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Section 4.1Maximum and Minimum Values

V63.0121.006/016, Calculus I

March 23, 2010

Announcements

Welcome back from Spring Break!

Quiz 3: April 2, Sections 2.63.5

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Announcements

Welcome back from Spring Break!

Quiz 3: April 2, Sections 2.63.5

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Computation of Midterm Letter Grades

HW

10%

WebAssign10%

Quizzes

15%

Midterm 25%

Final

40%

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HW

10%

WebAssign10%

Quizzes

15%

Midterm 25%

Final (to be determined)

40%

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HW

17%

WebAssign

17%Quizzes

25%

Midterm

41%

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Distribution of Midterm Averages and Letter Grades

Median = 81.35% (curved to B)

Average = 74.39% (curved to B-) Standard Deviation = 21%

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What can I do to improve my grade?

HW

10%

WebAssign

10%

Quizzes

15%

Midterm25%

Final

40%

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past HW

5%

future HW5% past WA

5% future WA

5%past Quizzes

6%future Quizzes9%

Midterm25%

Final

40%

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past HW

5%

past WA5% past Quizzes6%

Midterm

25%future HW 5%future WA 5%

future Quizzes

9%

Final

40%

59% of your grade is still in play!

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Help!

Free resources:

recitation

TAs office hours

my office hours

Math Tutoring Center(CIWW 524)

College Learning Center
http://learning.cas.nyu.edu/docs/CP/1928/FinalCLC_Schedule_jan25.pdfhttp://www.math.nyu.edu/degree/undergrad/tutor_schedule.html

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Outline

Introduction

The Extreme Value Theorem

Fermats Theorem (not the last one)

Tangent: Fermats Last Theorem

The Closed Interval Method

Examples

Challenge: Cubic functions

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Optimize

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Why go to the extremes?

Rationally speaking, it isadvantageous to find theextreme values of afunction (maximizeprofit, minimize costs,

etc.)

Pierre-Louis Maupertuis(16981759)

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Design

Image credit: Jason Tromm
http://www.flickr.com/photos/trommetter/1338616263/

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etc.) Many laws of science

are derived fromminimizing principles.

Pierre-Louis Maupertuis(16981759)

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Optics

Image credit: jacreative

h h

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etc.) Many laws of science

are derived fromminimizing principles.

Maupertuis principle:Action is minimizedthrough the wisdom ofGod. Pierre-Louis Maupertuis

(16981759)

O li

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Outline

Introduction





Examples


E i d l

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Extreme points and values

Definition

Letfhave domainD.

Image credit: Patrick Q

E t i t d l

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Definition

Letfhave domainD. The functionfhas anabsolute

maximum(orglobal maximum)(respectively,absolute minimum) atc

iff(c) f(x)(respectively,f(c) f(x))for allxinD


E t i t d l

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Definition



iff(c) f(x)(respectively,f(c) f(x))for allxinD The numberf(c)is called the

maximum value(respectively,minimum value) offonD.


E treme points and al es

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Definition



iff(c) f(x)(respectively,f(c) f(x))for allxinD The numberf(c)is called the

maximum value(respectively,minimum value) offonD.

Anextremumis either a maximum ora minimum. Anextreme valueiseither a maximum value or minimumvalue.


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Theorem (The Extreme Value Theorem)Let f be a function which is continuous on the closed interval[a,b]. Then f attains an absolute maximum value f(c)and an

absolute minimum value f(d)at numbers c and d in[a,b].

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a

b

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a

b

cmaximum

maximum

value

f(c)

d

minimum

minimum

value

f(d)

No proof of EVT forthcoming

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No proof of EVT forthcoming

This theorem is very hard to prove without using technical

facts about continuous functions and closed intervals. But we can show the importance of each of the hypotheses.

Bad Example #1

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Bad Example #1

Example

Consider the function

f(x) =x 0 x

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Bad Example #1

Example


f(x) =x 0 x

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Bad Example #1

Example


f(x) =x 0 x

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Bad Example #1

Example


f(x) =x 0 x

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Bad Example #2

Example

Consider the functionf(x) =xrestricted to the interval[0, 1).

Bad Example #2

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ad a p e

Example


|1

Bad Example #2

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p

Example


|1

There is still no maximum value (values get arbitrarily close to 1but do not achieve it).

Bad Example #2

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p

Example


|1

There is still no maximum value (values get arbitrarily close to 1but do not achieve it). This does not violate EVT because thedomain is not closed.

Final Bad Example

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p

ExampleConsider the functionf(x) =

1

xis continuous on the closed

interval [1,).

Final Bad Example

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p


1


interval [1,).

1

Final Bad Example

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p


1


interval [1,).

1

There is no minimum value (values get arbitrarily close to 0 butdo not achieve it).

Final Bad Example

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1


interval [1,).

1

There is no minimum value (values get arbitrarily close to 0 butdo not achieve it). This does not violate EVT because the domainis not bounded.

