lesson 16: maximum and minimum values

. . . . . .

Section4.1MaximumandMinimumValues

V63.0121.006/016, CalculusI

March23, 2010

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Announcements

I WelcomebackfromSpringBreak!I Quiz3: April2, Sections2.6–3.5

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ComputationofMidtermLetterGrades

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DistributionofMidtermAveragesandLetterGrades

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WhatcanI dotoimprovemygrade?

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. .HW.10%

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59%ofyourgradeisstillinplay!

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Help!

Freeresources:I recitationI TAs’officehoursI myofficehoursI MathTutoringCenter(CIWW 524)

I CollegeLearningCenter

http://www.math.nyu.edu/degree/undergrad/tutor_schedule.html

http://learning.cas.nyu.edu/docs/CP/1928/FinalCLC_Schedule_jan25.pdf

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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

http://www.flickr.com/photos/trommetter/1338616263/

. . . . . .

Whygototheextremes?



I Maupertuis’principle:“ActionisminimizedthroughthewisdomofGod.”

Pierre-LouisMaupertuis(1698–1759)

. . . . . .

Optics

.

.Imagecredit: jacreative

http://www.flickr.com/photos/jacreative/110448225

. . . . . .

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

http://www.flickr.com/photos/patrick_q/204252734/

. . . . . .

Theorem(TheExtremeValueTheorem)Let f beafunctionwhichiscontinuousontheclosedinterval[a,b]. Then f attainsanabsolutemaximumvalue f(c) andanabsoluteminimumvalue f(d) atnumbers c and d in [a,b].

.

. . . . . .


...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. ThisdoesnotviolateEVT because f isnotcontinuous.

. . . . . .

BadExample#1

Example

Considerthefunction

f(x) =

{x 0 ≤ x < 1

x− 2 1 ≤ x ≤ 2.. .|

.1.

.

.

.


. . . . . .

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.

ThisdoesnotviolateEVT because f isnotcontinuous.

. . . . . .

BadExample#1

Example

Considerthefunction

f(x) =

{x 0 ≤ x < 1

x− 2 1 ≤ x ≤ 2.. .|

.1.

.

.

.


. . . . . .

BadExample#2

ExampleConsiderthefunction f(x) = x restrictedtotheinterval [0, 1).

. .|.1

.

.

Thereisstillnomaximumvalue(valuesgetarbitrarilycloseto1butdonotachieveit). ThisdoesnotviolateEVT becausethedomainisnotclosed.

. . . . . .

BadExample#2


. .|.1

.

.

Thereisstillnomaximumvalue(valuesgetarbitrarilycloseto1butdonotachieveit).

ThisdoesnotviolateEVT becausethedomainisnotclosed.

. . . . . .

BadExample#2


. .|.1

.

.

Thereisstillnomaximumvalue(valuesgetarbitrarilycloseto1butdonotachieveit). ThisdoesnotviolateEVT becausethedomainisnotclosed.

. . . . . .

FinalBadExample

Example

Considerthefunction f(x) =1xiscontinuousontheclosed

interval [1,∞).

. ..1

.

Thereisnominimumvalue(valuesgetarbitrarilycloseto 0 butdonotachieveit). ThisdoesnotviolateEVT becausethedomainisnotbounded.

. . . . . .

FinalBadExample

Example


interval [1,∞).

. ..1

.

Thereisnominimumvalue(valuesgetarbitrarilycloseto 0 butdonotachieveit).

ThisdoesnotviolateEVT becausethedomainisnotbounded.

. . . . . .

FinalBadExample

Example


interval [1,∞).

. ..1

.

Thereisnominimumvalue(valuesgetarbitrarilycloseto 0 butdonotachieveit). ThisdoesnotviolateEVT becausethedomainisnotbounded.

. . . . . .

Outline

Introduction




Examples


. . . . . .

Localextrema

Definition

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

.globalmax

.local and global

min

. . . . . .

Localextrema

I Soalocalextremummustbe inside thedomainof f (notontheend).

I A globalextremumthatisinsidethedomainisalocalextremum.

..|.a

.|.b

.

.

.

.local

maximum

.

.local

minimum

.globalmax

.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 x isslightlygreaterthan c, f(x) ≤ f(c). Thismeans

f(x)− f(c)x− c

≤ 0 =⇒ limx→c+

f(x)− f(c)x− c

≤ 0

I Thesamewillbetrueontheotherend: if x isslightlylessthan c, f(x) ≤ f(c). Thismeans

f(x)− f(c)x− c

≥ 0 =⇒ limx→c−

f(x)− f(c)x− c

≥ 0

I Sincethelimit f′(c) = limx→c

f(x)− f(c)x− c

exists, itmustbe 0.

