Question

Question #7801b

Answer 1

`f` is even.

Explanation:

We need to define even and odd functions:

• `f` is an even function if `f(-x)=f(x)`.
• `f` is an odd function if `f(-x)=-f(x)`.

It's also useful to know that `cos(x)` is an even function—that is, `cos(-x)=cos(x)`.

We have:

`f(x)=x^2cos(x)`

Test this function by finding `f(-x)`:

`f(-x)=(-x)^2cos(-x)`

We already know that `cos(-x)=cos(x)`, Furthermore, `(-x)^2=x^2`. You can think about this as `(-x)^2=(-1)^2x^2=x^2`.

`f(-x)=x^2cos(x)=f(x)`

Since `f(-x)=f(x)`, `f` is an even function.