Lesson18 -maximum_and_minimum_values_slides

Section 4.1
Maximum and Minimum Values

V63.0121.002.2010Su, Calculus I

New York University

June 8, 2010

Announcements

Exams not graded yet
Assignment 4 is on the website
Quiz 3 on Thursday covering 3.3, 3.4, 3.5, 3.7

Announcements

Exams not graded yet
Assignment 4 is on the
website
Quiz 3 on Thursday covering
3.3, 3.4, 3.5, 3.7

V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values June 8, 2010 2 / 32

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
V63.0121.002.2010Su, Calculus I (NYU) Section 4.1 Maximum and Minimum Values (1698–1759) 8, 2010
June 5 / 32

Design

Image credit: Jason Tromm


costs, etc.)
Many laws of science are
derived from minimizing
principles.

June 7 / 32

Optics

Image credit: jacreative


costs, etc.)
Many laws of science are
derived from minimizing
principles.
Maupertuis’ principle:
“Action is minimized
through the wisdom of
God.”
June 9 / 32

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



Deﬁnition
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 f (c)
value

minimum f (d)
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


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 [1, ∞).
x


Final Bad Example

Example
1
x

1


Final Bad Example

Example
1
x

1

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


Final Bad Example

Example
1
x

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 b global
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



Plenty of solutions to
x 2 + y 2 = z 2 among positive
whole numbers (e.g., x = 3,
y = 4, z = 5)



y = 4, z = 5)
No solutions to
whole numbers



y = 4, z = 5)
No solutions to
whole numbers
Fermat claimed no solutions
to x n + y n = z n but didn’t
write down his proof



y = 4, z = 5)
No solutions to
whole numbers
Fermat claimed no solutions
to x n + y n = z n 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 f
Is c an f is not
no diﬀ’ble at no
endpoint? diﬀ at c
c?

yes yes
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) = x 2 − 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) = 2x 3 − 3x 2 + 1 on [−1, 2].



Example

Solution
Since f (x) = 6x 2 − 6x = 6x(x − 1), we have critical points at x = 0 and
x = 1.



Example

Solution
x = 1. The values to check are
f (−1) =
f (0) =
f (1) =
f (2) =



Example

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



Example

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



Example

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



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) = x 2/3 (x + 2) on [−1, 2].


Example

Solution
Write f (x) = x 5/3 + 2x 2/3 , then

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


Example

Solution
Write f (x) = x 5/3 + 2x 2/3 , then

5 4 1
f (x) = x 2/3 + x −1/3 = x −1/3 (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) =
f (−4/5) =
f (0) =
f (2) =


Example

Solution
Write f (x) = x 5/3 + 2x 2/3 , then

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


Example

Solution
Write f (x) = x 5/3 + 2x 2/3 , then

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


Example

Solution
Write f (x) = x 5/3 + 2x 2/3 , then

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


Example

Solution
Write f (x) = x 5/3 + 2x 2/3 , then

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


Example

Solution
Write f (x) = x 5/3 + 2x 2/3 , then

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


Example

Solution
Write f (x) = x 5/3 + 2x 2/3 , then

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


Example

Solution
Write f (x) = x 5/3 + 2x 2/3 , then

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



Example
Find the extreme values of f (x) = 4 − x 2 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) =
f (0) =
f (1) =



Example

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



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!


Lesson18 -maximum_and_minimum_values_slides

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