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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 ) = 3 x 4 x 2 + 1 ,   [ 2 , 2 ]

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 the local maximum and minimum values and use the 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 values from step (i) and (ii) are the absolute maximum values; the smallest values are the absolute minimum values.

2) Given:

fx=3x - 4x2 + 1, [-2, 2]

3) Calculation:

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

f'x=(x2+1)*ddx3x-4-ddxx2+1*(3x-4)x2+12

f'x=(x2+1)*ddx3x-4-ddxx2+1*(3x-4)x2+12

f'x=(x2+1)*3x-2x*(3x-4)x2+12

f'x=-3x2+8x+3x2+12

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

-3x2+8x+3x2+12=0

-(3x2- 8x- 3) = 0

-x-33x+1=0

By using zero product property,

x=-13 or x= 3 but 3 is not in the given interval

f'x exists for all x in the given interval

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