# Question #7801b

Apr 4, 2017

$f$ is even.

#### Explanation:

We need to define even and odd functions:

• $f$ is an even function if $f \left(- x\right) = f \left(x\right)$.
• $f$ is an odd function if $f \left(- x\right) = - f \left(x\right)$.

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

We have:

$f \left(x\right) = {x}^{2} \cos \left(x\right)$

Test this function by finding $f \left(- x\right)$:

$f \left(- x\right) = {\left(- x\right)}^{2} \cos \left(- x\right)$

We already know that $\cos \left(- x\right) = \cos \left(x\right)$, Furthermore, ${\left(- x\right)}^{2} = {x}^{2}$. You can think about this as ${\left(- x\right)}^{2} = {\left(- 1\right)}^{2} {x}^{2} = {x}^{2}$.

$f \left(- x\right) = {x}^{2} \cos \left(x\right) = f \left(x\right)$

Since $f \left(- x\right) = f \left(x\right)$, $f$ is an even function.