lesson 19: maximum and minimum values
DESCRIPTION
The closed interval method tells us how to find the extreme values of a continuous function defined on a closed, bounded interval: we check the end points and the critical points.TRANSCRIPT
. . . . . .
Section4.1MaximumandMinimumValues
V63.0121.027, CalculusI
November6, 2009
Announcements
I Quiznextweekon§§3.1–3.5I FinalExamFriday, December18, 2:00–3:50pm
..Imagecredit: KarenwithaK
. . . . . .
Outline
Introduction
TheExtremeValueTheorem
Fermat’sTheorem(notthelastone)Tangent: Fermat’sLastTheorem
TheClosedIntervalMethod
Examples
Challenge: Cubicfunctions
Optimize
. . . . . .
. . . . . .
Whygototheextremes?
I Rationallyspeaking, itisadvantageoustofindtheextremevaluesofafunction(maximizeprofit, minimizecosts,etc.)
I Manylawsofsciencearederivedfromminimizingprinciples.
I Maupertuis’principle:“ActionisminimizedthroughthewisdomofGod.”
Pierre-LouisMaupertuis(1698–1759)
. . . . . .
Whygototheextremes?
I Rationallyspeaking, itisadvantageoustofindtheextremevaluesofafunction(maximizeprofit, minimizecosts,etc.)
I Manylawsofsciencearederivedfromminimizingprinciples.
I Maupertuis’principle:“ActionisminimizedthroughthewisdomofGod.”
Pierre-LouisMaupertuis(1698–1759)
. . . . . .
Whygototheextremes?
I Rationallyspeaking, itisadvantageoustofindtheextremevaluesofafunction(maximizeprofit, minimizecosts,etc.)
I Manylawsofsciencearederivedfromminimizingprinciples.
I Maupertuis’principle:“ActionisminimizedthroughthewisdomofGod.” Pierre-LouisMaupertuis
(1698–1759)
. . . . . .
Outline
Introduction
TheExtremeValueTheorem
Fermat’sTheorem(notthelastone)Tangent: Fermat’sLastTheorem
TheClosedIntervalMethod
Examples
Challenge: Cubicfunctions
. . . . . .
Extremepointsandvalues
DefinitionLet f havedomain D.
I Thefunction f hasan absolutemaximum (or globalmaximum)(respectively, absoluteminimum)at cif f(c) ≥ f(x) (respectively, f(c) ≤ f(x))forall x in D
I Thenumber f(c) iscalledthemaximumvalue (respectively,minimumvalue)of f on D.
I An extremum iseitheramaximumoraminimum. An extremevalue iseitheramaximumvalueorminimumvalue.
.
.Imagecredit: PatrickQ
. . . . . .
Extremepointsandvalues
DefinitionLet f havedomain D.
I Thefunction f hasan absolutemaximum (or globalmaximum)(respectively, absoluteminimum)at cif f(c) ≥ f(x) (respectively, f(c) ≤ f(x))forall x in D
I Thenumber f(c) iscalledthemaximumvalue (respectively,minimumvalue)of f on D.
I An extremum iseitheramaximumoraminimum. An extremevalue iseitheramaximumvalueorminimumvalue.
.
.Imagecredit: PatrickQ
. . . . . .
Extremepointsandvalues
DefinitionLet f havedomain D.
I Thefunction f hasan absolutemaximum (or globalmaximum)(respectively, absoluteminimum)at cif f(c) ≥ f(x) (respectively, f(c) ≤ f(x))forall x in D
I Thenumber f(c) iscalledthemaximumvalue (respectively,minimumvalue)of f on D.
I An extremum iseitheramaximumoraminimum. An extremevalue iseitheramaximumvalueorminimumvalue.
.
.Imagecredit: PatrickQ
. . . . . .
Theorem(TheExtremeValueTheorem)Let f beafunctionwhichiscontinuousontheclosedinterval[a,b]. Then f attainsanabsolutemaximumvalue f(c) andanabsoluteminimumvalue f(d) atnumbers c and d in [a,b].
.
. . . . . .
Theorem(TheExtremeValueTheorem)Let f beafunctionwhichiscontinuousontheclosedinterval[a,b]. Then f attainsanabsolutemaximumvalue f(c) andanabsoluteminimumvalue f(d) atnumbers c and d in [a,b].
...a
..b
.
.
. . . . . .
