Lesson 18: Maximum and Minimum Vaues

. . . . . .

Section 4.1Maximum and Minimum Values

V63.0121, Calculus I

March 24, 2009

Announcements

I Homework due ThursdayI Quiz April 2, on Sections 2.5–3.5I Final Exam Friday, May 8, 2:00–3:50pm

..Image: Flickr user Karen with a K

. . . . . .

Outline

Introduction

The Extreme Value Theorem

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

The Closed Interval Method

Examples

Challenge: Cubic functions

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)

. . . . . .

Design

..Image credit: Jason Tromm

. . . . . .

Optics

.Image credit: jacreative

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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 if f(c) ≥ f(x) (respectively,f(c) ≤ f(x)) for all x in D

I The number f(c) is called themaximum value (respectively,minimum value) of f on D.

I An extremum is either a maximumor a minimum. An extreme value iseither a maximum value or minimumvalue.

.Image credit: Patrick Q

. . . . . .

Theorem (The Extreme Value Theorem)Let f be a function which is continuous on the closed interval [a, b]. Then fattains an absolute maximum value f(c) and an absolute minimum valuef(d) at numbers c and d in [a, b].

. . . . . .

.cmaximum

.maximum

minimum

.minimum

. . . . . .

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

ExampleConsider the function

f(x) =

{x 0 ≤ x < 1

x − 2 1 ≤ x ≤ 2.

. .|.1

Then although values of f(x) get arbitrarily close to 1 and neverbigger than 1, 1 is not the maximum value of f on [0, 1] because it isnever achieved.

. . . . . .

Bad Example #1

f(x) =

{x 0 ≤ x < 1

x − 2 1 ≤ x ≤ 2.

. .|.1

. . . . . .

Bad Example #1

f(x) =

{x 0 ≤ x < 1

x − 2 1 ≤ x ≤ 2.

. .|.1

. . . . . .

Bad Example #2

ExampleThe function f(x) = x restricted to the interval [0, 1) still has nomaximum value.

. .|.1

. . . . . .

Bad Example #2

ExampleThe function f(x) = x restricted to the interval [0, 1) still has nomaximum value.

. .|.1

. . . . . .

Final Bad Example

Example

The function f(x) =1x

is continuous on the closed interval [1,∞) but

has no minimum value.

. . . . . .

Final Bad Example

Example

The function f(x) =1x

is continuous on the closed interval [1,∞) but

has no minimum value.

. . . . . .

Outline

Introduction

Examples

. . . . . .

Local extremaDefinition

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

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

.local

maximum

.local

minimum

. . . . . .

Local extremaDefinition

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

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

.local

maximum

.local

minimum

. . . . . .

I So a local extremum must be inside the domain of f (not on theend).

I A global extremum that is inside the domain is a local extremum.

.globalmax

.localmax

.local and global

. . . . . .

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

.local

maximum

.local

minimum

. . . . . .

Sketch of proof of Fermat’s Theorem

Suppose that f has a local maximum at c.

I If h is close enough to 0 but greater than 0, f(c + h) ≤ f(c). Thismeans

f(c + h) − f(c)h

≤ 0 =⇒ limh→0+

f(c + h) − f(c)h

I The same will be true on the other end: if h is close enough to 0but less than 0, f(c + h) ≤ f(c). This means

f(c + h) − f(c)h

≥ 0 =⇒ limh→0−

f(c + h) − f(c)h

I Since the limit f′(c) = limh→0

f(c + h) − f(c)h

exists, it must be 0.

. . . . . .

Suppose that f has a local maximum at c.I If h is close enough to 0 but greater than 0, f(c + h) ≤ f(c). This

f(c + h) − f(c)h

=⇒ limh→0+

f(c + h) − f(c)h

≥ 0 =⇒ limh→0−

f(c + h) − f(c)h

. . . . . .

f(c + h) − f(c)h

≤ 0 =⇒ limh→0+

f(c + h) − f(c)h

≥ 0 =⇒ limh→0−

f(c + h) − f(c)h

. . . . . .

f(c + h) − f(c)h

≤ 0 =⇒ limh→0+

f(c + h) − f(c)h

=⇒ limh→0−

f(c + h) − f(c)h

. . . . . .

f(c + h) − f(c)h

≤ 0 =⇒ limh→0+

f(c + h) − f(c)h

≥ 0 =⇒ limh→0−

f(c + h) − f(c)h

. . . . . .

f(c + h) − f(c)h

≤ 0 =⇒ limh→0+

f(c + h) − f(c)h

≥ 0 =⇒ limh→0−

f(c + h) − f(c)h

. . . . . .

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

. . . . . .

