Lesson 18: Maximum and Minimum Values (Section 021 slides)

Section 4.1
Maximum and Minimum Values

V63.0121.021, Calculus I

New York University

November 9, 2010

Announcements

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

. . . . . .

Announcements

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

. . . . . .

V63.0121.021, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 9, 2010 2 / 34

Objectives

Understand and be able to
explain the statement of
the Extreme Value
Theorem.
Understand and be able to
explain the statement of
Fermat’s Theorem.
Use the Closed Interval
Method to find the extreme
values of a function defined
on 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

. . . . . .


Why go to the extremes?

Rationally speaking, it is
advantageous to find the
extreme values of a
function (maximize profit,
minimize costs, etc.)

Pierre-Louis Maupertuis
V63.0121.021, Calculus I (NYU) Section 4.1 Maximum and Minimum Values
(1698–1759) 9, 2010
November 6 / 34

Design

.
Image credit: Jason Tromm .


extreme values of a
Many laws of science are
derived from minimizing
principles.

(1698–1759) 9, 2010
November 8 / 34

Optics

.
Image credit: jacreative .


extreme values of a
Many laws of science are
derived from minimizing
principles.
Maupertuis’ principle:
“Action is minimized
through the wisdom of
God.”
(1698–1759) 9, 2010
November 10 / 34

Outline

Introduction




Examples


Extreme points and values

Definition
Let f have domain D.

.

.
Image credit: Patrick Q


Definition
The function f has an absolute
maximum (or global maximum)
(respectively, absolute minimum) at c if
f(c) ≥ f(x) (respectively, f(c) ≤ f(x)) for all
x in D

.

.


Definition
x in D
The number f(c) is called the maximum
value (respectively, minimum value) of f
on D.
.

.


Definition
x in D
The number f(c) is called the maximum
value (respectively, minimum value) of f
on D.
.
An extremum is either a maximum or a
minimum. An extreme value is either a
maximum value or minimum value.

.

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].



.

.

. .
a
. b
.


.
maximum .(c)
f
.
value

. .
minimum .(d)
f
.
value
. . ..
a
. d c
b
.
minimum maximum

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

Example

Consider the function
{
x 0≤x<1
f(x) =
x − 2 1 ≤ x ≤ 2.


Bad Example #1

Example

Consider the function .
{
x 0≤x<1
f(x) = . .
| .
x − 2 1 ≤ x ≤ 2. 1
.
.


Bad Example #1

Example

{
x 0≤x<1
f(x) = . .
| .
x − 2 1 ≤ x ≤ 2. 1
.
.
Then although values of f(x) get arbitrarily close to 1 and never bigger
than 1, 1 is not the maximum value of f on [0, 1] because it is never
achieved.


Bad Example #1

Example

{
x 0≤x<1
f(x) = . .
| .
x − 2 1 ≤ x ≤ 2. 1
.
.
Then although values of f(x) get arbitrarily close to 1 and never bigger
than 1, 1 is not the maximum value of f on [0, 1] because it is never
achieved. 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).


Bad Example #2

Example

.

. .
|
1
.


Bad Example #2

Example

.

. .
|
1
.

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


Bad Example #2

Example

.

. .
|
1
.

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


Final Bad Example

Example
1
Consider the function f(x) = is continuous on the closed interval
x
[1, ∞).


Final Bad Example

Example
1
x
[1, ∞).

.

. .
1
.


Final Bad Example

Example
1
x
[1, ∞).

.

. .
1
.

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


Final Bad Example

Example
1
x
[1, ∞).

.

. .
1
.

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


Outline

Introduction




Examples


Local extrema
.

Definition

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

.

Local extrema
.

Definition

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

.

.
.
.
.|.
. . .
|
a
. local local b
.
maximum minimum
.

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

.

.
.
.
.|.
. . ..
|
a
. local local and global . global
b
maximum min max
.

Fermat's Theorem

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

.

