Question

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

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

Short answer

The$f\left(x\right)=x^2\sin x$is an odd 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 $f\left(x\right)=x^2\sin x$and simplify.

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

Step 3. Simplify.

Note: $\sin \left(-t\right)=-\sin t$, that is., sine function, in general, is an odd function.

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

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

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