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