# Find the increasing interval for the given function.

### 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 3.3, Problem 33E
To determine

## Find the increasing interval for the given function.

Expert Solution

The function increasing in the interval (π3,5π3) .

### Explanation of Solution

Given:

The given functionis f(x)=x2sinx , 0x2π .

Calculation:

For increasing function f'(x)>0 .

f(x)=x2sinx

Apply difference rule.

(fg)'=f'g'

f'(x)=ddx(x)2ddx(sinx)

Use derivative rule ddx(xn)=nxn1 and ddx(sinx)=cosx

f'(x)=12cosx

Find the critical points.

f'(x)=12cosx=012cosx=012cosx1=012cosx=12cosx=122cosx=12cosx=12πx=π3,5π3

Now check the increasing interval in the intervals

(0,π3) , (π3,5π3) , (5π3,2π)

In the interval (0,π3) .

Substitute x=π6 .

f'(π6)=12cos(π6)=1232=11.732=0.732(negative)

In the interval (π3,5π3) .

Substitute x=π2 .

f'(π2)=12cos(π2)=120=1(positive)

In the interval (5π3,2π) .

Substitute x=23π12 .

f'(23π12)=12cos(23π12)=120.966=11.932=0.932(negative)

Hence thefunction increasing in the interval (π3,5π3) .

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