Theorem(TheExtremeValueTheorem)Let f beafunctionwhichiscontinuousontheclosedinterval[a,b]. Then f attainsanabsolutemaximumvalue f(c) andanabsoluteminimumvalue f(d) atnumbers c and d in [a,b].
...a
..b
.
.
.cmaximum
.maximum
value
.f(c)
.
.d
minimum
.minimum
value
.f(d)
. . . . . .
NoproofofEVT forthcoming
I Thistheoremisveryhardtoprovewithoutusingtechnicalfactsaboutcontinuousfunctionsandclosedintervals.
I Butwecanshowtheimportanceofeachofthehypotheses.
. . . . . .
BadExample#1
Example
Considerthefunction
f(x) =
{x 0 ≤ x < 1
x− 2 1 ≤ x ≤ 2.
. .|.1
.
.
.
.
Thenalthoughvaluesof f(x) getarbitrarilycloseto 1 andneverbiggerthan 1, 1 isnotthemaximumvalueof f on [0,1] becauseitisneverachieved.
. . . . . .
BadExample#1
Example
Considerthefunction
f(x) =
{x 0 ≤ x < 1
x− 2 1 ≤ x ≤ 2.. .|
.1.
.
.
.
Thenalthoughvaluesof f(x) getarbitrarilycloseto 1 andneverbiggerthan 1, 1 isnotthemaximumvalueof f on [0,1] becauseitisneverachieved.
. . . . . .
BadExample#1
Example
Considerthefunction
f(x) =
{x 0 ≤ x < 1
x− 2 1 ≤ x ≤ 2.. .|
.1.
.
.
.
Thenalthoughvaluesof f(x) getarbitrarilycloseto 1 andneverbiggerthan 1, 1 isnotthemaximumvalueof f on [0,1] becauseitisneverachieved.
. . . . . .
BadExample#2
ExampleThefunction f(x) = x restrictedtotheinterval [0, 1) stillhasnomaximumvalue.
. .|.1
.
.
. . . . . .
BadExample#2
ExampleThefunction f(x) = x restrictedtotheinterval [0, 1) stillhasnomaximumvalue.
. .|.1
.
.
. . . . . .
FinalBadExample
Example
Thefunction f(x) =1xiscontinuousontheclosedinterval [1,∞)
buthasnominimumvalue.
. ..1
.
. . . . . .
FinalBadExample
Example
Thefunction f(x) =1xiscontinuousontheclosedinterval [1,∞)
buthasnominimumvalue.
. ..1
.
. . . . . .
Outline
Introduction
TheExtremeValueTheorem
Fermat’sTheorem(notthelastone)Tangent: Fermat’sLastTheorem
TheClosedIntervalMethod
Examples
Challenge: Cubicfunctions
. . . . . .
LocalextremaDefinition
I A function f hasa localmaximum or relativemaximum at cif f(c) ≥ f(x) when x isnear c. Thismeansthat f(c) ≥ f(x) forall x insomeopenintervalcontaining c.
I Similarly, f hasa localminimum at c if f(c) ≤ f(x) when x isnear c.
..|.a
.|.b
.
.
.
.local
maximum
.
.local
minimum
. . . . . .
LocalextremaDefinition
I A function f hasa localmaximum or relativemaximum at cif f(c) ≥ f(x) when x isnear c. Thismeansthat f(c) ≥ f(x) forall x insomeopenintervalcontaining c.
I Similarly, f hasa localminimum at c if f(c) ≤ f(x) when x isnear c.
..|.a
.|.b
.
.
.
.local
maximum
.
.local
minimum
. . . . . .
I Soalocalextremummustbe inside thedomainof f (notontheend).
I A globalextremumthatisinsidethedomainisalocalextremum.
..|.a
.|.b
.
.
.
.globalmax
.localmax
.
.local and global
min
. . . . . .
Theorem(Fermat’sTheorem)Suppose f hasalocalextremumat c and f isdifferentiableat c.Then f′(c) = 0.
..|.a
.|.b
.
.
.
.local
maximum
.
.local
minimum
. . . . . .
SketchofproofofFermat’sTheorem
Supposethat f hasalocalmaximumat c.
I If h iscloseenoughto 0 butgreaterthan 0, f(c + h) ≤ f(c).Thismeans
f(c + h) − f(c)h
≤ 0 =⇒ limh→0+
f(c + h) − f(c)h
≤ 0
I Thesamewillbetrueontheotherend: if h iscloseenoughto 0 butlessthan 0, f(c + h) ≤ f(c). Thismeans
f(c + h) − f(c)h
≥ 0 =⇒ limh→0−
f(c + h) − f(c)h
≥ 0
I Sincethelimit f′(c) = limh→0
f(c + h) − f(c)h
exists, itmustbe 0.
