Q 43. - Precalculus Enhanced with Graphing Utilities

Question

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

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

Step-by-Step Solution

Answer

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

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 $-h\left(x\right)$,

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

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