Outline

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Introduction





Examples


Local extrema

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Definition

A functionfhas alocal maximumorrelative maximumatciff(c) f(x)whenxis nearc. This means thatf(c) f(x)forallxin some open interval containingc.

Similarly,fhas alocal minimumatciff(c) f(x)whenxisnearc.

Local extrema

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Definition

A functionfhas alocal maximumorrelative maximumatciff(c) f(x)whenxis nearc. This means thatf(c) f(x)forallxin some open interval containingc.

Similarly,fhas alocal minimumatciff(c) f(x)whenxisnearc.

|a |b

localmaximum

local

minimum

Local extrema

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So a local extremum must beinsidethe domain off(not onthe end).

A global extremum that is inside the domain is a localextremum.

|a |b

localmaximum

globalmax

local and global

min

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Theorem (Fermats Theorem)Suppose f has a local extremum at c and f is differentiable at c.

Then f(c) =0.

|a

|b

localmaximum

local

minimum

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Theorem (Fermats Theorem)Suppose f has a local extremum at c and f is differentiable at c.

Then f(c) =0.

|a

|b

localmaximum

local

minimum

Sketch of proof of Fermats Theorem

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Suppose thatfhas a local maximum atc.


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Ifxis slightly greater thanc,f(x) f(c). This meansf(x) f(c)

x c 0


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x c 0 = limxc+f(x) f(c)

x c 0


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x c 0

The same will be true on the other end: ifxis slightly lessthanc,f(x) f(c). This means

f(x) f(c)x

c 0


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x c 0


f(x) f(c)x

c 0 = lim

xc

f(x) f(c)x

c 0


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x c 0


f(x) f(c)x

c 0 = lim

xc

f(x) f(c)x

c 0

Since the limitf(c) =limxc

f(x) f(c)x c exists, it must be 0.

Meet the Mathematician: Pierre de Fermat

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16011665

Lawyer and number

theorist Proved many theorems,

didnt quite prove hislast one


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Plenty of solutions tox2 +y2 =z2 amongpositive whole numbers(e.g.,x=3,y=4,z=5)


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No solutions to

x3 +y3 =z3 amongpositive whole numbers


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No solutions to


Fermat claimed nosolutions toxn +yn =zn

but didnt write downhis proof


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No solutions to


Fermat claimed nosolutions toxn +yn =zn

but didnt write downhis proof

Not solved until 1998!(TaylorWiles)

Outline

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Introduction





Examples


Flowchart for placing extremaThanks to Fermat

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Supposefis a continuous function on the closed, boundedinterval [a, b], andcis a global maximum point.

start

Iscan

endpoint?

c =aorc =b

cis a

local max

Isfdiffble

atc?

fis notdiff atc

f(c) =0

no

yes

no

yes


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This means to find the maximum value offon[a,b], we need to:

Evaluatefat theendpointsaandb

Evaluatefat thecritical pointsorcritical numbersxwhereeitherf(x) =0 orfis not differentiable atx.

The points with the largest function value are the globalmaximum points

The points with the smallest or most negative function valueare the global minimum points.

Outline

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Introduction





Examples


E l

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Example

Find the extreme values off(x) =2x 5 on[1, 2].

Example

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Example


SolutionSince f(x) =2, which is never zero, we have no critical pointsand we need only investigate the endpoints:

f(

1) =2(

1)

5=

7

f(2) =2(2) 5= 1

Example

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Example


SolutionSince f(x) =2, which is never zero, we have no critical pointsand we need only investigate the endpoints:

f(

1) =2(

1)

5=

7

f(2) =2(2) 5= 1So

The absolute minimum (point) is at1; the minimum valueis

7.

The absolute maximum (point) is at2; the maximum value is1.

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Example

Find the extreme values off(x) =x2 1 on[1,2].

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Example


SolutionWe have f(x) =2x, which is zero when x =0.

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Example


SolutionWe have f(x) =2x, which is zero when x =0. So our points to

check are: f(1) = f(0) =

f(2) =

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Example



check are: f(1) =0 f(0) =

f(2) =

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Example



check are: f(1) =0 f(0) = 1 f(2) =

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Example


SolutionWe have f(x) =2x, which is zero when x =0. So our points tocheck are:

f(1) =0 f(0) = 1 f(2) =3

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Example



f(1) =0 f(0) = 1(absolute min) f(2) =3

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Example



f(1) =0 f(0) = 1(absolute min) f(2) =3(absolute max)

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Example

Find the extreme values off(x) =2x3 3x2 +1 on[1,2].

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Example

Find the extreme values off(x) =2x3 3x2 +1 on[1,2].SolutionSince f(x) =6x2 6x=6x(x 1), we have critical points atx=0and x=1.

