Answer
At $x=-1,3$, $y=0$ local and absolute minimum.
At $x=1$, $y=2$ local and absolute maximum.
Work Step by Step
Step 1. Given the function $y=\sqrt {3+2x-x^2}=(3+2x-x^2)^{1/2}$, we can find its derivative as $y′=\frac{1}{2}(3+2x-x^2)^{-1/2}(2-2x)=\frac{1-x}{\sqrt {(3+2x-x^2)^4}}$
Step 2. The critical points can be found when $y'=0$ or undefined which happens when $x=1$ or $3+2x-x^2=(3-x)(1+x)=0$ and we have $x=-1,1, 3$
Step 3. At $x=-1$, $y=0$ and we can identify this is a local and absolute minimum (use local test points) as shown in the figure.
Step 4. At $x=3$, $y=0$ and we can identify this is a local and absolute minimum (use local test points) as shown in the figure.
Step 5. At $x=1$, $y=2$ and we can identify this is a local and absolute maximum (use local test points) as shown in the figure.
