   Chapter 4.1, Problem 50E ### 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(x) = x3 − 6x2 + 5, [−3, 5]

To determine

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

Explanation

Given:

The function f(x)=x36x2+5 in the interval [3,5] .

Power Rule:

If n is positive integer, then ddx(xn)=nxn1 (1)

Calculation:

Obtain the first derivative of the given function.

f(x)=ddx(x36x2+5)=ddx(x3)ddx(6x2)+ddx(5)

Apply the power rule as shown in equation (1),

f(x)=3x316(2x21)+0=3x212x=3x(x4)

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

3x(x4)=0x=0,x4=0x=0,x=4

Thus, the critical numbers are 0 and 4, which lies in the given interval [3,5] .

Apply the extreme values of the given interval and the critical number values in f(x)

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