6.3 Newton’s s Method
127
waythatthecenterofthescissorsatAisfixed,andthepaperisalsofixed. Astheblades
areclosed(i.e.,theangle θ inthediagramis decreased),thedistance xbetweenAandC
increases,cuttingthepaper.
a. Expressxintermsofa,θ,andβ.
b. Expressdx/dtintermsofa,θ,β,anddθ/dt.
c. Suppose e that the distancea is 20cm,andtheangle β is5
. Further r suppose that θ
isdecreasingat50deg/sec. Attheinstantwhenθ=30
,findtherate(incm/sec)at
whichthepaperisbeingcut. 
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B
A
C
θ
Figure 6.16
Scissors.
Supposeyouhaveafunctionf(x),andyouwanttofindasaccuratelyaspossiblewhere
itcrossesthex-axis; inother r words,youwant tosolve f(x) =0. . Suppose e youknowof
nowaytofindanexactsolutionbyanyalgebraicprocedure,butyouareabletousean
approximation,provideditcanbemadequiteclosetothetruevalue.Newton’smethodis
awaytofindasolutiontotheequationtoasmanydecimalplacesasyouwant.Itiswhat
iscalled an “iterative procedure,” meaning that it can be repeated again andagain to
getananswerofgreaterandgreateraccuracy. IterativeprocedureslikeNewton’smethod
arewellsuitedtoprogrammingforacomputer. Newton’smethodusesthefactthatthe
tangentlinetoacurveisagoodapproximationtothecurvenearthepointoftangency.
EXAMPLE6.19
Approximate
3.Since
3isasolutiontox
2
=3orx
2
−3=0,we
usef(x)=x
2
−3. Westartbyguessingsomethingreasonablyclosetothetruevalue;this
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128
Chapter6 Applications s oftheDerivative
isusuallyeasytodo.Let’suse
3≈2.Nowusethetangentlinetothecurvewhenx=2
asanapproximationtothecurve,asshowninfigure6.17. Sincef
(x)=2x,theslopeof
thistangentlineis4anditsequationisy=4x−7. Thetangentlineisquiteclosetof(x),
soitcrossesthex-axisnearthepointatwhichf(x)crosses,that is,near
3. Itiseasy
tofindwherethetangentlinecrossesthex-axis: solve0=4x−7togetx=7/4=1.75.
Thisiscertainlyabetterapproximationthan2,butletussaynotcloseenough. Wecan
improveitbydoingthesamethingagain: findthetangent t lineat x=1.75,findwhere
thisnewtangentlinecrossesthex-axis,andusethatvalueasabetterapproximation.We
cancontinuethisindefinitely,thoughitgetsabittedious. Letsseeifwecanshortcutthe
process. Supposethebestapproximationtotheinterceptwehavesofarisx
i
. Tofinda
betterapproximationwewillalwaysdothesamething:findtheslopeofthetangentline
atx
i
,findtheequationofthetangentline,findthex-intercept.Theslopewillbe2x
i
.The
tangentlinewillbey=(2x
i
)(x−x
i
)+(x
2
i
−3),usingthepoint-slopeformulaforaline.
Finally,theinterceptisfoundbysolving0=(2x
i
)(x−x
i
)+(x
2
i
−3). Withalittlealgebra
thisturnsintox=(x
2
i
+3)/(2x
i
);thisisthenextapproximation,whichwenaturallycall
x
i+1
. Insteadofdoingthewholetangentlinecomputationeverytimewecansimplyuse
thisformula to get asmanyapproximationsaswe want. . Startingwith h x
0
=2, we e get
x
1
=(x
2
0
+3)/(2x
0
)=(2
2
+3)/4=7/4(thesameapproximationwegotabove,ofcourse),
x
2
=(x2
1
+3)/(2x
1
)=((7/4)2+3)/(7/2)=97/56≈1.73214,x
3
≈1.73205,andso on.
