Answer
$f(x)$ increasing on $\left( -\infty,0\right )$ and decreasing on $\left(0, \infty\right )$. $f(x)$ has a local maximum at $x=0$.
Work Step by Step
Given $$y=\frac{1}{x^{2}+1}$$
Since
$$f'(x) = -\frac{2x}{\left(x^2+1\right)^2} $$
Then $f'(x)=0$ for $x= 0 $. Choose $x=-1 $ and $x=1$. Then:
\begin{align*}
f'(-1)&> 0\\
f'(1)&<0
\end{align*}
Thus $f(x)$ is increasing on $\left( -\infty,0\right )$ and decreasing on $\left(0, \infty\right )$. Hence, $f(x)$ has a local maximum at $x= 0$.