. . . . . .
SketchofproofofFermat’sTheorem
Supposethat f hasalocalmaximumat c.I If h iscloseenoughto 0 butgreaterthan 0, f(c + h) ≤ f(c).Thismeans
f(c + h) − f(c)h
≤ 0
=⇒ limh→0+
f(c + h) − f(c)h
≤ 0
I Thesamewillbetrueontheotherend: if h iscloseenoughto 0 butlessthan 0, f(c + h) ≤ f(c). Thismeans
f(c + h) − f(c)h
≥ 0 =⇒ limh→0−
f(c + h) − f(c)h
≥ 0
I Sincethelimit f′(c) = limh→0
f(c + h) − f(c)h
exists, itmustbe 0.
. . . . . .
SketchofproofofFermat’sTheorem
Supposethat f hasalocalmaximumat c.I If h iscloseenoughto 0 butgreaterthan 0, f(c + h) ≤ f(c).Thismeans
f(c + h) − f(c)h
≤ 0 =⇒ limh→0+
f(c + h) − f(c)h
≤ 0
I Thesamewillbetrueontheotherend: if h iscloseenoughto 0 butlessthan 0, f(c + h) ≤ f(c). Thismeans
f(c + h) − f(c)h
≥ 0 =⇒ limh→0−
f(c + h) − f(c)h
≥ 0
I Sincethelimit f′(c) = limh→0
f(c + h) − f(c)h
exists, itmustbe 0.
. . . . . .
SketchofproofofFermat’sTheorem
Supposethat f hasalocalmaximumat c.I If h iscloseenoughto 0 butgreaterthan 0, f(c + h) ≤ f(c).Thismeans
f(c + h) − f(c)h
≤ 0 =⇒ limh→0+
f(c + h) − f(c)h
≤ 0
I Thesamewillbetrueontheotherend: if h iscloseenoughto 0 butlessthan 0, f(c + h) ≤ f(c). Thismeans
f(c + h) − f(c)h
≥ 0
=⇒ limh→0−
f(c + h) − f(c)h
≥ 0
I Sincethelimit f′(c) = limh→0
f(c + h) − f(c)h
exists, itmustbe 0.
. . . . . .
SketchofproofofFermat’sTheorem
Supposethat f hasalocalmaximumat c.I If h iscloseenoughto 0 butgreaterthan 0, f(c + h) ≤ f(c).Thismeans
f(c + h) − f(c)h
≤ 0 =⇒ limh→0+
f(c + h) − f(c)h
≤ 0
I Thesamewillbetrueontheotherend: if h iscloseenoughto 0 butlessthan 0, f(c + h) ≤ f(c). Thismeans
f(c + h) − f(c)h
≥ 0 =⇒ limh→0−
f(c + h) − f(c)h
≥ 0
I Sincethelimit f′(c) = limh→0
f(c + h) − f(c)h
exists, itmustbe 0.
. . . . . .
SketchofproofofFermat’sTheorem
Supposethat f hasalocalmaximumat c.I If h iscloseenoughto 0 butgreaterthan 0, f(c + h) ≤ f(c).Thismeans
f(c + h) − f(c)h
≤ 0 =⇒ limh→0+
f(c + h) − f(c)h
≤ 0
I Thesamewillbetrueontheotherend: if h iscloseenoughto 0 butlessthan 0, f(c + h) ≤ f(c). Thismeans
f(c + h) − f(c)h
≥ 0 =⇒ limh→0−
f(c + h) − f(c)h
≥ 0
I Sincethelimit f′(c) = limh→0
f(c + h) − f(c)h
exists, itmustbe 0.
. . . . . .
MeettheMathematician: PierredeFermat
I 1601–1665I Lawyerandnumbertheorist
I Provedmanytheorems,didn’tquiteprovehislastone
. . . . . .
Tangent: Fermat’sLastTheorem
I Plentyofsolutionstox2 + y2 = z2 amongpositivewholenumbers(e.g., x = 3, y = 4,z = 5)
I Nosolutionstox3 + y3 = z3 amongpositivewholenumbers
I Fermatclaimednosolutionsto xn + yn = zn
butdidn’twritedownhisproof
I Notsolveduntil1998!(Taylor–Wiles)
. . . . . .
