Chapter 2.2, Problem 70E

Solution

Prove that $g\left(x\right)=x^{-a}$ is an even if and only if $f\left(x\right)=x^a$ is even and $g\left(x\right)=x^{-a}$ is an odd if and only if $f\left(x\right)=x^a$ is odd.

Given:

  $g\left(x\right)=\frac{1}{f\left(x\right)}$

  $g\left(x\right)=x^{-a}$

Concept Used:

• A function $f(x)$ is an even function if $f\left(-x\right)=f\left(x\right)$ .
• A function $f(x)$ is an odd function if $f\left(-x\right)=-f\left(x\right)$ .

Proof:

The function is $g\left(x\right)=x^{-a}$ .

This can be rewritten as $g\left(x\right)=\frac{1}{x^a}$

Let $f\left(x\right)=x^a$ , then $g\left(x\right)=\frac{1}{f\left(x\right)}$

Now find,

  $g\left(-x\right)=\frac{1}{f\left(-x\right)}\mathrm{...}\left(i\right)$

If $f(x)$ is an even function then $f\left(-x\right)=f\left(x\right)$ .

  $\begin{array}{l}g\left(-x\right)=\frac{1}{f\left(x\right)}\\ g\left(-x\right)=g\left(x\right)\end{array}$

It is the condition for an even function.

Therefore, $g\left(x\right)$ is an even function.

Now,

If $f(x)$ is an odd function then $f\left(-x\right)=f\left(x\right)$ . So, equation (i) becomes

  $\begin{array}{l}g\left(-x\right)=\frac{1}{-f\left(x\right)}\\ g\left(-x\right)=-g\left(x\right)\end{array}$

It is the condition for an odd function.

Therefore, $g\left(x\right)$ is an odd function.

Hence, it has been concluded that $g\left(x\right)=x^{-a}$ is an even if and only if $f\left(x\right)=x^a$ is even and $g\left(x\right)=x^{-a}$ is an odd if and only if $f\left(x\right)=x^a$ is odd.