Question

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

Short answer

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

Step-by-step solution

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

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

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 role="math" localid="1645807726141" $h(-x)=-h\left(x\right),h$is an odd function.