# The number of g have an extreme value.

### 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 26RE

(a)

To determine

## To find:The number of g have an extreme value.

Expert Solution

g=0

whenx=0

### Explanation of Solution

Given:

The function

g(x)=f(x2)

f(x)<0forx<0andf(x)>0forx>0

Concept used:

The left derivative and right derivative of a function f at a point x=c are , equal

f'(c)=limh0f(ch)f(c)hf+'(c)=limh0+f(c+h)f(c)h

Calculation:

g(x)=f(x2)..................(1)

Differentiating equation (1) with respect to x

g(x)=2xf(x2).....................(2)

Differentiating equation (1) with respect to x by the product and chain rules

g(x)=2xf(x2)g(x)=(2x)f(x2)+2x[f(x2)]g(x)=2f(x2)+2x[2xf(x2)]g(x)=2f(x2)+4x2[f(x2)]

g=0

whenx=0

(b)

To determine

### To find:The number of g have the concavity .

Expert Solution

g is concave up for all real numbers.

### Explanation of Solution

Given:

The function

g(x)=f(x2)

f(x)<0forx<0andf(x)>0forx>0

Concept used:

The left derivative and right derivative of a function f at a point x=c are , equal

f'(c)=limh0f(ch)f(c)hf+'(c)=limh0+f(c+h)f(c)h

Calculation:

g(x)=f(x2)..................(1)

Differentiating equation (1) with respect to x

g(x)=2xf(x2).....................(2)

Differentiating equation (1) with respect to x by the product and chain rules

g(x)=2xf(x2)g(x)=(2x)f(x2)+2x[f(x2)]g(x)=2f(x2)+2x[2xf(x2)]g(x)=2f(x2)+4x2[f(x2)]

Given that

f(x)<0forx<0andf(x)>0forx>0

x2 is positive for x0,sof(x2)>0forx0

g=0 ;x=0

Since g is positive on either side that g is concave up for all real numbers.

### Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!