Answer
At $x=-1$, $y=-1/2$ local and absolute minimum.
At $x=1$, $y=1/2$ local and absolute maximum.
Work Step by Step
Step 1. Given the function $y=\frac{x}{x^2+1}$, we can find its derivative as $y′=\frac{x^2+1-x(2x)}{(x^2+1)^2}=\frac{1-x^2}{(x^2+1)^2}$
Step 2. The critical points can be found when $y'=0$ or undefined which happens when $x^2-1=0$ and we have $x=-1,1$
Step 3. At $x=-1$, $y=-1/2$ and we can identify this as a local and absolute minimum (use local test points) as shown in the figure.
Step 4. At $x=1$, $y=1/2$ and we can identify this as a local and absolute maximum (use local test points) as shown in the figure.