# The function f ( x ) = ln ( x + x 2 + 1 ) is an odd 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 1, Problem 12P

(a)

To determine

## To show: The function f(x)=ln(x+x2+1) is an odd function.

Expert Solution

Solution:

The function f(x)=ln(x+x2+1) is an odd function.

### Explanation of Solution

Result used:

f is an odd function if and only if f(x)=f(x) for all x in the domain of f.

logab=logalogb

Proof:

Consider the function f(x)=ln(x+x2+1)

Substitute x=x in f(x).

f(x)=ln(x+(x)2+1)=ln(x+x2+1)

Rationalize the numerator,

f(x)=ln[(x+x2+1)(xx2+1xx2+1)]=ln(x2(x2+1)xx2+1)=ln(1xx2+1)=ln(1x+x2+1)

Simplify further,

f(x)=ln(1x+x2+1)=ln(1)ln(x+x2+1)=ln(x+x2+1)            (ln(1)=0)=f(x)

Therefore, by definition it is concluded that the function is odd function.

(b)

To determine

### To find: The inverse function of f.

Expert Solution

Solution:

The inverse function of f(x) is, f1(x)=e2x12ex.

### Explanation of Solution

Calculation:

Consider the function y=ln(x+x2+1)

Interchange x and y,

x=ln(y+y2+1)

Take exponential on both sides,

ex=eln(y+y2+1)ex=y+y2+1exy=y2+1

Take square on both sides,

(exy)2=y2+1e2x2yex+y2=y2+1e2x1y2+y2=2yexe2x1=2yex

Simplify further,

y=e2x12ex=f1(x)

Therefore, it is concluded that the inverse function of f(x) is, f1(x)=e2x12ex.

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