Lesson 18: Maximum and Minimum Values (Section 041 slides)

Section 4.1
Maximum and Minimum Values
V63.0121.041, Calculus I
New York University
November 8, 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)
V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 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
(1698–1759)V63.0121.041, Calculus I (NYU) Section 4.1 Maximum and Minimum Values November 8, 2010 6 / 34

Design
..Image credit: Jason Tromm

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

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.”

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
.
.

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

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

Bad Example #1
Example
f(x) =
{
x 0 ≤ x < 1
x − 2 1 ≤ x ≤ 2.
. .|
.1
.
.
.
.

Bad Example #1
Example
f(x) =
{
x 0 ≤ x < 1
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
f(x) =
{
x 0 ≤ x < 1
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
Consider the function f(x) =
1
x
is continuous on the closed interval
[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
.|
.b
.
.
.
.
local
maximum
.
.
local
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
.|
.b
.
.
.
.
local
maximum
.
.
global
max
.
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. Then
f′
(c) = 0.
..|
.a
.|
.b
.
.
.
.
local
maximum
.
.
local
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)
x − c
≤ 0

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

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

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

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

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.
..
start
.
Is c an
endpoint?
.
c = a or
c = b
.
c is a
local max
.
Is f diff’ble
at c?
.
f is not
diff at c
.
f′
(c) = 0
.no
.yes
.no
.yes

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
− 1 on [−1, 2].
Solution
We have f′
(x) = 2x, which is zero when x = 0.

Example
− 1 on [−1, 2].
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
− 1 on [−1, 2].
Solution
We have f′
are:
f(−1) = 0
f(0) =
f(2) =

Example
− 1 on [−1, 2].
Solution
We have f′
are:
f(−1) = 0
f(0) = − 1
f(2) =

Example
− 1 on [−1, 2].
Solution
We have f′
are:
f(−1) = 0
f(0) = − 1
f(2) = 3

Example
− 1 on [−1, 2].
Solution
We have f′
are:
f(−1) = 0
f(0) = − 1 (absolute min)
f(2) = 3

Example
− 1 on [−1, 2].
Solution
We have f′
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
− 3x2
+ 1 on [−1, 2].
Solution
Since f′
(x) = 6x2
− 6x = 6x(x − 1), we have critical points at x = 0 and
x = 1.

Example
− 3x2
+ 1 on [−1, 2].
Solution
Since f′
(x) = 6x2
x = 1. The values to check are

Example
− 3x2
+ 1 on [−1, 2].
Solution
Since f′
(x) = 6x2
f(−1) = − 4

Example
− 3x2
+ 1 on [−1, 2].
Solution
Since f′
(x) = 6x2
f(−1) = − 4
f(0) = 1

Example
− 3x2
+ 1 on [−1, 2].
Solution
Since f′
(x) = 6x2
f(−1) = − 4
f(0) = 1
f(1) = 0

Example
− 3x2
+ 1 on [−1, 2].
Solution
Since f′
(x) = 6x2
f(−1) = − 4
f(0) = 1
f(1) = 0
f(2) = 5

Example
− 3x2
+ 1 on [−1, 2].
Solution
Since f′
(x) = 6x2
f(−1) = − 4 (global min)
f(0) = 1
f(1) = 0
f(2) = 5

Example
− 3x2
+ 1 on [−1, 2].
Solution
Since f′
(x) = 6x2
f(0) = 1
f(1) = 0
f(2) = 5 (global max)

Example
− 3x2
+ 1 on [−1, 2].
Solution
Since f′
(x) = 6x2
f(0) = 1 (local max)
f(1) = 0

Example
− 3x2
+ 1 on [−1, 2].
Solution
Since f′
(x) = 6x2
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
(x + 2) on [−1, 2].
Solution
Write f(x) = x5/3
+ 2x2/3
, then
f′
(x) =
5
3
x2/3
+
4
3
x−1/3
=
1
3
x−1/3
(5x + 4)
Thus f′
(−4/5) = 0 and f is not differentiable at 0.

