lesson 18: maximum and minimum values (section 041 slides)

Section 4.1Maximum and Minimum Values

V63.0121.041, Calculus I

New York University

November 8, 2010

Announcements

I Quiz 4 on Sections 3.3, 3.4, 3.5, and 3.7 next week (November16, 18, or 19)

. . . . . .

. . . . . .

Announcements

I Quiz 4 on Sections 3.3,3.4, 3.5, and 3.7 next week(November 16, 18, or 19)

V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 2 / 34

. . . . . .

Objectives

I Understand and be able toexplain the statement ofthe Extreme ValueTheorem.

I Understand and be able toexplain the statement ofFermat’s Theorem.

I Use the Closed IntervalMethod to find the extremevalues of a function definedon a closed interval.


. . . . . .

Outline

Introduction

The Extreme Value Theorem

Fermat’s Theorem (not the last one)Tangent: Fermat’s Last Theorem

The Closed Interval Method

Examples


.

.

Optimize

Why go to the extremes?

I Rationally speaking, it isadvantageous to find theextreme values of afunction (maximize profit,minimize costs, etc.)

I Many laws of science arederived from minimizingprinciples.

I Maupertuis’ principle:“Action is minimizedthrough the wisdom ofGod.”

Pierre-Louis Maupertuis(1698–1759)V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 6 / 34

Design

..Image credit: Jason TrommV63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 7 / 34

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

Optics

..Image credit: jacreativeV63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 9 / 34

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

Outline

Introduction




Examples


Extreme points and values

DefinitionLet f have domain D.

I The function f has an absolutemaximum (or global maximum)(respectively, absolute minimum) at c iff(c) ≥ f(x) (respectively, f(c) ≤ f(x)) for allx in D

I The number f(c) is called the maximumvalue (respectively, minimum value) of fon D.

I An extremum is either a maximum or aminimum. An extreme value is either amaximum value or minimum value.

.

.Image credit: Patrick Q


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


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

.





...a

..b

.

.





...a

..b

.

.

.cmaximum

.maximum

value.f(c)

.

.d

minimum

.minimum

value.f(d)


No proof of EVT forthcoming

I This theorem is very hard to prove without using technical factsabout continuous functions and closed intervals.

I But we can show the importance of each of the hypotheses.


Bad Example #1

Example

Consider the function

f(x) =

{x 0 ≤ x < 1x− 2 1 ≤ x ≤ 2.

. .|.1

.

.

.

.

Then although values of f(x) get arbitrarily close to 1 and never biggerthan 1, 1 is not the maximum value of f on [0,1] because it is neverachieved. This does not violate EVT because f is not continuous.


Bad Example #1

Example


f(x) =

{x 0 ≤ x < 1x− 2 1 ≤ x ≤ 2.

. .|.1

.

.

.

.

Then although values of f(x) get arbitrarily close to 1 and never biggerthan 1, 1 is not the maximum value of f on [0,1] because it is neverachieved.

This does not violate EVT because f is not continuous.


Bad Example #1

Example


f(x) =

{x 0 ≤ x < 1x− 2 1 ≤ x ≤ 2.

. .|.1

.

.

.

.

Then although values of f(x) get arbitrarily close to 1 and never biggerthan 1, 1 is not the maximum value of f on [0,1] because it is neverachieved. This does not violate EVT because f is not continuous.


Bad Example #2

Example

Consider the function f(x) = x restricted to the interval [0,1).

. .|.1

.

.

There is still no maximum value (values get arbitrarily close to 1 but donot achieve it). This does not violate EVT because the domain is notclosed.


Bad Example #2

Example


. .|.1

.

.

There is still no maximum value (values get arbitrarily close to 1 but donot achieve it).

This does not violate EVT because the domain is notclosed.


Bad Example #2

Example


. .|.1

.

.

There is still no maximum value (values get arbitrarily close to 1 but donot achieve it). This does not violate EVT because the domain is notclosed.