.
.
.
...
| . .
|
a
. local local b
.
maximum minimum


Sketch of proof of Fermat's Theorem

Suppose that f has a local maximum at c.



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

f(x) − f(c)
≤0
x−c




f(x) − f(c) f(x) − f(c)
≤ 0 =⇒ lim+ ≤0
x−c x→c x−c




f(x) − f(c) f(x) − f(c)
≤ 0 =⇒ lim+ ≤0
x−c x→c x−c

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)
≥0
x−c




f(x) − f(c) f(x) − f(c)
≤ 0 =⇒ lim+ ≤0
x−c x→c x−c


f(x) − f(c) f(x) − f(c)
≥ 0 =⇒ lim ≥0
x−c x→c − x−c




f(x) − f(c) f(x) − f(c)
≤ 0 =⇒ lim+ ≤0
x−c x→c x−c


f(x) − f(c) f(x) − f(c)
≥ 0 =⇒ lim ≥0
x−c x→c − x−c

f(x) − f(c)
Since the limit f′ (c) = lim exists, it must be 0.
x→c x−c


Meet the Mathematician: Pierre de Fermat

1601–1665
Lawyer and number
theorist
Proved many theorems,
didn’t quite prove his last
one


Tangent: Fermat's Last Theorem

Plenty of solutions to
x2 + y2 = z2 among
positive whole numbers
(e.g., x = 3, y = 4, z = 5)



x2 + y2 = z2 among
(e.g., x = 3, y = 4, z = 5)
No solutions to
x3 + y3 = z3 among



x2 + y2 = z2 among
(e.g., x = 3, y = 4, z = 5)
No solutions to
x3 + y3 = z3 among
Fermat claimed no
solutions to xn + yn = zn
but didn’t write down his
proof



x2 + y2 = z2 among
(e.g., x = 3, y = 4, z = 5)
No solutions to
x3 + y3 = z3 among
Fermat claimed no
solutions to xn + yn = zn
but didn’t write down his
proof
Not solved until 1998!
(Taylor–Wiles)


Outline

Introduction




Examples


Flowchart for placing extrema
Thanks to Fermat

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

. . .
Is c an Is f diff’ble f is not
n
.o n
.o
endpoint? at c? diff at c

y
. es y
. es
. .
c = a or
f′ (c) = 0
c = b



This means to find the maximum value of f on [a, b], we need to:
Evaluate f at the endpoints a and b
Evaluate f at the critical points or critical numbers x where
either f′ (x) = 0 or f is not differentiable at x.
The points with the largest function value are the global maximum
points
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].



Example

Solution
Since f′ (x) = 2, which is never zero, we have no critical points and we
need only investigate the endpoints:
f(−1) = 2(−1) − 5 = −7
f(2) = 2(2) − 5 = −1



Example

Solution
Since f′ (x) = 2, which is never zero, we have no critical points and we
need only investigate the endpoints:
f(−1) = 2(−1) − 5 = −7
f(2) = 2(2) − 5 = −1
So
The absolute minimum (point) is at −1; the minimum value is −7.
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].



Example

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



Example

Solution
We have f′ (x) = 2x, which is zero when x = 0. So our points to check
are:
f(−1) =
f(0) =
f(2) =



Example

Solution
are:
f(−1) = 0
f(0) =
f(2) =



Example

Solution
are:
f(−1) = 0
f(0) = − 1
f(2) =



Example

Solution
are:
f(−1) = 0
f(0) = − 1
f(2) = 3



Example

Solution
are:
f(−1) = 0
f(0) = − 1 (absolute min)
f(2) = 3



Example

Solution
are:
f(−1) = 0
f(0) = − 1 (absolute min)
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].



Example

Solution
Since f′ (x) = 6x2 − 6x = 6x(x − 1), we have critical points at x = 0 and
x = 1.