Thisisstillabittediousbyhand,butwithacalculatoror,evenbetter,agoodcomputer
program,itisquiteeasytogetmany,manyapproximations.Wemightguessalreadythat
1.73205isaccuratetotwodecimalplaces,andinfactitturnsoutthatitisaccurateto5
places.
Let’s think about this processin n more general terms. . We e want to o approximate a
solutiontof(x)=0. Westartwitharoughguess,whichwecallx
0
. Weusethetangent
line tof(x) to get a newapproximationthat we hope willbe closer to the true value.
Whatistheequationofthetangentlinewhenx=x
0
? Theslopeisf
(x
0
)andtheline
goesthrough(x
0
,f(x
0
)),sotheequationofthelineis
y=f
(x
0
)(x−x
0
)+f(x
0
).
Nowwefindwherethiscrossesthex-axisbysubstitutingy=0andsolvingforx:
x=
x
0
f
(x
0
)−f(x
0
)
f(x
0
)
=x
0
f(x
0
)
f(x
0
)
.
Wewilltypicallywant tocompute more thanone ofthese improvedapproximations,so
wenumberthemconsecutively;fromx
0
wehavecomputedx
1
:
x
1
=
x
0
f
(x
0
)−f(x
0
)
f(x
0
)
=x
0
f(x
0
)
f(x
0
)
,
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6.3 Newton’s s Method
129
1
2
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Figure 6.17
Newton’smethod.
andingeneralfromx
i
wecomputex
i+1
:
x
i+1
=
x
i
f
(x
i
)−f(x
i
)
f(x
i
)
=x
i
f(x
i
)
f(x
i
)
.
EXAMPLE6.20
Returningtothepreviousexample,f(x)=x2−3,f(x)=2x,and
theformulabecomesx
i+1
=x
i
−(x
2
i
−3)/(2x
i
)=(x
2
i
+3)/(2x
i
),asbefore.
In practice, , which h is s to say, , if f you u need to approximate a a value e in n the e course of
designingabridgeorabuildingoranairframe,youwillneedtohavesomeconfidencethat
theapproximationyousettleonisaccurate enough. . Asarule e ofthumb,once acertain
numberof decimalplacesstopchangingfromoneapproximation tothe next it islikely
thatthosedecimalplacesarecorrect. Still,thismaynot t beenoughassurance,inwhich
casewecantesttheresultforaccuracy.
EXAMPLE 6.21
Findthe xcoordinateoftheintersectionofthecurvesy=2x and
y = = tanx, , accurate to three decimal l places. . To o put this in the context of f Newton’s
method,wenotethatwewanttoknowwhere2x=tanxortanx−2x=0.Wecompute
f
(x)=sec
2
x−2andsetuptheformula:
x
i+1
=x
i
tanx
i
−2x
i
sec2x
i
−2
.
Fromthegraphinfigure6.18weguessx
0
=1asastartingpoint,thenusingtheformula
wecomputex
1
=1.310478030,x
2
=1.223929096,x
3
=1.176050900,x
4
=1.165926508,
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130
Chapter6 Applications s oftheDerivative
x
5
= 1.165561636. . So o we guessthat t the e first three placesare correct,but that t isnot
thesameassaying1.165 iscorrecttothreedecimalplaces—1.166might bethecorrect,
rounded approximation. . How w can we e tell? ? We e can substitute e 1.165, 1.1655 5 and 1.166
intotheoriginalfunctiontanx−2x;thisgives−0.002483652,−0.000271247,0.001948654.
Since the first two are negative andthe third ispositive, , tanx−2x x crosses s the e x axis
between1.1655and1.166,sothecorrectvaluetothreeplacesis1.166.
0
5
10
15
1
1.5
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Figure 6.18
y=tanxandy=2x.
Exercises
1. Approximatethefifthrootof7,usingx
0
=1.5asafirstguess.UseNewton’smethodtofind
x
3
asyourapproximation.