Tangent: Fermat’sLastTheorem
I Plentyofsolutionstox2 + y2 = z2 amongpositivewholenumbers(e.g., x = 3, y = 4,z = 5)
I Nosolutionstox3 + y3 = z3 amongpositivewholenumbers
I Fermatclaimednosolutionsto xn + yn = zn
butdidn’twritedownhisproof
I Notsolveduntil1998!(Taylor–Wiles)
. . . . . .
Tangent: Fermat’sLastTheorem
I Plentyofsolutionstox2 + y2 = z2 amongpositivewholenumbers(e.g., x = 3, y = 4,z = 5)
I Nosolutionstox3 + y3 = z3 amongpositivewholenumbers
I Fermatclaimednosolutionsto xn + yn = zn
butdidn’twritedownhisproof
I Notsolveduntil1998!(Taylor–Wiles)
. . . . . .
Tangent: Fermat’sLastTheorem
I Plentyofsolutionstox2 + y2 = z2 amongpositivewholenumbers(e.g., x = 3, y = 4,z = 5)
I Nosolutionstox3 + y3 = z3 amongpositivewholenumbers
I Fermatclaimednosolutionsto xn + yn = zn
butdidn’twritedownhisproof
I Notsolveduntil1998!(Taylor–Wiles)
. . . . . .
Outline
Introduction
TheExtremeValueTheorem
Fermat’sTheorem(notthelastone)Tangent: Fermat’sLastTheorem
TheClosedIntervalMethod
Examples
Challenge: Cubicfunctions
. . . . . .
FlowchartforplacingextremaThankstoFermat
Suppose f isacontinuousfunctionontheclosed, boundedinterval [a,b], and c isaglobalmaximumpoint.
..start
.Is c an
endpoint?
. c = a orc = b
.c is a
local max
.Is f diff’ble
at c?
.f is notdiff at c
.f′(c) = 0
.no
.yes
.no
.yes
. . . . . .
TheClosedIntervalMethod
Thismeanstofindthemaximumvalueof f on [a,b], weneedto:I Evaluate f atthe endpoints a and bI Evaluate f atthe criticalpoints or criticalnumbers x whereeither f′(x) = 0 or f isnotdifferentiableat x.
I Thepointswiththelargestfunctionvaluearetheglobalmaximumpoints
I Thepointswiththesmallestormostnegativefunctionvaluearetheglobalminimumpoints.
. . . . . .
Outline
Introduction
TheExtremeValueTheorem
Fermat’sTheorem(notthelastone)Tangent: Fermat’sLastTheorem
TheClosedIntervalMethod
Examples
Challenge: Cubicfunctions
. . . . . .
ExampleFindtheextremevaluesof f(x) = 2x− 5 on [−1, 2].
SolutionSince f′(x) = 2, whichisneverzero, wehavenocriticalpointsandweneedonlyinvestigatetheendpoints:
I f(−1) = 2(−1) − 5 = −7I f(2) = 2(2) − 5 = −1
SoI Theabsoluteminimum(point)isat −1; theminimumvalue
is −7.I Theabsolutemaximum(point)isat 2; themaximumvalueis
−1.
. . . . . .
ExampleFindtheextremevaluesof f(x) = 2x− 5 on [−1, 2].
SolutionSince f′(x) = 2, whichisneverzero, wehavenocriticalpointsandweneedonlyinvestigatetheendpoints:
I f(−1) = 2(−1) − 5 = −7I f(2) = 2(2) − 5 = −1
SoI Theabsoluteminimum(point)isat −1; theminimumvalue
is −7.I Theabsolutemaximum(point)isat 2; themaximumvalueis
−1.
. . . . . .
ExampleFindtheextremevaluesof f(x) = 2x− 5 on [−1, 2].
SolutionSince f′(x) = 2, whichisneverzero, wehavenocriticalpointsandweneedonlyinvestigatetheendpoints:
I f(−1) = 2(−1) − 5 = −7I f(2) = 2(2) − 5 = −1
SoI Theabsoluteminimum(point)isat −1; theminimumvalue
is −7.I Theabsolutemaximum(point)isat 2; themaximumvalueis
−1.
. . . . . .
ExampleFindtheextremevaluesof f(x) = x2 − 1 on [−1, 2].
SolutionWehave f′(x) = 2x, whichiszerowhen x = 0.
