   Chapter 3.9, Problem 40E

Chapter
Section
Textbook Problem

# 23-42 Find f. f ′ ′ ( x ) = 20 x 3 + 12 x 2 + 4 ,   f ( 0 ) = 8 ,   f ( 1 ) = 5

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.

2) Definition:

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

3) Given:

f''x=20x3+12x2+4, f0=8, f1=5

4) Calculations:

Here f''x=20x3+12x2+4

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

f'x=20x44+12x33+4x+C, Where C is the arbitrary constant

f'x=5x4+4x3+4x+C

Use power rules of antiderivative once more to find f(x)

fx=5x55+4x44+4x22+Cx+D Where,C & D are arbitrary constants

fx=

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