   Chapter 3.1, Problem 53E

Chapter
Section
Textbook Problem

# Find the absolute maximum and absolute minimum values of f on the given interval. f ( t ) = t − t 3 ,    [ − 1 , 4 ]

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:

ft= t-t3, -1, 4

3) Calculation:

Since f(t) is continuous on -1, 4, 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 power rule of derivative,

f't=1-13t23

Substitute f't =0, and solve for t.

0=1-13t23

13t23=1

3t23=1

t23=13

Take cube on both sides of the above equation.

t2 =127

Take square root on both sides of the above equation,

t= ±133

By rationalizing the denominator, multiply both the numerator and denominator by 3.

t= ±39

Hence,

t= 39, & t= -39

Since f't does not exist when t=0, t= 3 9 , 3 9 ,&0 are the critical points of ft

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