Question

If f and t are both even functions, is f 1 t even? If f and t are both odd functions, is f 1 t odd? What if f is even and t is odd? Justify your answers.

Answer 1

If the $f(x)$ and $t(x)$ are even function then $fo\ t\ (x)$ is an even function, if $f(x)$ and $t(x)$ are odd function then the function $fo\ t\ (x)$ is an odd function and if $f(x)$ is even and $t(x)$ is odd then the function $fo\ t\ (x)$ is an even function.

Further explanation:

An even functrion satisfies the property as shown below:

$\boxed{f(-x)=f(x)}$

An odd functrion satisfies the property as shown below:

$\boxed{f(-x)=-f(x)}$

Consider the given composite function as follows:

$\boxed{fo\ t\ (x)=f\left(t(x))\right}$

If both the function $f(x)$ and $t(x)$ are even function.

$\begin{aligned}fo\ t\ (-x)&=f\left(t(-x))\right\\&=f\left(t(x))\right\\&=fo\ t\ (x)\end{aligned}$

From the above calculation it is concluded that,

$\boxed{fo\ t\ (-x)=fo\ t\ (x)}$

This implies that the composite function $fo\ t\ (x)$ is an even function.

If both the function $f(x)$ and $t(x)$ are odd function.

$\begin{aligned}fo\ t\ (-x)&=f\left(t(-x))\right\\&=f\left(-t(x))\right\\&=-fo\ t\ (x)\end{aligned}$

From the above calculation it is concluded that,

$\boxed{fo\ t\ (-x)=-fo\ t\ (x)}$

This implies that the composite function $fo\ t\ (x)$ is an odd function.

If the function $f(x)$ is even and $t(x)$ is odd.

$\begin{aligned}fo\ t\ (-x)&=f\left(t(-x))\right\\&=f\left(-t(x))\right\\&=fo\ t\ (x)\end{aligned}$

From the above calculation it is concluded that,

$\boxed{fo\ t\ (-x)=fo\ t\ (x)}$

This implies that the composite function $fo\ t\ (x)$ is an even function.