# Tofind: the function using its second derivative.

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 4, Problem 55RE
To determine

## Tofind:the function using its second derivative.

Expert Solution

Thefunction using its second derivative f(x)=x22-x3+4x4+2x+1.

### Explanation of Solution

Given:

f(x)=1-6x+48x2, f(0)=1, f(0)=2 .

Concept used:

Antiderivative=integration

f(t)dt=f(t) . Or dydtdt=y .

Since, the multiplication of two operators ×ddt=1 .

Integration formula:

xndx=xn+1n+1 .

f(x)dx=f(x)dx=f(x) .

Calculation:

f(x)=1-6x+48x2, f(0)=1, f(0)=2 .

f(x)=1-6x+48x2 .

f(x)dx=f(x)dx=f(x) .

f(x)=(x-3x2+16x3+C)dx .

f(x)=(x-3x2+16x3+C) .

f(0)=2 .

2=(0-3(0)2+16(0)3+C) .

C=2.

f(x)=(x-3x2+16x3+C) .

f(x)=(x-3x2+16x3+2) .

f(x)=(x-3x2+16x3+2)dx .

f(x)=x22-x3+4x4+2x+C .

f(0)=(0)22-(0)3+4(0)4+2(0)+C .

C=1.

f(x)=x22-x3+4x4+2x+1.

Hence the function using its second derivative f(x)=x22-x3+4x4+2x+1.

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