. . . . . .


Supposethat f hasalocalmaximumat c.I If x isslightlygreaterthan c, f(x) ≤ f(c). Thismeans

f(x)− f(c)x− c

≤ 0

=⇒ limx→c+

f(x)− f(c)x− c

≤ 0


f(x)− f(c)x− c

≥ 0 =⇒ limx→c−

f(x)− f(c)x− c

≥ 0


f(x)− f(c)x− c

exists, itmustbe 0.

. . . . . .



f(x)− f(c)x− c

≤ 0 =⇒ limx→c+

f(x)− f(c)x− c

≤ 0


f(x)− f(c)x− c

≥ 0 =⇒ limx→c−

f(x)− f(c)x− c

≥ 0


f(x)− f(c)x− c

exists, itmustbe 0.

. . . . . .



f(x)− f(c)x− c

≤ 0 =⇒ limx→c+

f(x)− f(c)x− c

≤ 0


f(x)− f(c)x− c

≥ 0

=⇒ limx→c−

f(x)− f(c)x− c

≥ 0


f(x)− f(c)x− c

exists, itmustbe 0.

. . . . . .



f(x)− f(c)x− c

≤ 0 =⇒ limx→c+

f(x)− f(c)x− c

≤ 0


f(x)− f(c)x− c

≥ 0 =⇒ limx→c−

f(x)− f(c)x− c

≥ 0


f(x)− f(c)x− c

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

at c?

.f is notdiff at c

.f′(c) = 0

.no

.yes

.no

.yes

. . . . . .


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

−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. 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 (globalmin)

I f(0) =

1 (localmax)

I f(1) =

0 (localmin)

I f(2) =

5 (globalmax)

. . . . . .


SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), wehavecriticalpointsatx = 0 and x = 1.

ThevaluestocheckareI f(−1) =

− 4 (globalmin)

I f(0) =

1 (localmax)

I f(1) =

0 (localmin)

I f(2) =

5 (globalmax)

. . . . . .



I f(−1) =

− 4 (globalmin)

I f(0) =

1 (localmax)

I f(1) =

0 (localmin)

I f(2) =

5 (globalmax)

. . . . . .



I f(−1) = − 4

(globalmin)

I f(0) =

1 (localmax)

I f(1) =

0 (localmin)

I f(2) =

5 (globalmax)

. . . . . .



I f(−1) = − 4

(globalmin)

I f(0) = 1

(localmax)

I f(1) =

0 (localmin)

I f(2) =

5 (globalmax)

. . . . . .



I f(−1) = − 4

(globalmin)

I f(0) = 1

(localmax)

I f(1) = 0

(localmin)

I f(2) =

5 (globalmax)

. . . . . .



I f(−1) = − 4

(globalmin)

I f(0) = 1

(localmax)

I f(1) = 0

(localmin)

I f(2) = 5

(globalmax)

. . . . . .



I f(−1) = − 4 (globalmin)I f(0) = 1

(localmax)

I f(1) = 0

(localmin)

I f(2) = 5

(globalmax)

. . . . . .



I f(−1) = − 4 (globalmin)I f(0) = 1

(localmax)

I f(1) = 0

(localmin)

I f(2) = 5 (globalmax)

. . . . . .



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

(localmin)

I f(2) = 5 (globalmax)

. . . . . .



I f(−1) = − 4 (globalmin)I f(0) = 1 (localmax)I f(1) = 0 (localmin)I f(2) = 5 (globalmax)

. . . . . .

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) =

. . . . . .



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) = 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) =

√3

. . . . . .


√4− x2 on [−2, 1].



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

√3

. . . . . .


√4− x2 on [−2, 1].



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

√3

. . . . . .

Outline

Introduction




Examples


. . . . . .


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.

. . . . . .

Whathavewelearnedtoday?

I TheExtremeValueTheorem: acontinuousfunctiononaclosedintervalmustachieveitsmaxandmin

I Fermat’sTheorem: localextremaarecriticalpointsI TheClosedIntervalMethod: analgorithmforfindingglobalextrema

I ShowyourworkunlessyouwanttoenduplikeFermat!

lesson 16: maximum and minimum values

Education

minimumvalueof f

astwa f

f havedomain d

fc fxforall x

absoluteminimumat cif

optics imagecredit

whatcani dotoimprovemygrade

value minimum