Chapter 1.8, Problem 67E

To determine

To prove: product of two odd functions is an even function and the product of two even function is an even function

Explanation of Solution

Concept Used:

A function $f\left(x\right)$ is said to be an even function if for all x in the domain of f , $f\left(-x\right)=f\left(x\right)$ .

A function $f\left(x\right)$ is said to be an odd function if for all x in the domain of f , $f\left(-x\right)=-f\left(x\right)$ .

First prove that product of two odd functions is an even function.

Let $f\left(x\right)$ and $g\left(x\right)$ be two odd functions.

Then,

  $f\left(-x\right)=-f\left(x\right)$

And

  $g\left(-x\right)=-g\left(x\right)$

Now, consider the product of these two odd functions.

  $h\left(x\right)=f\left(x\right)\cdot g\left(x\right)$

Now check if this function is even or odd.

  $\begin{array}{c}h\left(-x\right)=f\left(-x\right)\cdot g\left(-x\right)\\ =\left(-f\left(x\right)\right)\cdot \left(-g\left(x\right)\right)\\ =f\left(x\right)\cdot g\left(x\right)\\ =h\left(x\right)\end{array}$

By definition, clearly $h\left(x\right)$ is an even function.

Thus, product of two odd functions is an even function.

Now prove that product of two even functions is an even function.

Let $p\left(x\right)$ and $q\left(x\right)$ be two even functions.

Then,

  $p\left(-x\right)=p\left(x\right)$

And

  $q\left(-x\right)=q\left(x\right)$

Now, consider the product of these two even functions.

  $r\left(x\right)=p\left(x\right)\cdot q\left(x\right)$

Now check if this function is even or odd.

  $\begin{array}{c}r\left(-x\right)=p\left(-x\right)\cdot q\left(-x\right)\\ =p\left(x\right)\cdot q\left(x\right)\\ =r\left(x\right)\end{array}$

By definition, clearly $r\left(x\right)$ is an even function.

Thus, product of two even functions is an even function.