Example

Solution
x = 1. The values to check are



Example

Solution
f(−1) = − 4



Example

Solution
f(−1) = − 4
f(0) = 1



Example

Solution
f(−1) = − 4
f(0) = 1
f(1) = 0



Example

Solution
f(−1) = − 4
f(0) = 1
f(1) = 0
f(2) = 5



Example

Solution
f(−1) = − 4 (global min)
f(0) = 1
f(1) = 0
f(2) = 5



Example

Solution
f(0) = 1
f(1) = 0
f(2) = 5 (global max)



Example

Solution
f(0) = 1 (local max)
f(1) = 0



Example

Solution
f(0) = 1 (local max)
f(1) = 0 (local min)


Extreme values of an algebraic function
Example
Find the extreme values of f(x) = x2/3 (x + 2) on [−1, 2].


Example

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

5 2/3 4 −1/3 1 −1/3
f′ (x) = x + x = x (5x + 4)
3 3 3
Thus f′ (−4/5) = 0 and f is not differentiable at 0.


Example

Solution

5 2/3 4 −1/3 1 −1/3
f′ (x) = x + x = x (5x + 4)
3 3 3
Thus f′ (−4/5) = 0 and f is not differentiable at 0. So our points to
check are:
f(−1) =


Example

Solution

5 2/3 4 −1/3 1 −1/3
f′ (x) = x + x = x (5x + 4)
3 3 3
check are:
f(−1) = 1
f(−4/5) =


Example

Solution

5 2/3 4 −1/3 1 −1/3
f′ (x) = x + x = x (5x + 4)
3 3 3
check are:
f(−1) = 1
f(−4/5) = 1.0341
f(0) =


Example

Solution

5 2/3 4 −1/3 1 −1/3
f′ (x) = x + x = x (5x + 4)
3 3 3
check are:
f(−1) = 1
f(−4/5) = 1.0341
f(0) = 0
f(2) =

Example

Solution

5 2/3 4 −1/3 1 −1/3
f′ (x) = x + x = x (5x + 4)
3 3 3
check are:
f(−1) = 1
f(−4/5) = 1.0341
f(0) = 0
f(2) = 6.3496

Example

Solution

5 2/3 4 −1/3 1 −1/3
f′ (x) = x + x = x (5x + 4)
3 3 3
check are:
f(−1) = 1
f(−4/5) = 1.0341
f(0) = 0 (absolute min)
f(2) = 6.3496

Example

Solution

5 2/3 4 −1/3 1 −1/3
f′ (x) = x + x = x (5x + 4)
3 3 3
check are:
f(−1) = 1
f(−4/5) = 1.0341
f(2) = 6.3496 (absolute max)

Example

Solution

5 2/3 4 −1/3 1 −1/3
f′ (x) = x + x = x (5x + 4)
3 3 3
check are:
f(−1) = 1
f(−4/5) = 1.0341 (relative max)
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].



Example
√

Solution
x
We have f′ (x) = − √ , which is zero when x = 0. (f is not
4 − x2
differentiable at ±2 as well.)



Example
√

Solution
x
4 − x2
differentiable at ±2 as well.) So our points to check are:
f(−2) =



Example
√

Solution
x
4 − x2
f(−2) = 0
f(0) =



Example
√

Solution
x
4 − x2
f(−2) = 0
f(0) = 2
f(1) =



Example
√

Solution
x
4 − x2
f(−2) = 0
f(0) = 2
√
f(1) = 3



Example
√

Solution
x
4 − x2
f(−2) = 0 (absolute min)
f(0) = 2
√
f(1) = 3



Example
√

Solution
x
4 − x2
f(−2) = 0 (absolute min)
f(0) = 2 (absolute max)
√
f(1) = 3


Summary

The Extreme Value Theorem: a continuous function on a closed
interval must achieve its max and min
Fermat’s Theorem: local extrema are critical points
The Closed Interval Method: an algorithm for finding global
extrema
Show your work unless you want to end up like Fermat!


Lesson 18: Maximum and Minimum Values (Section 021 slides)

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