3.1 Extrema on an Interval

A lot of effort goes into determining the behavior of a function f on an interval I. Does it have a maximum in the interval? Where is it increasing, decreasing?

p. 164

A function need not have a maximum or minimum in an interval.

You can see that continuity or discontinuity can affect the existence of an extremum on an interval. This suggests this theorem:

Notice that this theorem guarantees a minimum and maximum, but doesn’t help you find them!

p. 164

p. 165

Example 1, p. 165

Find the value of the derivative at each of the relative extrema shown:

a. 2

3

9 3( )

xf x

x

3 2 2 4 2 2

3 2 6

(9)(2 ) 9( 3)(3 ) 18 27 ( 3)'( )

( )

x x x x x x xf x

x x

2 2 2 2

6 4

9 (2 3 9) 9(9 )x x x x

x x

At the point (3, 2), f’(3) = 0

b. f(x) = |x|

0 0

( ) (0)lim lim 1

0x x

xf x f

x x

0 0

( ) (0)lim lim 1

0x x

xf x f

x x

These two one-sided limits disagree, so derivative doesn’t exist at (0, 0)

c. The derivative of f(x) = sin x is f’ = cos x. At the point (π/2, 1) the derivative is f’(π/2) = cos (π/2)=0At the point (3 π/2, -1), f’(3π/2) = cos (3π/2)= 0

We can see that at relative extrema, the derivative is either zero or does not exist. The x-values of these extrema are called critical numbers. There are two types.

p. 166

p. 166

Ex 2 p. 167 Find extrema of on [-1,2]

4 3( ) 3 4f x x x

1. Need derivative to find critical numbers:

What values of x make f’ = 0 or f’ be undefined?

3 2 2'( ) 12 12 12 ( 1)f x x x x x

2. Evaluate f at critical numbers3. Evaluate f at endpoints

212 ( 1) 0 so 0, 1 are critical numbersx x x

Left endpoint

Critical number

Critical number

Right endpoint

f(-1) = 7 f(0) = 0 f(1) = -1minimum

f(2) = 16maximum

4. So f(1) = -1 is minimum and f(2)=16 is maximum.

Notice that all critical numbers don’t have to produce extrema. Converse of Thm 3.2 is not necessarily true! In other words, if x=c is a critical number, f doesn’t have to have a relative max or min there.

Ex 3 p. 1682

3( ) 2 3f x x x Find extrema of on [-1, 3]

1. Find critical numbers. 132'( ) 2 (3) x3f x

13

1 13 3

2 x 1=2 =2

x x

Two critical numbers, x = 0 because that makes f’ undefined and it doesn’t exist,and x = 1 because that makes f’ = 0

2&3. Evaluate f at critical numbers and endpointsLeft endpoint

Critical number

Critical number

Right endpoint

f(-1) = -5Minimum

f(0) = 0Maximum

f(1) = -1 f(3) ≈-0.24

4. Determine max and min for interval.

Ex 4 p. 168 Finding Extrema on a closed intervalFind the extrema of f(x) = 2sin x – cos 2x on [0, 2π]

1. Critical numbers: '( ) 2cos 2sin 2 0f x x x 2cos 4cos sin 0x x x 2cos (1 2sin ) 0x x

32cos 0 when x = ,2 2x 7 1111 2sin 0 sin when x = ,2 6 6x x

2&3. Plug in endpoints and critical numbersLeft endpt

Critical # Critical # Critical # Critical # Right endpt

f(0) = -1 f(π/2) = 3Maximum

f(7π/6)= -1.5minimum

f(3π/2)=-1 f(11π/6) =-1.5mimumum

f(2π) = -1

4. Determine maximum and minimum

Assignment 3.1 p. 169/ 1-45 every other odd, 53-59 odd, 63-66 due Wednesday, Oct 19