   Chapter 4.1, Problem 52E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Find the absolute maximum and absolute minimum values of f on the given interval.f(t) = (t2 − 4)3, [−2, 3]

To determine

To find: The absolute maximum and absolute minimum of the given function.

Explanation

Given:

The function f(t)=(t24)3 in the interval [2,3] .

Chain Rule:

If two functions g(x) and h(x) are differentiable, then the derivative of f(x)=g(h(x)) is,

f(x)=g(h(x))(h(x)) (1)

Calculation:

Obtain the first derivative of the given function.

f(t)=ddx(t24)3

Apply the Chain Rule as shown in equation (1),

f(t)=3(t24)(2t)=6t(t+2)2(t2)2=6t(t+2)3(t2)

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

6t(t+2)3(t2)=06t=0,t+2=0,t2=0t=0,t=2,t=2

Thus, the critical numbers are 2,0 and 2 , which lies in the given interval [2,3]

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