š x )3)=sin( Š x3)= Š sin( x3).Theintegrationinterval isalso symmetric ,[ Š ].Thus,byproperty(5.25), Š sin( x3) dx =0 Remark .Inthepreviousexample,takeapartitionof[ Š ]by points xk= k x k = Š N, Š N +1 ,..., Š 1 0 1 ,...,N Š 1 ,N ,where x = /N .ConsidertheRiemannsumwithsamplepointsbeingthe midpoints.ItisthenstraightforwardtoshowthattheRiemannsum vanishesbecausesin( x 3 Š k)= Š sin( x 3 k)for k =1 2 ,...,N .35.2.Exercises.(1)Usethesuggestedsubstitutionto“ndtheinde“niteintegrals: (i) x3( x4+1)1 / 3dx,u = x4+1 (ii) sin( x ) / xdx,u = x (iii) sin xecos xdx,u =cos x (iv) x2 1 Š x2dx,u =sin x (v) (ln x )3/xdx,u =ln x (2)Useasubstitutionto“ndtheinde“niteintegrals: (i) x 1+2 xdx (ii) x2/ 2 Š 3 xdx (iii) x3 1+ xdx (iv) eŠ x/ xdx (v) x/ ( x4+2 x2+2) dx Hint :Completethesquaresinthe denominator

PAGE 188

1825.INTEGRATION (3)Useachangeofvariablesand/orsymmetrytoevaluatethe de“niteintegrals: (i) 1 0x 2+ x2dx (ii) 1 0tanŠ 1x/ (1+ x2) dx (iii) 0x cos( x2) dx (iv) 1 Š 1x3ex4dx (v) 2 Š 2x ( ex2Š eŠ x2) dx (vi) 2 Š 2( ex2+ eŠ x2) dx (vii) a Š ag ( x ) dx,g ( x )= x 0cos( t2) dt (viii) 2 1(2 x +1) x2+ x +3 dx (ix) / 2 0sin(2 x )3 1+cos2xdx (x) / 4 0(tan x )psec2xdx,p> 0 (xi) / 6 Š / 6tan3(3 x )sin5(2 x ) dx