Final Bad Example

Example

Consider the function f(x) =1xis continuous on the closed interval

[1,∞).

. ..1

.

There is no minimum value (values get arbitrarily close to 0 but do notachieve it). This does not violate EVT because the domain is notbounded.


Final Bad Example

Example


[1,∞).

. ..1

.

There is no minimum value (values get arbitrarily close to 0 but do notachieve it).

This does not violate EVT because the domain is notbounded.


Final Bad Example

Example


[1,∞).

. ..1

.

There is no minimum value (values get arbitrarily close to 0 but do notachieve it). This does not violate EVT because the domain is notbounded.


Outline

Introduction




Examples


Local extrema.

.

Definition

I A function f has a local maximum or relative maximum at c if f(c) ≥ f(x)when x is near c. This means that f(c) ≥ f(x) for all x in some open intervalcontaining c.

I Similarly, f has a local minimum at c if f(c) ≤ f(x) when x is near c.

..|.a

.|.b

.

.

.

.local

maximum

.

.local

minimum

.globalmax

.local and global

min


Local extrema.

.

I So a local extremum must be inside the domain of f (not on the end).I A global extremum that is inside the domain is a local extremum.

..|.a

.|.b

.

.

.

.local

maximum

.

.local

minimum

.globalmax

.local and global

min


Fermat's Theorem

Theorem (Fermat’s Theorem)

Suppose f has a local extremum at c and f is differentiable at c. Thenf′(c) = 0.

..|.a

.|.b

.

.

.

.local

maximum

.

.local

minimum


Sketch of proof of Fermat's Theorem

Suppose that f has a local maximum at c.

I If x is slightly greater than c, f(x) ≤ f(c). This means

f(x)− f(c)x− c

≤ 0 =⇒ limx→c+

f(x)− f(c)x− c

≤ 0

I The same will be true on the other end: if x is slightly less than c,f(x) ≤ f(c). This means

f(x)− f(c)x− c

≥ 0 =⇒ limx→c−

f(x)− f(c)x− c

≥ 0

I Since the limit f′(c) = limx→c

f(x)− f(c)x− c

exists, it must be 0.



Suppose that f has a local maximum at c.I If x is slightly greater than c, f(x) ≤ f(c). This means

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





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





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





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



Meet the Mathematician: Pierre de Fermat

I 1601–1665I Lawyer and number

theoristI Proved many theorems,

didn’t quite prove his lastone


Tangent: Fermat's Last Theorem

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

I No solutions tox3 + y3 = z3 amongpositive whole numbers

I Fermat claimed nosolutions to xn + yn = zn

but didn’t write down hisproof

I Not solved until 1998!(Taylor–Wiles)


Outline

Introduction




Examples


Flowchart for placing extremaThanks to Fermat

Suppose f is a continuous function on the closed, bounded interval[a,b], and c is a global maximum point.

..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



This means to find the maximum value of f on [a,b], we need to:I Evaluate f at the endpoints a and bI Evaluate f at the critical points or critical numbers x where

either f′(x) = 0 or f is not differentiable at x.I The points with the largest function value are the global maximum

pointsI The points with the smallest or most negative function value are

the global minimum points.


Outline

Introduction




Examples


Extreme values of a linear function

Example

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

SolutionSince f′(x) = 2, which is never zero, we have no critical points and weneed only investigate the endpoints:

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

SoI The absolute minimum (point) is at −1; the minimum value is −7.I The absolute maximum (point) is at 2; the maximum value is −1.


Extreme values of a quadratic function

Example

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

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

So our points to checkare:

I f(−1) =I f(0) =I f(2) =



Example


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

I f(−1) =I f(0) =I f(2) =



Example



I f(−1) = 0I f(0) =I f(2) =



Example



I f(−1) = 0I f(0) = − 1I f(2) =



Example



I f(−1) = 0I f(0) = − 1I f(2) = 3



Example



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



Example



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


Extreme values of a cubic function

Example

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

SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), we have critical points at x = 0 andx = 1.

