Question

Determine(algebraically) whethere the given function is even, odd,or neither.

$f\left(x\right)=\frac{x}{1+x^2}$

Short answer

The given function$f\left(x\right)=\frac{x}{1+x^2}$is odd function

Step-by-step solution

Step 1.Given information

The given function$f\left(x\right)=\frac{x}{1+x^2}$

Step 2.Determine the given function is even,odd or neither.

The function is even if $f(-x)=f\left(x\right)$and odd if $f(-x)=f(-x)$.If none of these conditions are satisfied.

Find $f(-x)$ by replacing -x for x in the given function.

$$\begin{gathered} f(-x)=\frac{-x}{1+{(-x)}^2} \\ =\frac{-x}{1+x^2} \\ =-\left(\frac{x}{1+x^2}\right) \\ =-f\left(x\right) \end{gathered}$$

Therefore,the given function is odd since$f(-x)=-f\left(x\right)$