Soourpointstocheckare:
I f(−1) =
I f(0) =
I f(2) =
. . . . . .
ExampleFindtheextremevaluesof f(x) = x2 − 1 on [−1, 2].
SolutionWehave f′(x) = 2x, whichiszerowhen x = 0.
Soourpointstocheckare:
I f(−1) =
I f(0) =
I f(2) =
. . . . . .
ExampleFindtheextremevaluesof f(x) = x2 − 1 on [−1, 2].
SolutionWehave f′(x) = 2x, whichiszerowhen x = 0. Soourpointstocheckare:
I f(−1) =
I f(0) =
I f(2) =
. . . . . .
ExampleFindtheextremevaluesof f(x) = x2 − 1 on [−1, 2].
SolutionWehave f′(x) = 2x, whichiszerowhen x = 0. Soourpointstocheckare:
I f(−1) = 0I f(0) =
I f(2) =
. . . . . .
ExampleFindtheextremevaluesof f(x) = x2 − 1 on [−1, 2].
SolutionWehave f′(x) = 2x, whichiszerowhen x = 0. Soourpointstocheckare:
I f(−1) = 0I f(0) = − 1I f(2) =
. . . . . .
ExampleFindtheextremevaluesof f(x) = x2 − 1 on [−1, 2].
SolutionWehave f′(x) = 2x, whichiszerowhen x = 0. Soourpointstocheckare:
I f(−1) = 0I f(0) = − 1I f(2) = 3
. . . . . .
ExampleFindtheextremevaluesof f(x) = x2 − 1 on [−1, 2].
SolutionWehave f′(x) = 2x, whichiszerowhen x = 0. Soourpointstocheckare:
I f(−1) = 0I f(0) = − 1 (absolutemin)I f(2) = 3
. . . . . .
ExampleFindtheextremevaluesof f(x) = x2 − 1 on [−1, 2].
SolutionWehave f′(x) = 2x, whichiszerowhen x = 0. Soourpointstocheckare:
I f(−1) = 0I f(0) = − 1 (absolutemin)I f(2) = 3 (absolutemax)
. . . . . .
ExampleFindtheextremevaluesof f(x) = 2x3 − 3x2 + 1 on [−1, 2].
SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), wehavecriticalpointsatx = 0 and x = 1. Thevaluestocheckare
I f(−1) =
− 4 (absolutemin)
I f(0) =
1 (localmax)
I f(1) =
0 (localmin)
I f(2) =
5 (absolutemax)
. . . . . .
ExampleFindtheextremevaluesof f(x) = 2x3 − 3x2 + 1 on [−1, 2].
SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), wehavecriticalpointsatx = 0 and x = 1.
ThevaluestocheckareI f(−1) =
− 4 (absolutemin)
I f(0) =
1 (localmax)
I f(1) =
0 (localmin)
I f(2) =
5 (absolutemax)
. . . . . .
ExampleFindtheextremevaluesof f(x) = 2x3 − 3x2 + 1 on [−1, 2].
SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), wehavecriticalpointsatx = 0 and x = 1. Thevaluestocheckare
I f(−1) =
− 4 (absolutemin)
I f(0) =
1 (localmax)
I f(1) =
0 (localmin)
I f(2) =
5 (absolutemax)
. . . . . .
ExampleFindtheextremevaluesof f(x) = 2x3 − 3x2 + 1 on [−1, 2].
SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), wehavecriticalpointsatx = 0 and x = 1. Thevaluestocheckare
I f(−1) = − 4
(absolutemin)
I f(0) =
1 (localmax)
I f(1) =
0 (localmin)
I f(2) =
5 (absolutemax)
. . . . . .
ExampleFindtheextremevaluesof f(x) = 2x3 − 3x2 + 1 on [−1, 2].
SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), wehavecriticalpointsatx = 0 and x = 1. Thevaluestocheckare
I f(−1) = − 4
(absolutemin)
I f(0) = 1
(localmax)
I f(1) =
0 (localmin)
I f(2) =
5 (absolutemax)
. . . . . .
ExampleFindtheextremevaluesof f(x) = 2x3 − 3x2 + 1 on [−1, 2].
SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), wehavecriticalpointsatx = 0 and x = 1. Thevaluestocheckare
I f(−1) = − 4
(absolutemin)
I f(0) = 1
(localmax)
I f(1) = 0
(localmin)
I f(2) =
5 (absolutemax)
. . . . . .
