## University Calculus: Early Transcendentals (3rd Edition)

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)