The values to check are

I f(−1) =

− 4 (global min)

I f(0) =

1 (local max)

I f(1) =

0 (local min)

I f(2) =

5 (global max)



Example


SolutionSince f′(x) = 6x2 − 6x = 6x(x− 1), we have critical points at x = 0 andx = 1. The values to check are

I f(−1) =

− 4 (global min)

I f(0) =

1 (local max)

I f(1) =

0 (local min)

I f(2) =

5 (global max)



Example



I f(−1) = − 4

(global min)I f(0) =

1 (local max)

I f(1) =

0 (local min)

I f(2) =

5 (global max)



Example



I f(−1) = − 4

(global min)

I f(0) = 1

(local max)I f(1) =

0 (local min)

I f(2) =

5 (global max)



Example



I f(−1) = − 4

(global min)

I f(0) = 1

(local max)

I f(1) = 0

(local min)I f(2) =

5 (global max)



Example



I f(−1) = − 4

(global min)

I f(0) = 1

(local max)

I f(1) = 0

(local min)

I f(2) = 5

(global max)



Example



I f(−1) = − 4 (global min)I f(0) = 1

(local max)

I f(1) = 0

(local min)

I f(2) = 5

(global max)



Example



I f(−1) = − 4 (global min)I f(0) = 1

(local max)

I f(1) = 0

(local min)

I f(2) = 5 (global max)



Example



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

(local min)

I f(2) = 5 (global max)



Example



I f(−1) = − 4 (global min)I f(0) = 1 (local max)I f(1) = 0 (local min)I f(2) = 5 (global max)


Extreme values of an algebraic function

Example

Find the extreme values of 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 is not differentiable at 0.

So our points tocheck are:

I f(−1) =I f(−4/5) =

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



Example



f′(x) =53x2/3 +

43x−1/3 =

13x−1/3(5x+ 4)

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

I f(−1) =

I f(−4/5) =

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



Example



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) =



Example



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) =



Example



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) =



Example



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



Example



f′(x) =53x2/3 +

43x−1/3 =

13x−1/3(5x+ 4)


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



Example



f′(x) =53x2/3 +

43x−1/3 =

13x−1/3(5x+ 4)


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



Example



f′(x) =53x2/3 +

43x−1/3 =

13x−1/3(5x+ 4)


I f(−1) = 1I f(−4/5) = 1.0341 (relative max)I f(0) = 0 (absolute min)I f(2) = 6.3496 (absolute max)


Extreme values of another algebraic function

Example

Find the extreme values of f(x) =√4− x2 on [−2,1].

SolutionWe have f′(x) = − x√

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

differentiable at ±2 as well.)

So our points to check are:

I f(−2) =I f(0) =I f(1) =



Example




differentiable at ±2 as well.) So our points to check are:I f(−2) =

I f(0) =I f(1) =



Example




differentiable at ±2 as well.) So our points to check are:I f(−2) = 0I f(0) =

I f(1) =



Example




differentiable at ±2 as well.) So our points to check are:I f(−2) = 0I f(0) = 2I f(1) =



Example




differentiable at ±2 as well.) So our points to check are:I f(−2) = 0I f(0) = 2I f(1) =

√3



Example




differentiable at ±2 as well.) So our points to check are:I f(−2) = 0 (absolute min)I f(0) = 2I f(1) =

√3



Example




differentiable at ±2 as well.) So our points to check are:I f(−2) = 0 (absolute min)I f(0) = 2 (absolute max)I f(1) =

√3


Summary

I The Extreme Value Theorem: a continuous function on a closedinterval must achieve its max and min

I Fermat’s Theorem: local extrema are critical pointsI The Closed Interval Method: an algorithm for finding global

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


lesson 18: maximum and minimum values (section 041 slides)

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