ExampleFindtheextremevaluesof f(x) = 2x3 − 3x2 + 1 on [−1, 2].
SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), wehavecriticalpointsatx = 0 and x = 1. Thevaluestocheckare
I f(−1) = − 4
(absolutemin)
I f(0) = 1
(localmax)
I f(1) = 0
(localmin)
I f(2) = 5
(absolutemax)
. . . . . .
ExampleFindtheextremevaluesof f(x) = 2x3 − 3x2 + 1 on [−1, 2].
SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), wehavecriticalpointsatx = 0 and x = 1. Thevaluestocheckare
I f(−1) = − 4 (absolutemin)I f(0) = 1
(localmax)
I f(1) = 0
(localmin)
I f(2) = 5
(absolutemax)
. . . . . .
ExampleFindtheextremevaluesof f(x) = 2x3 − 3x2 + 1 on [−1, 2].
SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), wehavecriticalpointsatx = 0 and x = 1. Thevaluestocheckare
I f(−1) = − 4 (absolutemin)I f(0) = 1
(localmax)
I f(1) = 0
(localmin)
I f(2) = 5 (absolutemax)
. . . . . .
ExampleFindtheextremevaluesof f(x) = 2x3 − 3x2 + 1 on [−1, 2].
SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), wehavecriticalpointsatx = 0 and x = 1. Thevaluestocheckare
I f(−1) = − 4 (absolutemin)I f(0) = 1 (localmax)I f(1) = 0
(localmin)
I f(2) = 5 (absolutemax)
. . . . . .
ExampleFindtheextremevaluesof f(x) = 2x3 − 3x2 + 1 on [−1, 2].
SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), wehavecriticalpointsatx = 0 and x = 1. Thevaluestocheckare
I f(−1) = − 4 (absolutemin)I f(0) = 1 (localmax)I f(1) = 0 (localmin)I f(2) = 5 (absolutemax)
. . . . . .
ExampleFindtheextremevaluesof f(x) = x2/3(x + 2) on [−1, 2].
SolutionWrite f(x) = x5/3 + 2x2/3, then
f′(x) =53x2/3 +
43x−1/3 =
13x−1/3(5x + 4)
Thus f′(−4/5) = 0 and f isnotdifferentiableat 0.
Soourpointstocheckare:
I f(−1) =
I f(−4/5) =
I f(0) =
I f(2) =
. . . . . .
ExampleFindtheextremevaluesof f(x) = x2/3(x + 2) on [−1, 2].
SolutionWrite f(x) = x5/3 + 2x2/3, then
f′(x) =53x2/3 +
43x−1/3 =
13x−1/3(5x + 4)
Thus f′(−4/5) = 0 and f isnotdifferentiableat 0.
Soourpointstocheckare:
I f(−1) =
I f(−4/5) =
I f(0) =
I f(2) =
. . . . . .
ExampleFindtheextremevaluesof f(x) = x2/3(x + 2) on [−1, 2].
SolutionWrite f(x) = x5/3 + 2x2/3, then
f′(x) =53x2/3 +
43x−1/3 =
13x−1/3(5x + 4)
Thus f′(−4/5) = 0 and f isnotdifferentiableat 0. Soourpointstocheckare:
I f(−1) =
I f(−4/5) =
I f(0) =
I f(2) =
. . . . . .
ExampleFindtheextremevaluesof f(x) = x2/3(x + 2) on [−1, 2].
SolutionWrite f(x) = x5/3 + 2x2/3, then
f′(x) =53x2/3 +
43x−1/3 =
13x−1/3(5x + 4)
Thus f′(−4/5) = 0 and f isnotdifferentiableat 0. Soourpointstocheckare:
I f(−1) = 1I f(−4/5) =
I f(0) =
I f(2) =
. . . . . .
ExampleFindtheextremevaluesof f(x) = x2/3(x + 2) on [−1, 2].
SolutionWrite f(x) = x5/3 + 2x2/3, then
f′(x) =53x2/3 +
43x−1/3 =
13x−1/3(5x + 4)
Thus f′(−4/5) = 0 and f isnotdifferentiableat 0. Soourpointstocheckare:
I f(−1) = 1I f(−4/5) = 1.0341I f(0) =
I f(2) =
. . . . . .
ExampleFindtheextremevaluesof f(x) = x2/3(x + 2) on [−1, 2].
