Q 42. - Precalculus Enhanced with Graphing Utilities

Question

In Problems 33– 44, determine algebraically whether each function is even, odd, or neither.
$h\left(x\right)=\frac{x}{x^2-1}$

Step-by-Step Solution

Answer

The given function $h\left(x\right)=\frac{x}{x^2-1}$ is odd.

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

The given function is: $h\left(x\right)=\frac{x}{x^2-1}$

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,

$$\begin{gathered} h(-x)=\frac{-x}{{(-x)}^2-1} \\ =\frac{-x}{x^2-1} \\ =-\frac{x}{x^2-1} \end{gathered}$$

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

Step 3. Determine if the function is odd.

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

$-h\left(x\right)=-\frac{x}{x^2-1}$

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