2. UseNewton’sMethodtoapproximatethecuberootof10totwodecimalplaces. 
3. Thefunctionf(x)=x
3
−3x
2
−3x+6hasarootbetween3and4,becausef(3)=−3and
f(4)=10.Approximatetheroottotwodecimalplaces. 
4. Arectangular r piece e of cardboard of dimensions 8×17 is used to make e an n open-top box
by cuttingout a smallsquare of side x fromeachcorner and d bendingup thesides. . (See
exercise6.1.20.) Ifx=2,thenthevolumeoftheboxis2·4·13=104. . UseNewton’smethod
tofindavalueofxforwhichtheboxhasvolume100,accurateto3significantfigures. 
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6.4 Linear r Approximations
131
Newton’smethodisoneexampleoftheusefulnessofthetangentlineasanapproximation
toacurve. Hereweexploreanothersuchapplication.
Recallthatthetangentlinetof(x)atapointx=aisgivenbyL(x)=f(a)(x−a)+
f(a).Thetangentlineinthiscontextisalsocalledthelinearapproximationtofata.
IffisdifferentiableatathenLisagoodapproximationoffsolongasxis“nottoo
far”froma. Putanotherway,if f f isdifferentiable atathenundera microscope f will
lookverymuchlike astraightline. . Figure 6.19showsatangentlineto y=x
2
at three
differentmagnifications.
If we e want to approximate f(b), , because e computing g it exactly is difficult, , we e can
approximatethe valueusingalinear approximation,providedthat wecancompute the
tangentlineatsomeaclosetob.
Figure6.19
Thelinearapproximationtoy=x
2
.
EXAMPLE 6.22
Let f(x) ) =
x+4. Then n f
(x) = 1/(2
x+4). The e linear ap-
proximationtof atx=5 5 isL(x)=1/(2
5+4)(x−5)+
5+4=(x−5)/6+3. As
animmediateapplication wecanapproximate square rootsofnumbersnear9 byhand.
To estimate
10,wesubstitute6into thelinear approximationinsteadofinto f(x),so
6+4≈(6−5)/6+3=19/6≈3.1
6. Thisroundsto3.17whilethesquarerootof10is
actually3.16totwodecimalplaces,sothisestimateisonlyaccuratetoonedecimalplace.
Thisisnottoosurprising,as10isreallynotverycloseto9;ontheotherhand,formany
calculations,3.2wouldbeaccurateenough.
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132
Chapter6 Applications s oftheDerivative
Withmoderncalculatorsandcomputingsoftwareitmaynotappearnecessarytouse
linearapproximations. But t infact theyarequite useful. . Incasesrequiring g anexplicit
numericalapproximation,theyallowustogetaquickroughestimatewhichcanbeused
asa“realitycheck”onamorecomplexcalculation.Insomecomplexcalculationsinvolving
functions,the linear approximationmakesan otherwise intractablecalculationpossible,
withoutseriouslossofaccuracy.
EXAMPLE 6.23
Considerthetrigonometricfunctionsinx. Itslinearapproximation
atx=0issimplyL(x)=x. Whenxissmallthisisquiteagoodapproximationandis
usedfrequentlybyengineersandscientiststosimplifysomecalculations.
DEFINITION 6.24
Let y=f(x) be adifferentiablefunction. . Wedefinea a newin-
dependentvariabledx,andanewdependentvariabledy=f
(x)dx. Noticethatdyisa
functionbothofx(sincef
(x)isafunctionofx)andofdx. Wesaythatdxanddy are
differentials.
Let∆x =x−aand ∆y =f(x)−f(a). . If f xisnear athen∆xissmall. . If f weset
dx=∆xthen
dy=f
(a)dx≈
∆y
∆x
∆x=∆y.
Thus,dycanbeusedtoapproximate∆y,theactualchangeinthefunctionf betweena
andx.Thisisexactlytheapproximationgivenbythetangentline:
dy=f
(a)(x−a)=f
(a)(x−a)+f(a)−f(a)=L(x)−f(a).