SolutionWrite f(x) = x5/3 + 2x2/3, then
f′(x) =53x2/3 +
43x−1/3 =
13x−1/3(5x + 4)
Thus f′(−4/5) = 0 and f isnotdifferentiableat 0. Soourpointstocheckare:
I f(−1) = 1I f(−4/5) = 1.0341I f(0) = 0I f(2) =
. . . . . .
ExampleFindtheextremevaluesof f(x) = x2/3(x + 2) on [−1, 2].
SolutionWrite f(x) = x5/3 + 2x2/3, then
f′(x) =53x2/3 +
43x−1/3 =
13x−1/3(5x + 4)
Thus f′(−4/5) = 0 and f isnotdifferentiableat 0. Soourpointstocheckare:
I f(−1) = 1I f(−4/5) = 1.0341I f(0) = 0I f(2) = 6.3496
. . . . . .
ExampleFindtheextremevaluesof f(x) = x2/3(x + 2) on [−1, 2].
SolutionWrite f(x) = x5/3 + 2x2/3, then
f′(x) =53x2/3 +
43x−1/3 =
13x−1/3(5x + 4)
Thus f′(−4/5) = 0 and f isnotdifferentiableat 0. Soourpointstocheckare:
I f(−1) = 1I f(−4/5) = 1.0341I f(0) = 0 (absolutemin)I f(2) = 6.3496
. . . . . .
ExampleFindtheextremevaluesof f(x) = x2/3(x + 2) on [−1, 2].
SolutionWrite f(x) = x5/3 + 2x2/3, then
f′(x) =53x2/3 +
43x−1/3 =
13x−1/3(5x + 4)
Thus f′(−4/5) = 0 and f isnotdifferentiableat 0. Soourpointstocheckare:
I f(−1) = 1I f(−4/5) = 1.0341I f(0) = 0 (absolutemin)I f(2) = 6.3496 (absolutemax)
. . . . . .
ExampleFindtheextremevaluesof f(x) = x2/3(x + 2) on [−1, 2].
SolutionWrite f(x) = x5/3 + 2x2/3, then
f′(x) =53x2/3 +
43x−1/3 =
13x−1/3(5x + 4)
Thus f′(−4/5) = 0 and f isnotdifferentiableat 0. Soourpointstocheckare:
I f(−1) = 1I f(−4/5) = 1.0341 (relativemax)I f(0) = 0 (absolutemin)I f(2) = 6.3496 (absolutemax)
. . . . . .
ExampleFindtheextremevaluesof f(x) =
√4− x2 on [−2, 1].
SolutionWehave f′(x) = − x√
4− x2, whichiszerowhen x = 0. (f isnot
differentiableat ±2 aswell.)
Soourpointstocheckare:I f(−2) =
I f(0) =
I f(1) =
. . . . . .
ExampleFindtheextremevaluesof f(x) =
√4− x2 on [−2, 1].
SolutionWehave f′(x) = − x√
4− x2, whichiszerowhen x = 0. (f isnot
differentiableat ±2 aswell.)
Soourpointstocheckare:I f(−2) =
I f(0) =
I f(1) =
. . . . . .
ExampleFindtheextremevaluesof f(x) =
√4− x2 on [−2, 1].
SolutionWehave f′(x) = − x√
4− x2, whichiszerowhen x = 0. (f isnot
differentiableat ±2 aswell.) Soourpointstocheckare:I f(−2) =
I f(0) =
I f(1) =
. . . . . .
ExampleFindtheextremevaluesof f(x) =
√4− x2 on [−2, 1].
SolutionWehave f′(x) = − x√
4− x2, whichiszerowhen x = 0. (f isnot
differentiableat ±2 aswell.) Soourpointstocheckare:I f(−2) = 0I f(0) =
I f(1) =
. . . . . .
ExampleFindtheextremevaluesof f(x) =
√4− x2 on [−2, 1].
SolutionWehave f′(x) = − x√
4− x2, whichiszerowhen x = 0. (f isnot
differentiableat ±2 aswell.) Soourpointstocheckare:I f(−2) = 0I f(0) = 2I f(1) =
. . . . . .
ExampleFindtheextremevaluesof f(x) =
√4− x2 on [−2, 1].
SolutionWehave f′(x) = − x√
4− x2, whichiszerowhen x = 0. (f isnot
differentiableat ±2 aswell.) Soourpointstocheckare:I f(−2) = 0I f(0) = 2I f(1) =
√3
. . . . . .
