Q. 77 - Precalculus Enhanced with Graphing Utilities

Question

If $f$ is an odd function and $g$is an even function, show that the composite functions$f\circ g$ and $g\circ f$ are both even. 

Step-by-Step Solution

Answer

The, $f\circ g$ and $g\circ f$ both are even function.

Step 1. Given Information

Given that $f$ is odd function and $g$ is even function.

Step 2. Solution

A function $f$ is said to be odd if $f(-x)=-f\left(x\right) \forall x.$

A function $g$ is said to be even if $g(-x)=g\left(x\right) \forall x.$

Since

$f\circ g\left(x\right)=f\left(g\right(x\left)\right)$

Now,

$$\begin{gathered} f\circ g(-x)=f\left(g\right(-x\left)\right) \\ f\circ g(-x)=f\left(g\right(x\left)\right) \{g is even function.\} \\ f\circ g(-x)=f\circ g\left(x\right) \forall x \end{gathered}$$

So, $f\circ g$ is an even function.

For $g\circ f\left(x\right)=g\left(f\right(x\left)\right).$

$$\begin{gathered} g\circ f(-x)=g\left(f\right(-x\left)\right) \\ g\circ f(-x)=g(-f(x\left)\right) \{f is odd function\} \\ g\circ f(-x)=g\left(f\right(x\left)\right) \{g is even function\} \\ g\circ f(-x)=g\circ f\left(x\right) \forall x \end{gathered}$$

So $g\circ f$ is an even function.