Answer
the critical points are: (0,0), (1,1), (2,0)
Work Step by Step
Given $y=({(2x)}-{x^2})^{(\frac{1}{2})}$
The critical points are where the first derivative is zero:
$\frac{dy}{dx}=\frac{d{(({(2x)}-{x^2})^{(\frac{1}{2})})}}{dx}$
${\frac{dy}{dx}}=\frac{1}{2({(2x)}-{x^2})^{(\frac{1}{2})}}{({2}-{{2x}})}$
so the critical point is ${\frac{dy}{dx}}=0$
$f'(x)=\frac{1}{2({(2x)}-{x^2})^{(\frac{1}{2})}}{({2}-{{2x}})}=0$
$1-x=0$ and ${(2x)}-{x^2})^{(\frac{1}{2})}=0$
$x=1,x=0,x=2$
the critical value of y can be obtained by putting in the value of x:
$y=({(2x)}-{x^2})^{(\frac{1}{2})}=0$
$y=({(2x)}-{x^2})^{(\frac{1}{2})}=({(2)}-{1^2})^{(\frac{1}{2})}=1$
$y=({(2x)}-{x^2})^{(\frac{1}{2})}=({(4)}-{4})^{(\frac{1}{2})}=0$
thus the critical points are: (0,0), (1,1), (2,0)
