# 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 56RE
To determine

## Tofind:the function using its second derivative.

Expert Solution

Thefunction using its second derivative f(x)=x510+x442x33+5x224.1833x+2.

### Explanation of Solution

Given:

f(x)=2x3+3x2-4x+5, f(0)=2, f(1)=0 .

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)=2x3+3x2-4x+5, f(0)=2, f(1)=0 .

f(x)=2x3+3x2-4x+5 .

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

f(x)=(x42+x32x2+5x+C1)dx .

f(x)=(x42+x32x2+5x+C) .

f(x)=(x42+x32x2+5x+C1)dx .

f(x)=x510+x442x33+5x22+C1x+C2

f(0)=(0)510+(0)442(0)33+5(0)22+C1(0)+C2 .

C2=2.

f(1)=(1)510+(1)442(1)33+5(1)22+C1(1)+C2 .

f(1)=(1)510+(1)442(1)33+5(1)22+C1(1)+2

0=4.1833+C1(1) .

C1=-4.1833 .

f(x)=x510+x442x33+5x224.1833x+2.

Hence the function using its second derivative f(x)=x510+x442x33+5x224.1833x+2.

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