Question

In Problems 33– 44, determine algebraically whether each function is even, odd, or neither.

$h\left(x\right)=\frac{-x^3}{3x^2-9}$

Short answer

The given function$h\left(x\right)=\frac{-x^3}{3x^2-9}$is an odd function.

Step-by-step solution

Step 1. Write the given information and use the definition of odd and even function.

The given function is:

$h\left(x\right)=\frac{-x^3}{3x^2-9}$

A function $f$is even if, for every number $x$in its domain, the number $-x$is also in the domain and $f(-x)=f\left(x\right)$.

A function $f$is odd if, for every number$x$in its domain, the number $-x$is also in the domain and$f(-x)=-f\left(x\right)$.

Step 2. Determine if the function is even.

Replace $x$by $-x$in the given function,

$$\begin{gathered} h(-x)=\frac{-{(-x)}^3}{3{(-x)}^2-9} \\ =\frac{-{(-x)}^3}{3{\left(x\right)}^2-9} \\ =\frac{x^3}{3x^2-9} \end{gathered}$$

Since$h(-x)\ne h\left(x\right),h$is not an even function.

Step 3. Determine if the function is odd.

Find the function $-h\left(x\right)$,

role="math" localid="1645811470834" $$\begin{gathered} -h\left(x\right)=-\left(\frac{-{\mathrm{x}}^3}{3{\mathrm{x}}^2-9}\right) \\ =\frac{{\mathrm{x}}^3}{3{\mathrm{x}}^2-9} \end{gathered}$$

Since$h(-x)=-h\left(x\right),h$is an odd function.