WhileL(x) approximatesf(x),dy approximateshowf(x) haschangedfromf(a). . Fig-
ure6.20illustratestherelationships.
a
x
.
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....................................................................................................................................................................................
←−−
dx=∆x
−−→
↑|
∆y
|
|
dy
|
Figure6.20
Differentials.
6.5 TheMeanValueTheorem
133
Exercises
1. Letf(x)=x
4
.Ifa=1anddx=∆x=1/2,whatare∆yanddy? 
2. Letf(x)=
x. Ifa=1anddx=∆x=1/10,whatare∆yanddy? 
3. Letf(x)=sin(2x). . Ifa=πanddx=∆x=π/100,whatare∆yanddy?
4. Usedifferentials s toestimate theamount of paint neededtoapplya coat ofpaint 0.02 cm
thicktoaspherewithdiameter40meters. (Recallthatthevolumeofasphereofradiusris
V =(4/3)πr
3
.Noticethatyouaregiventhatdr=0.02.) 
5. Show w indetailthat thelinearapproximationofsinxatx =0is L(x) =xandthelinear
approximationofcosxatx=0isL(x)=1.
Herearetwointerestingquestionsinvolvingderivatives:
1. Supposetwodifferentfunctionshavethesamederivative;whatcanyousayabout
therelationshipbetweenthetwofunctions?
2. Supposeyoudrivea a carfromtollboothonatollroadto another tollboothat
anaveragespeedof70milesperhour.Whatcanbeconcludedaboutyouractual
speedduringthetrip? Inparticular,didyouexceedthe65mileperhourspeed
limit?
Whilethesesoundverydifferent,itturnsoutthatthetwoproblemsareveryclosely
related. Weknowthat“speed”isreallythederivativebyadifferentname;let’sstartby
translatingthesecondquestionintosomethingthatmaybeeasier tovisualize. . Suppose
thatthefunctionf(t)givesthepositionofyourcaronthetollroadattimet. Yourchange
inpositionbetweenonetollboothandthenextisgivenbyf(t
1
)−f(t
0
),assumingthat
at time t
0
you were at thefirst boothandat time t
1
youarrivedat the second booth.
Youraveragespeedforthetripis(f(t
1
)−f(t
0
))/(t
1
−t
0
). Ifwethinkaboutthegraph
off(t),theaveragespeedistheslopeofthelinethatconnectsthetwopoints(t
0
,f(t
0
))
and(t
1
,f(t
1
)). Yourspeedat t anyparticulartime tbetweent
0
andt
1
isf(t),theslope
ofthecurve.Nowquestion(2)becomesaquestionaboutslope. Inparticular,iftheslope
betweenendpointsis70,whatcanbesaidoftheslopesatpointsbetweentheendpoints?
Asageneralrule,whenfacedwithanewproblemitisoftenagoodideatoexamine
one or more simplified versions of the problem, , in the e hope that this will l lead to o an
understanding of the original l problem. . In n thiscase,the probleminits“slope” formis
somewhateasiertosimplifythantheoriginal,butequivalent,problem.
Here isa a specialinstance e of the e problem. . Suppose e that t f(t
0
) = = f(t
1
). Then n the
two endpointshave the same height and the slope ofthe lineconnectingtheendpoints
is zero. . What t can n we say y about the e slope e between the endpoints? ? It t shouldn’t t take
134
Chapter6 Applications s oftheDerivative
muchexperimentationbeforeyouareconvincedofthetruthofthisstatement:Somewhere
betweent
0
andt
1
theslopeisexactlyzero,thatis,somewherebetweent
0
andt
1
theslope
isequaltotheslopeofthelinebetweentheendpoints.Thissuggeststhatperhapsthesame
istrueeveniftheendpointsareatdifferentheights,andagainabitofexperimentationwill
probablyconvinceyouthatthisisso. Butwecandobetterthan“experimentation”—we
canprovethatthisisso.
