   Chapter 3.9, Problem 35E

Chapter
Section
Textbook Problem

# 23-42 Find f. f ′ ′ ( x ) = − 2 + 12 x + 12 x 2 ,    f ( 0 ) = 4 ,    f ' ( 0 ) = 12

To determine

To find:

The function fx

Explanation

1) Concept:

If F is an antiderivative of f on an interval I, then the most general antiderivative of f on I is, Fx+c where c is an arbitrary constant.

Definition:

A function F  is called an antiderivative of f on an interval I if F'x=fx  f for all x in I.

2) Given:

f''x=-2+12x-12x2, f0=4 and f'0=12

3) Calculations:

Here f''x=-2+12x-12x2

The general antiderivative of f'x using rules of antiderivative is,

f'x=-2x+12x22-12x33+C, Where C is the arbitrary constant

f'x=-2x+6x2-4x3+C

It is given that f'0=12, therefore, substitute x=0 and f'0=12 to find the value of C

f'0=-2(0)+602-403+C

12=0+C

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