Section 2.4 Using Derivatives to Find Absolute Maximum and ...

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2.4

Copyright © 2016 Pearson Education, Inc.

Section 2.4

2.4

Copyright © 2016 Pearson Education, Inc.

Using Derivatives to Find Absolute

Maximum and Minimum

Values

OBJECTIVE

• Find absolute extrema using Maximum-Minimum Principle 1.

• Find absolute extrema using Maximum-Minimum Principle 2.

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Slide 2- 3Copyright © 2016 Pearson Education, Inc.

DEFINITION:

Suppose that f is a function with domain I.

f (c) is an absolute minimum if f (c) ≤ f (x) for all x in I.

f (c) is an absolute maximum if f (c) ≥ f (x) for all x in I.

2.4 Using Derivatives to Find Absolute Maximum

and Minimum Values


THEOREM 7: The Extreme Value Theorem

A continuous function f defined over a closed interval

[a, b] must have an absolute maximum value and an

absolute minimum value over [a, b].


and Minimum Values

3


THEOREM 8: Maximum-Minimum Principle 1

Suppose that f is a continuous function defined over a

closed interval [a, b]. To find the absolute maximum

and minimum values over [a, b]:

a) First find f ′(x).

b) Then determine all critical values in [a, b]. That is,

find all c in [a, b] for which

f ′(c) = 0 or f ′(c) does not exist.


and Minimum Values



(continued)

c) List the values from step (b) and the endpoints of the

interval:

a, c1, c2, …., cn, b.

d) Evaluate f (x) for each value in step (c):

f (a), f (c1), f (c2), …., f (cn), f (b).

The largest of these is the absolute maximum of f

over [a, b]. The smallest of these is the absolute

minimum of f over [a, b].


and Minimum Values

4


Example 1: Find the absolute maximum and

minimum values of over the

interval [–2, ].

a)

b) Note that f ′(x) exists for all real numbers.

( ) 3 3 2= − +f x x x

( ) 23 3′ = −f x x

3

2


and Minimum Values

23 3 0x − =

23 3x =

21x =

1x = ±


Example 1 (concluded):

c) –2, –1, 1,

d)

2

3


and Minimum Values

( 2)f −

( 1)f −

(1)f

3

2

f

=3

( 2) 3( 2) 2− − − +3

( 1) 3( 1) 2− − − +3

(1) 3(1) 2− +

33 3

3 22 2

− +

0

4

7

8

=

=

=

=

=

=

=

0

5



Thus, the absolute maximum value of f (x) on [–2, ]

is 4, which occurs at x = 1. The absolute minimum

value of f (x) on [–2, ] is 0, which occurs at x = 2

and x = 1.

2

3

2

3


and Minimum Values



and Minimum Values

Quick Check 1

Find the absolute maximum and minimum values of the function

given in Example 1 over the interval

Recall that the function in Example 1 was

So, . Set this equal to zero and you get:

[ ]0,3 .

3( ) 3 2.= − +f x x x

2( ) 3 3f x x′ = −

23 3 0x − =2

3 3x =2

1x =1x = ±

6



and Minimum Values

Quick Check 1 Concluded

Next we check the equation at to find the

minimum and maximum:

The absolute maximum is at

The absolute minimum is at

1,0,1,3x = −

3( 1) ( 1) 3( 1) 2f − = − − − + 1 3 2= − + + 4=

3(0) (0) 3(0) 2f = − + 0 0 2= + + 2=

3(1) (1) 3(1) 2f = − + 1 3 2= − + 0=

3(3) (3) 3(3) 2f = − + 27 9 2= − + 20=

20 3.x =

0 1.x =



Suppose that f is a function such that f ′(x) exists for

every x in an interval I, and that there is exactly one

(critical) value c in I, for which f ′(c) = 0. Then

f (c) is the absolute maximum value over I if f ′′(c) < 0

or

f (c) is the absolute minimum value over I if f ′′(c) > 0.


and Minimum Values

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minimum values of

When no interval is specified, we consider the entire

domain of the function. In this case, the domain is the

set of all real numbers.

a) ( ) 4 2′ = −f x x

( ) 24 .= −f x x x


and Minimum Values


Example 2 (continued):

b)

c) Since there is only one critical value, we can apply

Maximum-Minimum Principle 2 using the second

derivative.

( ) 2′′ = −f x


and Minimum Values

4 2 0x− =

2 4x− = −

2x =

8



Since f ′′(x) < 0 and f(x) has an

absolute maximum value of 4 at x = 2.

f(x) has no absolute minimum value.

