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