Chapter 3

Applications of Differentiation

Definition of Extrema

Figure 3.1

Theorem 3.1 The Extreme Value Theorem

Extreme Value Theorem

4

2

-2

-4

-5 5

If f is continuous on [a,b] then f must have both an absolute maximum and absolute minimum value in the interval.

a b

4

2

-2

-4

-5 5

a b

4

2

-2

-4

-5 5

a b

Examples:

Extreme Value Theorem

4

2

-2

-4

-5 5

If f is continuous on [a,b] then f must have both an absolute maximum and absolute minimum value in the interval.

a b

Absolute Maximum

4

2

-2

-4

-5 5

a b

4

2

-2

-4

-5 5

a b

Examples:

Absolute Minimum

Abs Max

Abs Min

Abs Max

Abs Min

Extreme values can be in the interior or the end points of a function.

2y x

,D Absolute Minimum

No AbsoluteMaximum

2y x

0,2D No Minimum

NoMaximum

Definition of Relative Extrema

Local Extreme Values:

A local maximum is the maximum value within some open interval.

A local minimum is the minimum value within some open interval.

Figure 3.2

Absolute minimum(also local minimum)

Local maximum

Local minimum

Absolute maximum

(also local maximum)

Local minimum

Local extremes are also called relative extremes.

Absolute minimum(also local minimum)

Local maximum

Local minimum

Absolute maximum

(also local maximum)

Local minimum

Local extremes are also called relative extremes.

Local maximum

Local minimum

Notice that local extremes in the interior of the function

occur where is zero or is undefined.f f

Absolute maximum

(also local maximum)

Definition of a Critical Number and Figure 3.4

Theorem 3.2 Relative Extrema Occur Only at Critical Numbers

Note:Maximum and minimum points in the interior of a function always occur at critical points, but critical points are not always maximum or minimum values.

Guidelines for Finding Extrema on a Closed Interval

EXAMPLE 3 FINDING ABSOLUTE EXTREMA

Find the absolute maximum and minimum values of on the interval . 2/3f x x 2,3

2/3f x x

1

32

3f x x

3

2

3f x

x

There are no values of x that will makethe first derivative equal to zero.

The first derivative is undefined at x=0,so (0,0) is a critical point.

Because the function is defined over aclosed interval, we also must check theendpoints.

0 0f To determine if this critical point isactually a maximum or minimum, wetry points on either side, withoutpassing other critical points.

2/3f x x

1 1f 1 1f

Since 0<1, this must be at least a local minimum, and possibly a global minimum.

2,3D

At: 0x

At: 2x 2

32 2 1.5874f

At: 3x 2

33 3 2.08008f

0 0f To determine if this critical point isactually a maximum or minimum, wetry points on either side, withoutpassing other critical points.

2/3f x x

1 1f 1 1f

Since 0<1, this must be at least a local minimum, and possibly a global minimum.

2,3D

At: 0x

At: 2x 2

32 2 1.5874f

At: 3x

Absoluteminimum:

Absolutemaximum:

0,0

3,2.08

2

33 3 2.08008f

Critical points are not always extremes!

3y x

0f (not an extreme)

1/3y x

is undefined.f

(not an extreme)

Locate the extrema of the function on [-5, 10]

3 25( ) 24

3 2

x xf x x

1. Find f’(x) = 0

2'( ) 5 24

0 8 3

. . 8, 3

f x x x

x x

C P at x

2. Test the critical points and the endpoints in the original function.

( 5) 15.833

( 3) 40.5

(8) 181.3

(10) 156.7

f

f

f

f

3. Identify the highest and lowest points.

Absolute max at (3, 40.5)

Absolute min at (8, -181.3)

max

min

3 2( ) 5 9 45f x x x x Find the extrema on [-5, -2]

1. Find f’(x) = 02

2

'( ) 3 10 9

0 3 10 9

.407 .737

f x x x

x x

x and

2. Test the critical points and the endpoints in the original function.

f(-5) = 0

f(-4.07) = 7.035

f(.737) = -48.52f(.737) = -48.52

f(-2) = -15

We don’t use x = .737 since it’s not in the interval.

3. Identify the highest and lowest points.

Absolute max (-4.07. 7.035)

Absolute min (-2, -15)

max

min

• Find the extrema of ( ) 2sin cos 2 on the interval [0,2 ]f x x x x

'( ) 2cos 2sin 2 0 2cos 4cos sin 0 2cos (1 2sin ) 0

1 3 7 11cos 0 or sin , , , ,

2 2 2 6 6(0) 1

( ) 3 ----- MAXIMUM27 3

( ) ----- MINIMUM6 2

3( ) 1

211 3

( ) ----- MINIMUM6 2

(2 )

f x x x x x x x x

x x So x

f

f

f

f

f

f

1

2

2

2 8 6 3 0( )

2 6 0 3

x x xf x

x x x

Find the extrema of the function.

1. Find f’(x) = 0

4 8 3 0'( )

2 2 0 3

'( 2) 0

' 1 0

x xf x

x x

f

f

2. Test the critical points and the endpoints in the original function.

0

( 3) 0 ( 2) 2 (0) 6

lim (0) 6 (1) 7 (3) 3x

f f f

f f f

3. Identify the highest and lowest points.

Absolute max (-2, 2)

Absolute min (1, -7)

max

min

Or you can set the equation equal to 0 and look for the critical number!!!!!!