   Chapter 3.1, Problem 47E

Chapter
Section
Textbook Problem

# Find the absolute maximum and absolute minimum values of f on the given interval. f ( x ) = 2 x 3 − 3 x 2 − 12 x + 1 ,    [ − 2 , 3 ]

To determine

To find:

The absolute maximum and absolute minimum values of f(x) on the given interval.

Explanation

1) Concept:

Use the Closed Interval Method to find the absolute maximum and minimum values of f(x).

The Closed Interval Method:

To find the absolute maximum and minimum values of a continuous function f on a closed interval a, b:

i. Find the values of f at the critical numbers of f in a, b.

ii. Find the values of f at the end points of the interval .

iii. The largest of the values from step (i) and (ii) is the absolute maximum value; the smallest of these values is the absolute minimum value.

2) Given:

fx=2x3-3x2-12x+1, -2, 3

3) Calculation:

Since f(x) is continuous on -2, 3, use the Closed Interval Method.

Differentiate f(x) with respect to x, and then find the values of x, where f'x=0 and f'x doesn’t exists.

By using the power rule of derivative,

f'x=6x2-6x-12

f'x=6x2-x-2

f'x=6(x+1)(x-2)

Substitute f'x =0, and solve for x.

0=6(x+1)(x-2)

Hence, x= -1 & x=2

Since f'x exists for all x,x= -1 & x=2 are the critical points of fx.

Also, the critical numbers x= -1 & x=2 lie in the interval -2, 3

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