Answer
Critical points $x=\pm\sqrt 2, \pm2$, endpoints $x=\pm2$.
At $x=-2$, $y=0$, local maximum.
At $x=2$, $y=0$, local minimum.
At $x=-\sqrt 2$, $y=-2$, local and absolute minimum.
At $x=\sqrt 2$, $y=2$, local and absolute maximum.
Work Step by Step
Step 1. Given the function $y=x\sqrt {4-x^2}=x(4-x^2)^{1/2}$, we can find its derivative as $y'=(4-x^2)^{1/2}+\frac{1}{2}x(4-x^2)^{-1/2}(-2x)=\frac{4-x^2}{\sqrt {4-x^2}}-\frac{x^2}{\sqrt {4-x^2}}=\frac{4-2x^2}{\sqrt {4-x^2}}$
Step 2. The critical points can be found when $y'=0$ or undefined which happens when $x=\pm\sqrt 2, \pm2$. The domain of the function is $(-2,2)$ and the endpoints are $x=\pm2$.
Step 3. At $x=-2$, $y=0$ and we can identify this is a local maximum (use local test points as necessary) as shown in the figure.
Step 4. At $x=2$, $y=0$ and we can identify this as a local minimum (use local test points as necessary) as shown in the figure.
Step 5. At $x=-\sqrt 2$, $y=-2$ and we can identify this as a local and absolute minimum (use local test points as necessary) as shown in the figure.
Step 6. At $x=\sqrt 2$, $y=2$ and we can identify this as a local and absolute maximum (use local test points as necessary) as shown in the figure.