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Example

Find the extreme values off(x) =2x3 3x2 +1 on[1,2].SolutionSince f(x) =6x2 6x=6x(x 1), we have critical points atx=0and x=1. The values to check are

f(1) = f(0) =

f(1) =

f(2) =

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Example


f(1) = 4 f(0) =

f(1) =

f(2) =

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Example


f(1) = 4 f(0) =1

f(1) =

f(2) =

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Example


f(1) = 4 f(0) =1

f(1) =0

f(2) =

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Example


f(1) = 4 f(0) =1

f(1) =0

f(2) =5

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Example


f(1) = 4(global min) f(0) =1

f(1) =0

f(2) =5

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Example


f(1) = 4(global min) f(0) =1

f(1) =0

f(2) =5(global max)

l

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Example


f(1) = 4(global min) f(0) =1(local max)

f(1) =0

f(2) =5(global max)

l

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Example


f(1) = 4(global min) f(0) =1(local max)

f(1) =0(local min)

f(2) =5(global max)

Example

Find the extreme values of f(x) = x2/3(x + 2) on [1,2].

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Find the extreme values off(x) x (x+2)on[ 1,2].

Example

Find the extreme values off(x) =x2/3(x+2)on[1,2].

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d t e e t e e a ues o ( ) ( + ) o [ , ].

SolutionWrite f(x) =x5/3 +2x2/3, then

f(x) =5

3x2/3 +

4

3x1/3 =

1

3x1/3(5x+4)

Thus f(4/5) =0and f is not differentiable at0.

Example


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( ) ( + ) [ , ]


f(x) =5

3x2/3 +

4

3x1/3 =

1

3x1/3(5x+4)

Thus f(4/5) =0and f is not differentiable at0. So our points tocheck are:

f(1) = f(

4/5) =

f(0) =

f(2) =

Example


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( ) ( ) [ , ]


f(x) =5

3x2/3 +

4

3x1/3 =

1

3x1/3(5x+4)


f(1) =1 f(

4/5) =

f(0) =

f(2) =

Example


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( ) ( ) [ ]


f(x) =5

3x2/3 +

4

3x1/3 =

1

3x1/3(5x+4)


f(1) =1 f(

4/5) =1.0341

f(0) =

f(2) =

Example


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f(x) =5

3x2/3 +

4

3x1/3 =

1

3x1/3(5x+4)


f(1) =1 f(

4/5) =1.0341

f(0) =0

f(2) =

Example


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f(x) =5

3x2/3 +

4

3x1/3 =

1

3x1/3(5x+4)


f(1) =1 f(

4/5) =1.0341

f(0) =0

f(2) =6.3496

Example


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f(x) =5

3x2/3 +

4

3x1/3 =

1

3x1/3(5x+4)


f(1) =1 f(

4/5) =1.0341

f(0) =0(absolute min)

f(2) =6.3496

Example


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f(x) =5

3x2/3 +

4

3x1/3 =

1

3x1/3(5x+4)


f(1) =1 f(

4/5) =1.0341


f(2) =6.3496(absolute max)

Example


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f(x) =5

3x2/3 +

4

3x1/3 =

1

3x1/3(5x+4)


f(1) =1 f(

4/5) =1.0341(relative max)


f(2) =6.3496(absolute max)

Example

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Example

Find the extreme values off(x) =

4x2 on[2,1].

Example

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Example


4x2 on[2,1].SolutionWe have f(x) = x

4x2, which is zero when x=0. (f is not

differentiable at

2as well.)

Example

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Example




differentiable at

2as well.) So our points to check are:

f(2) = f(0) =

f(1) =

Example

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Example




differentiable at


f(2) =0 f(0) =

f(1) =

Example

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Example




differentiable at


f(2) =0 f(0) =2

f(1) =

Example

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Example




differentiable at


f(2) =0 f(0) =2

f(1) =

3

Example

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Example




differentiable at


f(2) =0(absolute min) f(0) =2

f(1) =

3

Example

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Example




differentiable at


f(2) =0(absolute min) f(0) =2(absolute max)

f(1) =

3

Outline

Introduction

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Fermats Theorem (not the last one)Tangent: Fermats Last Theorem


Examples



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Example

How many critical points can a cubic function

f(x) =ax3

+bx2

+cx+d

have?

SolutionIf f(x) =0, we have

3ax2 +2bx+c=0,

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and so

x=2b

4b2 12ac6a

=b

b2 3ac

3a ,

and so we have three possibilities:


3ax2 +2bx+c=0,

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and so

x=2b

4b2 12ac6a

=b

b2 3ac

3a ,

and so we have three possibilities: b2 3ac>0, in which case there are two distinct critical

points. An example would be f(x) =x3 +x2, where a=1,b=1, and c =0.


3ax2 +2bx+c=0,

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and so

x=2b

4b2 12ac6a

=b

b2 3ac

3a ,



b2

3ac

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and so

x=2b

4b2 12ac6a

=b

b2 3ac

3a ,



b2

3ac

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The Extreme Value Theorem: a continuous function on aclosed interval must achieve its max and min

Fermats Theorem: local extrema are critical points

The Closed Interval Method: an algorithm for finding globalextrema

Show your work unless you want to end up like Fermat!

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