Explanation Given information: f ( x ) = x 3 − 6 x 2 − 15 x + 20 Calculation: f ( x ) = x 3 − 6 x 2 − 15 x + 20 Differentiating with respect to ‘x’, we get f ' ( x ) = d d x ( x 3 − 6 x 2 − 15 x + 20 ) = d d x ( x 3 ) − d d x ( 6 x 2 ) − d d x ( 15 x ) + d d x ( 20 ) = d d x ( x 3 ) − 6 d d x ( x 2 ) − 15 d d x ( x ) + d d x ( 20 ) = ( 3 x 2 ) − 6 ( 2 x ) − 15 ( 1 ) + 0 = 3 x 2 − 12 x − 15 ⇒ f ' ( x ) = 3 x 2 − 12 x − 15 Put f

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ISBN: 9781119275565

Author: Hughes-Hallett

Publisher: WILEY

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Chapter 3.1, Problem 78P

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f'(x) and all values of x for which f'(x)=0. And explain the relation between the values of x and the graph of f(x).

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Chapter 3.1, Problem 77P

Chapter 3.1, Problem 79P

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f ' ( x ) and all values of x for which f ' ( x ) = 0. And explain the relation between the values of x and the graph of f ( x ) .

Solution Summary: The author explains the relation between the values of x and the graph of f(x).

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f'(x) and all values of x for which f'(x)=0. And explain the relation between the values of x and the graph of f(x).

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f ' ( x ) and all values of x for which f ' ( x ) = 0. And explain the relation between the values of x and the graph of f ( x ) .

Solution Summary: The author explains the relation between the values of x and the graph of f(x).

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To find: f'(x) and all values of x for which f'(x)=0. And explain the relation between the values of x and the graph of f(x).

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Chapter 3 Solutions

To find:

f'(x) and all values of x for which f'(x)=0. And explain the relation between the values of x and the graph of f(x).