Let’s put this together logically. Let f be a continuous functiondefined on a closed interval [a, b]. We are in search of its globalmaximum, call it c. Then:

I Either the maximumoccurs at an endpoint ofthe interval, i.e., c = a orc = b,

I Or the maximum occursinside (a, b). In this case, cis also a local maximum.

I Either f is differentiableat c, in which casef′(c) = 0 by Fermat’sTheorem.

I Or f is notdifferentiable at c.

This means to find themaximum value of f on [a, b],we need to check:

I a and bI Points x where f′(x) = 0I Points x where f is not

differentiable.

The latter two are both calledcritical points of f. Thistechnique is called the ClosedInterval Method.

. . . . . .

differentiable.

. . . . . .

differentiable.

. . . . . .

differentiable.

. . . . . .

differentiable.

. . . . . .

differentiable.

. . . . . .

I a and b

I Points x where f′(x) = 0I Points x where f is not

differentiable.

. . . . . .

I a and bI Points x where f′(x) = 0

I Points x where f is notdifferentiable.

. . . . . .

differentiable.

. . . . . .

differentiable.

. . . . . .

Outline

Introduction

Examples

. . . . . .

ExampleFind 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.

. . . . . .

ExampleFind 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.

. . . . . .

ExampleFind 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) =

. . . . . .

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

I f(−1) =

I f(0) =

I f(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) =

. . . . . .

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 (absolute min)I f(2) = 3

. . . . . .

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

. . . . . .

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

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

f′(x) =53

x2/3 +43

x−1/3 =13

x−1/3(5x + 4)

Thus f′(−4/5) = 0 and f is not differentiable at 0.

I f(−1) =

I f(−4/5) =

I f(0) =

I f(2) =

. . . . . .

f′(x) =53

x2/3 +43

x−1/3 =13

x−1/3(5x + 4)

Thus f′(−4/5) = 0 and f is not differentiable at 0.

I f(−1) =

I f(−4/5) =

I f(0) =

I f(2) =

. . . . . .

f′(x) =53

x2/3 +43

x−1/3 =13

x−1/3(5x + 4)

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

I f(−1) =

I f(−4/5) =

I f(0) =

I f(2) =

. . . . . .

f′(x) =53

x2/3 +43

x−1/3 =13

x−1/3(5x + 4)

I f(−1) = 1I f(−4/5) =

I f(0) =

I f(2) =

. . . . . .

f′(x) =53

x2/3 +43

x−1/3 =13

x−1/3(5x + 4)

I f(−1) = 1I f(−4/5) = 1.0341I f(0) =

I f(2) =

. . . . . .

f′(x) =53

x2/3 +43

x−1/3 =13

x−1/3(5x + 4)

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

. . . . . .

f′(x) =53

x2/3 +43

x−1/3 =13

x−1/3(5x + 4)

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

. . . . . .

f′(x) =53

x2/3 +43

x−1/3 =13

x−1/3(5x + 4)

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

. . . . . .

f′(x) =53

x2/3 +43

x−1/3 =13

x−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)

. . . . . .

f′(x) =53

x2/3 +43

x−1/3 =13

x−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)

. . . . . .

ExampleFind 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) =

. . . . . .

√4 − x2 on [−2, 1].

differentiable at ±2 as well.)

So our points to check are:I f(−2) =

I f(0) =

I f(1) =

. . . . . .

√4 − x2 on [−2, 1].

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

I f(0) =

I f(1) =

. . . . . .

√4 − x2 on [−2, 1].

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

I f(1) =

. . . . . .

√4 − x2 on [−2, 1].

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

. . . . . .

√4 − x2 on [−2, 1].

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

. . . . . .

√4 − x2 on [−2, 1].

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

. . . . . .

√4 − x2 on [−2, 1].

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) =

. . . . . .

Outline

Introduction

Examples

. . . . . .

ExampleHow many critical points can a cubic function

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

. . . . . .

SolutionIf f′(x) = 0, we have

3ax2 + 2bx + c = 0,

and so

x =−2b ±

√4b2 − 12ac6a

=−b ±

√b2 − 3ac

and so we have three possibilities:I 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.I b2 − 3ac < 0, in which case there are no real roots to the quadratic,

hence no critical points. An example would be f(x) = x3 + x2 + x,where a = b = c = 1.

I b2 − 3ac = 0, in which case there is a single critical point. Example:x3, where a = 1 and b = c = 0.

. . . . . .

Review

I Concept: absolute (global) and relative (local) maxima/minimaI Fact: Fermat’s theorem: f′(x) = 0 at local extremaI Technique: the Closed Interval Method

Lesson 18: Maximum and Minimum Vaues

minimum value of f

absolute maximum value

function f

themaximum value

absolute minimum value

extreme value iseither

maximum minimum

absolute minimumat c

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