Question

Determine whether the function is even, odd, or neither.

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

Short answer

The$f\left(x\right)=x^2\cos 2x$ is an even function.

Step-by-step solution

Step 1. State the definitions of even and odd functions.

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

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

Step 2. Calculate f−x. 

Substitute $-x$for $x$into the function $f\left(x\right)=x^2\cos 2x$ and simplify.

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

Step 3. Simplify.

Note: $\cos \left(-t\right)=\cos t$, that is., cosine function, in general, is an even function.

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

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

Since, $f\left(-x\right)=f\left(x\right)$, therefore, $f\left(x\right)=x^2\cos 2x$ is an even function.