Example
(x + 2) on [−1, 2].
Solution
Write f(x) = x5/3
+ 2x2/3
, then
f′
(x) =
5
3
x2/3
+
4
3
x−1/3
=
1
3
x−1/3
(5x + 4)
Thus f′
(−4/5) = 0 and f is not differentiable at 0. So our points to
check are:
f(−1) =

Example
(x + 2) on [−1, 2].
Solution
Write f(x) = x5/3
+ 2x2/3
, then
f′
(x) =
5
3
x2/3
+
4
3
x−1/3
=
1
3
x−1/3
(5x + 4)
Thus f′
check are:
f(−1) = 1
f(−4/5) =

Example
(x + 2) on [−1, 2].
Solution
Write f(x) = x5/3
+ 2x2/3
, then
f′
(x) =
5
3
x2/3
+
4
3
x−1/3
=
1
3
x−1/3
(5x + 4)
Thus f′
check are:
f(−1) = 1
f(−4/5) = 1.0341
f(0) =

Example
(x + 2) on [−1, 2].
Solution
Write f(x) = x5/3
+ 2x2/3
, then
f′
(x) =
5
3
x2/3
+
4
3
x−1/3
=
1
3
x−1/3
(5x + 4)
Thus f′
check are:
f(−1) = 1
f(−4/5) = 1.0341
f(0) = 0
f(2) =

Example
(x + 2) on [−1, 2].
Solution
Write f(x) = x5/3
+ 2x2/3
, then
f′
(x) =
5
3
x2/3
+
4
3
x−1/3
=
1
3
x−1/3
(5x + 4)
Thus f′
check are:
f(−1) = 1
f(−4/5) = 1.0341
f(0) = 0
f(2) = 6.3496

Example
(x + 2) on [−1, 2].
Solution
Write f(x) = x5/3
+ 2x2/3
, then
f′
(x) =
5
3
x2/3
+
4
3
x−1/3
=
1
3
x−1/3
(5x + 4)
Thus f′
check are:
f(−1) = 1
f(−4/5) = 1.0341
f(0) = 0 (absolute min)
f(2) = 6.3496

Example
(x + 2) on [−1, 2].
Solution
Write f(x) = x5/3
+ 2x2/3
, then
f′
(x) =
5
3
x2/3
+
4
3
x−1/3
=
1
3
x−1/3
(5x + 4)
Thus f′
check are:
f(−1) = 1
f(−4/5) = 1.0341
f(2) = 6.3496 (absolute max)

Example
(x + 2) on [−1, 2].
Solution
Write f(x) = x5/3
+ 2x2/3
, then
f′
(x) =
5
3
x2/3
+
4
3
x−1/3
=
1
3
x−1/3
(5x + 4)
Thus f′
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
√
4 − x2 on [−2, 1].
Solution
We have f′
(x) = −
x
√
4 − x2
, which is zero when x = 0. (f is not
differentiable at ±2 as well.)

Example
√
4 − x2 on [−2, 1].
Solution
We have f′
(x) = −
x
√
4 − x2
differentiable at ±2 as well.) So our points to check are:
f(−2) =

Example
√
4 − x2 on [−2, 1].
Solution
We have f′
(x) = −
x
√
4 − x2
f(−2) = 0
f(0) =

Example
√
4 − x2 on [−2, 1].
Solution
We have f′
(x) = −
x
√
4 − x2
f(−2) = 0
f(0) = 2
f(1) =

Example
√
4 − x2 on [−2, 1].
Solution
We have f′
(x) = −
x
√
4 − x2
f(−2) = 0
f(0) = 2
f(1) =
√
3

Example
√
4 − x2 on [−2, 1].
Solution
We have f′
(x) = −
x
√
4 − x2
f(−2) = 0 (absolute min)
f(0) = 2
f(1) =
√
3

Example
√
4 − x2 on [−2, 1].
Solution
We have f′
(x) = −
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 041 slides)

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