   Chapter 3.1, Problem 50E

Chapter
Section
Textbook Problem

# Find the absolute maximum and absolute minimum values of f on the given interval. f ( t ) = ( t 2 − 4 ) 3 ,    [ − 2 , 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 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:

ft= (t2-4)3, -2, 3

3) Calculation:

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

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

By using the chain rule of derivative,

f't=3(t2-4)2*2t

f't=6t(t2-4)2

Substitute  f't =0, and then solve for t.

0=6t(t2-4)2

Hence, t= 0, t= -2,  & t= 2

Since f't exists for all t,t= 0, t= -2, & t= 2 are the only critical points of ft

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