   Chapter 3.9, Problem 39E

Chapter
Section
Textbook Problem

# 23-42 Find f. f ′ ′ ( x ) = 4 + 6 x + 24 x 2 ,    f ( 0 ) = 3 ,    f ( 1 ) = 10

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=4+6x+24x2, f0=3, f1=10

3) Calculations:

Here f''x=4+6x+24x2

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

f'x=4x+6x22+24x33+C , where C is the arbitrary constant

f'x=4x+3x2+8x3+C

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

fx=4x22+3x33+8x44+Cx+D where,C & D are the arbitrary constants

fx=2x2

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

#### Multiply: 68005200

Elementary Technical Mathematics

#### In Problems 5-14, solve for x by writing the equation in exponential form. 10.

Mathematical Applications for the Management, Life, and Social Sciences

#### For y = 3x, y = a) 3x log3 e b) 3x ln 3 c) 3xlog3e d) 3xln3

Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th 