   Chapter 3.1, Problem 52E

Chapter
Section
Textbook Problem

# Find the absolute maximum and absolute minimum values of f on the given interval. f ( x ) = x x 2 − x + 1 ,    [ 0 , 3 ]

To determine

To find:

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

Explanation

1) Concept:

Use 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 endpoints 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=xx2-x+1, 0, 3

3) Calculation:

Since f(x) is continuous on 0, 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 exist.

By using power rule of derivative,

f'x= x2-x+11-(x)(2x-1)(x2-x+1)2

f'x= x2-x+1-2x2+x(x2-x+1)2

f'x= 1-x2(x2-x+1)2

Substitute f'x =0, then solve for x

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