   Chapter 3.9, Problem 36E

Chapter
Section
Textbook Problem

# 23-42 Find f. f ′ ′ ( x ) = 8 x 3 + 5 ,    f ( 1 ) = 0 ,    f ′ ( 1 ) = 8

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=8x3+5, f1=0 and f'1=8

3) Calculations:

Here f''x=8x3+5

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

f'x=8x44+5x+C , Where C is the arbitrary constant

f'x=2x4+5x+C

It is given that f'1=8, therefore, substitute x=1 and f'1=8 to find the value of C

f'1=214+5(1)+C

8=2+5+C

C=1

Substitute C=1 in f'x<

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