   Chapter 3.R, Problem 57E

Chapter
Section
Textbook Problem

# 55-58 Find f. f ″ ( x ) = 1 − 6 x + 48 x 2 ,     f ( 0 ) = 1 ,     f ′ ( 0 ) = 2

To determine

To find:

The function f

Solution:fx=x22-x3+4x4+2x+1

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) Formula:

Power rule of antiderivative;ddxxn+1n+1=xn

3) Given:

f''x=1-6x+48x2, f0=1, f'0=2

4) Calculations:

Here, f''x=1-6x+48x2, f0=1, f'0=2

The general antiderivative of f''x using power rule of antiderivative is

f'x=x-6*x22+48*x33+C, Where C  is arbitrary constant

f'x=x-3x2+16x3+C

It is given that f'0=2

Substitute x=0 in f'x

f'0=0-3<

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