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Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621

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Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621
Textbook Problem

1-6 Find the local and absolute extreme values of the function on the given interval.

f ( x ) = x 3 9 x 2 + 24 x 2 ,   [ 0 , 5 ]

To determine

To find:

The local and absolute extreme values of the function on the given interval

Explanation

1) Concept:

Use the first derivative test to find local maximum and minimum values, and use closed interval method to find the absolute maximum and minimum values of f(x).

i. First derivative test:

Suppose that c is a critical number of a continuous function f.

a) If f' changes from positive to negative at c, then f has a local maximum at c

b) If f' changes from negative to positive at c, then f has a local minimum at c

c) If f' is positive to the left and right of c or negative to the left and right of c then f has no local maximum or minimum at c.

ii. 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=x3-9x2+24x-2 ,[0, 5]

3) Calculation:

Differentiate fx with respect to x by using power rule of derivative.

f'x=3x2-18x+24

To find critical numbers, set f'x=0 and solve x.

3(x2-6x+8)=0

x2-6x+8=0

x-2x-4=0

By using zero product property,

x = 2 or x=4

f'x exists for all x in the given interval.

So the critical numbers are x = 2 and x=4

Split the given domain using above critical points

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