Lesson 19: Maximum and Minimum Values

. . . . . .

Section4.1MaximumandMinimumValues

V63.0121.027, CalculusI

November6, 2009

Announcements

I Quiznextweekon§§3.1–3.5I FinalExamFriday, December18, 2:00–3:50pm

..Imagecredit: KarenwithaK

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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)

. . . . . .

Design

..Imagecredit: JasonTromm

. . . . . .

Whygototheextremes?

I Maupertuis’principle:“ActionisminimizedthroughthewisdomofGod.”

Pierre-LouisMaupertuis(1698–1759)

. . . . . .

Optics

.Imagecredit: jacreative

. . . . . .

Whygototheextremes?

I Maupertuis’principle:“ActionisminimizedthroughthewisdomofGod.” Pierre-LouisMaupertuis

(1698–1759)

. . . . . .

Outline

Introduction

Examples

. . . . . .

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].

. . . . . .

.cmaximum

.maximum

minimum

.minimum

. . . . . .

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.. .|

. . . . . .

BadExample#1

Example

Considerthefunction

f(x) =

{x 0 ≤ x < 1

x− 2 1 ≤ x ≤ 2.. .|

. . . . . .

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.

. . . . . .

FinalBadExample

Example

Thefunction f(x) =1xiscontinuousontheclosedinterval [1,∞)

buthasnominimumvalue.

. . . . . .

Outline

Introduction

Examples

. . . . . .

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.

.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.

.local

maximum

.local

minimum

. . . . . .

I Soalocalextremummustbe inside thedomainof f (notontheend).

I A globalextremumthatisinsidethedomainisalocalextremum.

.globalmax

.localmax

.local and global

. . . . . .

Theorem(Fermat’sTheorem)Suppose f hasalocalextremumat c and f isdifferentiableat c.Then f′(c) = 0.

.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

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

I Sincethelimit f′(c) = limh→0

f(c + h) − f(c)h

exists, itmustbe 0.

. . . . . .

Supposethat f hasalocalmaximumat c.I If h iscloseenoughto 0 butgreaterthan 0, f(c + h) ≤ f(c).Thismeans

f(c + h) − f(c)h

=⇒ limh→0+

f(c + h) − f(c)h

≥ 0 =⇒ limh→0−

f(c + h) − f(c)h

exists, itmustbe 0.

. . . . . .

f(c + h) − f(c)h

≤ 0 =⇒ limh→0+

f(c + h) − f(c)h

≥ 0 =⇒ limh→0−

f(c + h) − f(c)h

exists, itmustbe 0.

. . . . . .

f(c + h) − f(c)h

≤ 0 =⇒ limh→0+

f(c + h) − f(c)h

=⇒ limh→0−

f(c + h) − f(c)h

exists, itmustbe 0.

. . . . . .

f(c + h) − f(c)h

≤ 0 =⇒ limh→0+

f(c + h) − f(c)h

≥ 0 =⇒ limh→0−

f(c + h) − f(c)h

exists, itmustbe 0.

. . . . . .

f(c + h) − f(c)h

≤ 0 =⇒ limh→0+

f(c + h) − f(c)h

≥ 0 =⇒ 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)

. . . . . .

Outline

Introduction

Examples

. . . . . .

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

.f is notdiff at c

.f′(c) = 0

. . . . . .

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

Examples

. . . . . .

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

. . . . . .

I f(−1) = 2(−1) − 5 = −7I f(2) = 2(2) − 5 = −1

. . . . . .

I f(−1) = 2(−1) − 5 = −7I f(2) = 2(2) − 5 = −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) =

. . . . . .

SolutionWehave f′(x) = 2x, whichiszerowhen x = 0.

I f(−1) =

I f(0) =

I f(2) =

. . . . . .

SolutionWehave f′(x) = 2x, whichiszerowhen x = 0. Soourpointstocheckare:

I f(−1) =

I f(0) =

I f(2) =

. . . . . .

I f(−1) = 0I f(0) =

I f(2) =

. . . . . .

I f(−1) = 0I f(0) = − 1I f(2) =

. . . . . .

I f(−1) = 0I f(0) = − 1I f(2) = 3

. . . . . .

I f(−1) = 0I f(0) = − 1 (absolutemin)I f(2) = 3

. . . . . .

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)

. . . . . .

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)

. . . . . .

I f(−1) =

− 4 (absolutemin)

I f(0) =

1 (localmax)

I f(1) =

0 (localmin)

I f(2) =

5 (absolutemax)

. . . . . .

I f(−1) = − 4

(absolutemin)

I f(0) =

1 (localmax)

I f(1) =

0 (localmin)

I f(2) =

5 (absolutemax)

. . . . . .

I f(−1) = − 4

(absolutemin)

I f(0) = 1

(localmax)

I f(1) =

0 (localmin)

I f(2) =

5 (absolutemax)

. . . . . .