Westartwiththesimplifiedversion:
THEOREM 6.25
Rolle’s Theorem
Suppose that t f(x) ) has s a a derivative on n the
interval(a,b),iscontinuousontheinterval[a,b],andf(a)=f(b). Then n at some value
c∈(a,b),f
(c)=0.
Proof. We e know w that t f(x) hasa maximumand minimumvalue on [a,b](because it
is continuous), and d we e also o know that t the maximum and d minimum m must occur at an
endpoint,atapointatwhichthederivativeiszero,oratapointwherethederivativeis
undefined. Sincethederivativeisneverundefined,thatpossibilityisremoved.
If the maximum or minimum occurs at a point c, , other than n an n endpoint, , where
f
(c)=0,thenwehavefoundthepointweseek.Otherwise,themaximumandminimum
bothoccuratanendpoint,andsincetheendpointshavethesameheight,themaximum
andminimumarethesame. Thismeansthat t f(x)=f(a)=f(b)at everyx∈[a,b],so
thefunctionisahorizontalline,andithasderivativezeroeverywherein(a,b). Thenwe
maychooseanycatalltogetf
(c)=0.
Perhapsremarkably,thisspecialcaseisallweneedtoprovethemoregeneraloneas
well.
THEOREM 6.26
Mean ValueTheorem
Supposethatf(x)hasaderivativeon
theinterval(a,b)andiscontinuousontheinterval[a,b]. Thenatsomevaluec∈(a,b),
f
(c)=
f(b)−f(a)
b−a
.
Proof. Letm=
f(b)−f(a)
b−a
,andconsideranewfunctiong(x)=f(x)−m(x−a)−f(a).
Weknowthatg(x)hasaderivativeeverywhere,sinceg(x)=f(x)−m. Wecancompute
g(a)=f(a)−m(a−a)−f(a)=0and
g(b)=f(b)−m(b−a)−f(a)=f(b)−
f(b)−f(a)
b−a
(b−a)−f(a)
=f(b)−(f(b)−f(a))−f(a)=0.
6.5 TheMeanValueTheorem
135
Sotheheightofg(x)isthesameatbothendpoints.Thismeans,byRolle’sTheorem,that
atsomec,g
(c)=0. Butweknowthatg
(c)=f
(c)−m,so
0=f
(c)−m=f
(c)−
f(b)−f(a)
b−a
,
whichturnsinto
f
(c)=
f(b)−f(a)
b−a
,
exactlywhatwewant.
Returning to the e original formulationof f question(2), we see that t if f f(t) givesthe
positionofyourcar attimet,thenthe MeanValue Theoremsaysthatatsometime c,
f(c) =70, , thatis,at t some timeyoumust have been traveling atexactlyyour average
speedforthetrip,andthatindeedyouexceededthespeedlimit.
Nowlet’sreturntoquestion(1).Suppose,forexample,thattwofunctionsareknownto
havederivativeequalto5everywhere,f
(x)=g
(x)=5.Itiseasytofindsuchfunctions:
5x,5x+47,5x−132,etc. Arethereother,morecomplicated,examples? No—theonly
functionsthatworkarethe“obvious”ones,namely,5xplussomeconstant. Howcanwe
seethatthisistrue?
Although“5”isaverysimplederivative,let’slookatanevensimplerone. Suppose
thatf
(x)=g
(x)=0. Againwecanfindexamples: f(x)=0,f(x)=47,f(x)=−511
allhavef
(x)=0. Aretherenon-constantfunctionsf f withderivative0? ? No,andhere’s
why:Supposethatf(x)isnotaconstantfunction. Thismeansthattherearetwopoints
onthe functionwithdifferentheights,sayf(a) =f(b). . TheMeanValueTheoremtells
usthatatsomepointc,f
(c)=(f(b)−f(a))/(b−a)=0. Soanynon-constantfunction
doesnothaveaderivativethatiszeroeverywhere;thisisthesameassayingthattheonly
functionswithzeroderivativearetheconstantfunctions.