,4)2()2(4)2( 2 =−=f


and Minimum Values



and Minimum Values

Quick Check 2

Find the absolute maximum and minimum values of

over each interval:

a.)

b.)

Looking at the function, we can see that the absolute minimum over

is located at Since is in both intervals, the

absolute minimum for both intervals will be located at

2( ) 10f x x x= −

[ ]0,6

[4,10]

( , )−∞ ∞ 25− 5.x = 5x =

25− 5.x =

9



and Minimum Values


Now we need to check both endpoints for both intervals to find the

absolute maximum:

a.)

So the absolute maximum over is located at

b.)

So the absolute maximum over is 0 located at

2(0) 0 10(0)f = −

2(6) 6 10(6)f = −

[0,6]

24= −36 60= −

0 0= − 0=

0 0.x =

2(4) 4 10(4)f = − 16 40= − 24= −

2(10) 10 10(10)f = − 100 100= − 0=

[4,10] 10.x =


A Strategy for Finding Absolute Maximum and

Minimum Values:

To find absolute maximum and minimum values of a

continuous function over an interval:

a) Find f ′(x).

b) Find the critical values.

c) If the interval is closed and there is more than one

critical value, use Maximum-Minimum Principle 1.


and Minimum Values

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Minimum Values (continued):

d) If the interval is closed and there is exactly one

critical value, use either Maximum-Minimum

Principle 1 or Maximum-Minimum Principle 2. If it

is easy to find f ′′(x), use Maximum-Minimum

Principle 2.


and Minimum Values



Minimum Values (concluded):

e) If the interval is not closed, such as (–∞, ∞), (0, ∞),

or (a, b), and the function has only one critical value,

use Maximum-Minimum Principle 2. In such a case,

if the function has a maximum, it will have no

minimum; and if it has a minimum, it was have no

maximum.


and Minimum Values

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minimum values of over the interval

(0, ∞).

a)

b) Since f ′(x) exists for all values of x in (0, ∞), the

only critical values are those for which f ′(x) = 0.

( ) 2

355′ = −f x

x

( )35

5= +f x xx


and Minimum Values



c) The interval (0, ∞) is not closed, and the only

critical value is Thus, we can use Maximum-

Minimum Principle 2 using the second derivative.

.7


and Minimum Values

2

355 0

x− =

2

355

x=

25 35x =

27x =

7x =

12



Thus f (x) has an absolute minimum at 7.=x


and Minimum Values

( ) 3

70f x

x′′ =

( )( )

3

707 0

7

f ′′ = >

( ) 357 5 7

7f = ⋅ +

( ) 35 77 5 7

7f = +

( )7 5 7 5 7 10 7f = + =



Thus, the absolute minimum of f (x) is which

occurs at

f (x) has no absolute maximum value.

7.=x

10 7,


and Minimum Values

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and Minimum Values

Quick Check 3

Find the absolute maximum and minimum values of

over the interval .

First we want to find :

22 18( )

xg x

x

+=

(0, )∞

( )g x′

2

2

(4 ) (2 18)( )

x x xg x

x

− +′ =

2 2

2

4 2 18x x

x

− −=

2

2

2 18x

x

−=



and Minimum Values

Quick Check 3 Continued

Next we find out where

Since is not in the interval, the only critical value is at

( ) 0.g x′ =

2 2

2 2 2 2

2 18 2 18 182 0

x x

x x x x

−= − = − =

2

182

x=

22 18x =

2 9x =

3x = ±

3x = − 3.x =

14



and Minimum Values


We can now apply the Maximum-Minimum Principle 2:

So an absolute minimum occurs at , and is 12.

Since this is on an open interval, there is no absolute maximum.

3

36( )f x

x′′ =

3

36(3)

3f ′′ = 0>

3x =



and Minimum Values

Section Summary

• An absolute minimum of a function is a value such that

for all in the domain of .

• An absolute maximum of a function is a value such that

for all in the domain of .

• If the domain of is a closed interval and is continuous over

that domain, then the Extreme-Value Theorem guarantees the

existence of both an absolute minimum and an absolute maximum.

f ( )f c

( ) ( )f c f x≤ x f

f ( )f c

( ) ( )f c f x≥ x f

f f

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and Minimum Values

Section Summary Concluded

• Endpoints of a closed interval may be absolute extrema, but not

relative extrema.

• If there is exactly one critical value such that in the

domain of , then Maximum-Minimum Principle 2 may be used.

Otherwise, Maximum-Minimum Principle 1 has to be used.

c ( ) 0f c′ =

f

Section 2.4 Using Derivatives to Find Absolute Maximum and ...

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