ExampleFindtheextremevaluesof f(x) =
√4− x2 on [−2, 1].
SolutionWehave f′(x) = − x√
4− x2, whichiszerowhen x = 0. (f isnot
differentiableat ±2 aswell.) Soourpointstocheckare:I f(−2) = 0 (absolutemin)I f(0) = 2I f(1) =
√3
. . . . . .
ExampleFindtheextremevaluesof f(x) =
√4− x2 on [−2, 1].
SolutionWehave f′(x) = − x√
4− x2, whichiszerowhen x = 0. (f isnot
differentiableat ±2 aswell.) Soourpointstocheckare:I f(−2) = 0 (absolutemin)I f(0) = 2 (absolutemax)I f(1) =
√3
. . . . . .
Outline
Introduction
TheExtremeValueTheorem
Fermat’sTheorem(notthelastone)Tangent: Fermat’sLastTheorem
TheClosedIntervalMethod
Examples
Challenge: Cubicfunctions
. . . . . .
Challenge: Cubicfunctions
ExampleHowmanycriticalpointscanacubicfunction
f(x) = ax3 + bx2 + cx + d
have?
. . . . . .
SolutionIf f′(x) = 0, wehave
3ax2 + 2bx + c = 0,
andso
x =−2b±
√4b2 − 12ac6a
=−b±
√b2 − 3ac3a
,
andsowehavethreepossibilities:
I b2 − 3ac > 0, inwhichcasetherearetwodistinctcriticalpoints. Anexamplewouldbe f(x) = x3 + x2, where a = 1,b = 1, and c = 0.
I b2 − 3ac < 0, inwhichcasetherearenorealrootstothequadratic, hencenocriticalpoints. Anexamplewouldbef(x) = x3 + x2 + x, where a = b = c = 1.
I b2 − 3ac = 0, inwhichcasethereisasinglecriticalpoint.Example: x3, where a = 1 and b = c = 0.
. . . . . .
SolutionIf f′(x) = 0, wehave
3ax2 + 2bx + c = 0,
andso
x =−2b±
√4b2 − 12ac6a
=−b±
√b2 − 3ac3a
,
andsowehavethreepossibilities:I b2 − 3ac > 0, inwhichcasetherearetwodistinctcritical
points. Anexamplewouldbe f(x) = x3 + x2, where a = 1,b = 1, and c = 0.
I b2 − 3ac < 0, inwhichcasetherearenorealrootstothequadratic, hencenocriticalpoints. Anexamplewouldbef(x) = x3 + x2 + x, where a = b = c = 1.
I b2 − 3ac = 0, inwhichcasethereisasinglecriticalpoint.Example: x3, where a = 1 and b = c = 0.
. . . . . .
SolutionIf f′(x) = 0, wehave
3ax2 + 2bx + c = 0,
andso
x =−2b±
√4b2 − 12ac6a
=−b±
√b2 − 3ac3a
,
andsowehavethreepossibilities:I b2 − 3ac > 0, inwhichcasetherearetwodistinctcritical
points. Anexamplewouldbe f(x) = x3 + x2, where a = 1,b = 1, and c = 0.
I b2 − 3ac < 0, inwhichcasetherearenorealrootstothequadratic, hencenocriticalpoints. Anexamplewouldbef(x) = x3 + x2 + x, where a = b = c = 1.
I b2 − 3ac = 0, inwhichcasethereisasinglecriticalpoint.Example: x3, where a = 1 and b = c = 0.
. . . . . .
SolutionIf f′(x) = 0, wehave
3ax2 + 2bx + c = 0,
andso
x =−2b±
√4b2 − 12ac6a
=−b±
√b2 − 3ac3a
,
andsowehavethreepossibilities:I b2 − 3ac > 0, inwhichcasetherearetwodistinctcritical
points. Anexamplewouldbe f(x) = x3 + x2, where a = 1,b = 1, and c = 0.
I b2 − 3ac < 0, inwhichcasetherearenorealrootstothequadratic, hencenocriticalpoints. Anexamplewouldbef(x) = x3 + x2 + x, where a = b = c = 1.
I b2 − 3ac = 0, inwhichcasethereisasinglecriticalpoint.Example: x3, where a = 1 and b = c = 0.
. . . . . .
Review
I Concept: absolute(global)andrelative(local)maxima/minima
I Fact: Fermat’stheorem: f′(x) = 0 atlocalextremaI Technique: the ClosedIntervalMethod