Let’sgobacktotheslightlylesseasyexample: supposethatf
(x)=g
(x)=5. Then
(f(x)−g(x)) =f(x)−g(x)=5−5=0. Sousingwhatwediscoveredintheprevious
paragraph,weknowthatf(x)−g(x)=k,forsomeconstantk.Soanytwofunctionswith
derivative5mustdifferbyaconstant;since5xisknowntowork,theonlyotherexamples
mustlooklike5x+k.
Now we e can extend this to more complicated functions, , without t any y extra a work.
Suppose that f
(x) = = g
(x). Then n as before (f(x)−g(x))
= f
(x)−g
(x) = 0, so
f(x)−g(x)=k.Againthismeansthatifwefindjustasinglefunctiong(x)withacertain
derivative,theneveryotherfunctionwiththesamederivativemustbeoftheformg(x)+k.
136
Chapter6 Applications s oftheDerivative
EXAMPLE 6.27
Describeallfunctionsthathavederivative5x−3. It’seasytofind
one: g(x) ) =(5/2)x
2
−3x hasg
(x) =5x−3. . The e onlyother functionswiththe same
derivativearethereforeoftheformf(x)=(5/2)x2−3x+k.
Alternately,thoughnotobviously,youmighthavefirstnoticedthatg(x)=(5/2)x
2
3x+47 hasg
(x) =5x−3. . Theneveryother r function with thesame derivative must
have the e formf(x) = (5/2)x−3x+47+k. . This s looks different, but t it really y isn’t.
The functionsoftheformf(x)=(5/2)x
2
−3x+kare exactlythesameastheonesof
theformf(x)=(5/2)x
2
−3x+47+k. Forexample,(5/2)x
2
−3x+10isthe sameas
(5/2)x−3x+47+(−37),andthe first isofthefirst“form”while thesecondhasthe
secondform.
Thisisworthcallingatheorem:
THEOREM 6.28
If f
(x) = = g
(x) for every x x ∈ ∈ (a,b), , then n for some constant k,
f(x)=g(x)+kontheinterval(a,b).
EXAMPLE 6.29
Describeallfunctionswithderivativesinx+e
x
. Onesuchfunction
is−cosx+e
x
,soallsuchfunctionshavetheform−cosx+e
x
+k.
Exercises
1. Letf(x)=x
2
.Findavaluec∈(−1,2)sothatf
(c)equalstheslopebetweentheendpoints
off(x)on[−1,2]. 
2. Verify y that f(x) = x/(x+2) satisfies the hypotheses of the Mean n Value Theoremonthe
interval[1,4]andthenfindallofthevalues,c,thatsatisfytheconclusionofthetheorem.
3. Verify y that f(x)=3x/(x+7) satisfies thehypotheses ofthe MeanValue Theoremonthe
interval[−2,6]andthenfindallofthevalues,c,thatsatisfytheconclusionofthetheorem.
4. Letf(x)=tanx.Showthatf(π)=f(2π)=0butthereisnonumberc∈(0,2π)suchthat
f
(c)=0.WhydoesthisnotcontradictRolle’stheorem?
5. Letf(x)=(x−3)
−2
. Showthatthereisnovaluecintheinterval(1,4)suchthatf
(c)=
(f(4)−f(1))/(4−1).WhyisthisnotacontradictionoftheMeanValueTheorem?
6. Describeallfunctionswithderivativex
2
+47x−5.
7. Describeallfunctionswithderivative
1
1+x2
8. Describeallfunctionswithderivativex
3
1
x
9. Describeallfunctionswithderivativesin(2x). 
10. Showthattheequation
6x
4
−7x+1=0
doesnothavemorethantwodistinctrealroots.
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