Lesson 16: Maximum and Minimum Values

Section 4.1
Maximum and Minimum Values

V63.0121.006/016, Calculus I

March 23, 2010

Announcements
Welcome back from Spring Break!
Quiz 3: April 2, Sections 2.6–3.5

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Announcements

Welcome back from Spring Break!
Quiz 3: April 2, Sections 2.6–3.5

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Computation of Midterm Letter Grades
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Distribution of Midterm Averages and Letter Grades

Median = 81.35% (curved to B)
Average = 74.39% (curved to B-)
Standard Deviation = 21%

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What can I do to improve my grade?
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59% of your grade is still in play!

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Help!

Free resources:
recitation
TAs’ ofﬁce hours
my ofﬁce hours
Math Tutoring Center
(CIWW 524)
College Learning Center

. . . . . .

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?

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

Pierre-Louis Maupertuis
(1698–1759)
. . . . . .

Design

.
Image credit: Jason Tromm
.
. . . . . .


extreme values of a
function (maximize
etc.)
Many laws of science
are derived from
minimizing principles.

Pierre-Louis Maupertuis
(1698–1759)
. . . . . .

Optics

.
.
Image credit: jacreative
. . . . . .


extreme values of a
function (maximize
etc.)
Many laws of science
are derived from
minimizing principles.
Maupertuis’ principle:
“Action is minimized
through the wisdom of
God.” Pierre-Louis Maupertuis
(1698–1759)
. . . . . .

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

.

.
. . . . . .


Deﬁnition
for all 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
x
interval [1, ∞).

. . . . . .

Final Bad Example

Example
1
x
interval [1, ∞).

.

. .
1
.

. . . . . .

Final Bad Example

Example
1
x
interval [1, ∞).

.

. .
1
.

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

. . . . . .

Final Bad Example

Example
1
x
interval [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.

. . . . . .

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
. . . . . .

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

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


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


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
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)

. . . . . .

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

. . . . . .

The Closed Interval Method

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.

. . . . . .

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.

. . . . . .

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
check are:
f(−1) = 0
f(0) =
f(2) =

. . . . . .

Example

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

. . . . . .

Example

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

. . . . . .

Example

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

. . . . . .

Example

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

. . . . . .

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 = 0 and 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)

. . . . . .

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) =
f(−4/5) =
f(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) =
f(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) =
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) =

. . . . . .

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)

. . . . . .

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) =
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

. . . . . .

Challenge: Cubic functions

Example
How many critical points can a cubic function

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

have?

. . . . . .

Solution
If f′ (x) = 0, we have

3ax2 + 2bx + c = 0,

and so
√ √
−2b ± 4b2 − 12ac −b ± b2 − 3ac
x= = ,
6a 3a
and so we have three possibilities:

. . . . . .

Solution

3ax2 + 2bx + c = 0,

and so
√ √
−2b ± 4b2 − 12ac −b ± b2 − 3ac
x= = ,
6a 3a
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.

. . . . . .

Solution

3ax2 + 2bx + c = 0,

and so
√ √
−2b ± 4b2 − 12ac −b ± b2 − 3ac
x= = ,
6a 3a
b = 1, and c = 0.
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.

. . . . . .

Solution

3ax2 + 2bx + c = 0,

and so
√ √
−2b ± 4b2 − 12ac −b ± b2 − 3ac
x= = ,
6a 3a
b = 1, and c = 0.
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.
b2 − 3ac = 0, in which case there is a single critical point.
Example: x3 , where a = 1 and b = c = 0.
. . . . . .

What have we learned today?

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 16: Maximum and Minimum Values

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Lesson 16: Maximum and Minimum Values