# The local maximum and minimum and the absolute maximum and absolute minimum of the function f ( x ) = 3 x − 4 x 2 + 1 on the interval [ − 2 , 2 ] . ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 4, Problem 3RE
To determine

## To find: The local maximum and minimum and the absolute maximum and absolute minimum of the function f(x)=3x−4x2+1 on the interval [−2,2] .

Expert Solution

The local minimum occurs at x=13 .

The absolute maximum occurs at x=2 .

The absolute minimum is x=13 .

### Explanation of Solution

Given:

The function is, f(x)=3x4x2+1 and the interval is, [2,2] .

Calculation:

Obtain the first derivative of the given function.

f'(x)=(x2+1)(3)(3x4)(2x)(x2+1)2=3x2+36x2+8x(x2+1)2=3x2+8x+3(x2+1)2

Set f(x)=0 and obtain the critical numbers.

3x2+8x+3(x2+1)2=03x2+8x+3=0

Apply quadratic formula and find the value of x.

x=8±824(3)(3)2(3)=8±64+366=8±106=186 or 26

Thus, the value of x=13,3 .

Thus, the critical numbers are x=13,3 in which 3 does not lie on the given interval [2,2] .

Apply the extreme values of the given interval and the critical number x=13 in f(x) .

Substitute x=2 in f(x) ,

f(2)=3(2)4(2)2+1=644+1=105=2

Substitute x=13 in f(x) ,

f(13)=3(13)4(13)2+1=1419+1=5109=92

Substitute x=2 in f(x) ,

f(2)=3(2)4(2)2+1=644+1=25

Since the largest functional value is the absolute maximum and the smallest functional value is the absolute minimum, the absolute maximum of f(x) is approximately 25 and the absolute minimum of f(x) is −4.5.

Therefore, the absolute maximum of f(x) occurs at x=2 and the absolute minimum of f(x) is occurs at x=13 .

To find the local maximum and minimum obtain the second derivative of the function.

f''(x)=ddx(3x2+8x+3(x2+1)2)=6x+82(x2+1)(2x)=2(3x+4)4x(x2+1)=3x+42x(x2+1) .

Substitute the critical numbers in the second derivative and obtain the local maximum and minimum as follows.

Substitute x=13 in f''(x) ,

f''(13)=3(13)+42(13)((13)2+1)=1+4(23)(19+1)=5(23)(109)=274<0

Thus, the local minimum occurs at x=13 .

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