Chapter 1.3, Problem 64E

To determine

To check: Whether $h=f\circ g$ is always an odd function if g is odd; also check what would be the result if f is an odd function and if f is even function.

Answer to Problem 64E

The composition function $h=f\circ g$ is not always an odd function.

Iff is an odd function, then the composition function $h=f\circ g$ is always an odd function.

If f is an even function, then the composition function $h=f\circ g$ is always an even function.

Explanation of Solution

Given:

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

Therefore, $g\left(x\right)=-g\left(x\right)$.

The composite function is, $h\left(x\right)=\left(f\circ g\right)\left(x\right)$

That is $h\left(x\right)=f\left(g\left(x\right)\right)$.

Verification:

The composition function $h\left(x\right)$ is said to be an odd function if $h\left(-x\right)=-h\left(x\right)$, is said to be an even function if $h\left(-x\right)=h\left(x\right)$.

So, first obtain the value of $h\left(-x\right)$ as shown below.

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

Thus, $h\left(-x\right)=f\left(-g\left(x\right)\right)$ (1)

Here, it cannot be concluded that whether the composition function $h\left(x\right)$ is an even function or an odd function as it depends on the nature of the function $f\left(x\right)$. Thus, $h\left(x\right)$ is not always an odd function.

There exist two cases which are explained as follows.

Case 1:

If $f\left(x\right)$ is an odd function, then the equation (1) becomes,

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

Thus, the composite function $h\left(x\right)=\left(f\circ g\right)\left(x\right)$ is an odd function when the functions $f\left(x\right)$ and $g\left(x\right)$ are odd functions.

Case 2:

If $f\left(x\right)$ is an even function, then the equation (1) becomes,

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

Thus, the composite function $h\left(x\right)=\left(f\circ g\right)\left(x\right)$ is an even function when the function $f\left(x\right)$ is even and the function $g\left(x\right)$ is odd functions.