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.