Answer
$0,1,2$
Work Step by Step
Step 1. Given the function $g(x)=\sqrt {2x-x^2}=(2x-x^2)^{1/2}$, take the derivative to get:
$g'(x)=\frac{1}{2}(2x-x^2)^{-1/2}(2-2x)=\frac{1-x}{\sqrt {2x-x^2}}$
Step 2. The critical points can be found when $g'(x)=0$ or is undefined; we have $x=1$ and $2x-x^2=0$ which gives $x=0,2$
Step 3. We conclude that the critical points are $x=0,1,2$