I f(−1) = − 4

(absolutemin)

I f(0) = 1

(localmax)

I f(1) = 0

(localmin)

I f(2) =

5 (absolutemax)

. . . . . .

I f(−1) = − 4

(absolutemin)

I f(0) = 1

(localmax)

I f(1) = 0

(localmin)

I f(2) = 5

(absolutemax)

. . . . . .

I f(−1) = − 4 (absolutemin)I f(0) = 1

(localmax)

I f(1) = 0

(localmin)

I f(2) = 5

(absolutemax)

. . . . . .

I f(−1) = − 4 (absolutemin)I f(0) = 1

(localmax)

I f(1) = 0

(localmin)

I f(2) = 5 (absolutemax)

. . . . . .

I f(−1) = − 4 (absolutemin)I f(0) = 1 (localmax)I f(1) = 0

(localmin)

I f(2) = 5 (absolutemax)

. . . . . .

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.

I f(−1) =

I f(−4/5) =

I f(0) =

I f(2) =

. . . . . .

f′(x) =53x2/3 +

43x−1/3 =

13x−1/3(5x + 4)

Thus f′(−4/5) = 0 and f isnotdifferentiableat 0.

I f(−1) =

I f(−4/5) =

I f(0) =

I f(2) =

. . . . . .

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) =

. . . . . .

f′(x) =53x2/3 +

43x−1/3 =

13x−1/3(5x + 4)

I f(−1) = 1I f(−4/5) =

I f(0) =

I f(2) =

. . . . . .

f′(x) =53x2/3 +

43x−1/3 =

13x−1/3(5x + 4)

I f(−1) = 1I f(−4/5) = 1.0341I f(0) =

I f(2) =

. . . . . .

f′(x) =53x2/3 +

43x−1/3 =

13x−1/3(5x + 4)

I f(−1) = 1I f(−4/5) = 1.0341I f(0) = 0I f(2) =

. . . . . .

f′(x) =53x2/3 +

43x−1/3 =

13x−1/3(5x + 4)

I f(−1) = 1I f(−4/5) = 1.0341I f(0) = 0I f(2) = 6.3496

. . . . . .

f′(x) =53x2/3 +

43x−1/3 =

13x−1/3(5x + 4)

I f(−1) = 1I f(−4/5) = 1.0341I f(0) = 0 (absolutemin)I f(2) = 6.3496

. . . . . .

f′(x) =53x2/3 +

43x−1/3 =

13x−1/3(5x + 4)

I f(−1) = 1I f(−4/5) = 1.0341I f(0) = 0 (absolutemin)I f(2) = 6.3496 (absolutemax)

. . . . . .

f′(x) =53x2/3 +

43x−1/3 =

13x−1/3(5x + 4)

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) =

. . . . . .

√4− x2 on [−2, 1].

differentiableat ±2 aswell.)

Soourpointstocheckare:I f(−2) =

I f(0) =

I f(1) =

. . . . . .

√4− x2 on [−2, 1].

differentiableat ±2 aswell.) Soourpointstocheckare:I f(−2) =

I f(0) =

I f(1) =

. . . . . .

√4− x2 on [−2, 1].

differentiableat ±2 aswell.) Soourpointstocheckare:I f(−2) = 0I f(0) =

I f(1) =

. . . . . .

√4− x2 on [−2, 1].

differentiableat ±2 aswell.) Soourpointstocheckare:I f(−2) = 0I f(0) = 2I f(1) =

. . . . . .

√4− x2 on [−2, 1].

differentiableat ±2 aswell.) Soourpointstocheckare:I f(−2) = 0I f(0) = 2I f(1) =

. . . . . .

√4− x2 on [−2, 1].

differentiableat ±2 aswell.) Soourpointstocheckare:I f(−2) = 0 (absolutemin)I f(0) = 2I f(1) =

. . . . . .

√4− x2 on [−2, 1].

differentiableat ±2 aswell.) Soourpointstocheckare:I f(−2) = 0 (absolutemin)I f(0) = 2 (absolutemax)I f(1) =

. . . . . .

Outline

Introduction

Examples

. . . . . .

ExampleHowmanycriticalpointscanacubicfunction

f(x) = ax3 + bx2 + cx + d

. . . . . .

SolutionIf f′(x) = 0, wehave

3ax2 + 2bx + c = 0,

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.

. . . . . .

3ax2 + 2bx + c = 0,

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.

. . . . . .

3ax2 + 2bx + c = 0,

x =−2b±

√4b2 − 12ac6a

=−b±

√b2 − 3ac3a

. . . . . .

3ax2 + 2bx + c = 0,

x =−2b±

√4b2 − 12ac6a

=−b±

√b2 − 3ac3a

. . . . . .

Review

I Concept: absolute(global)andrelative(local)maxima/minima

I Fact: Fermat’stheorem: f′(x) = 0 atlocalextremaI Technique: the ClosedIntervalMethod

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