Answer
the critical points are: (0,0), (4,-48)
Work Step by Step
Given $y={(x)^2}-32\sqrt{x}$
The critical point occurs where the first derivative is zero:
$\frac{dy}{dx}=\frac{d{({(x)^2}-32\sqrt{x})}}{dx}$
${\frac{dy}{dx}}={{2x}-\frac{32}{2\sqrt{x}}}$
${\frac{dy}{dx}}=\frac{(4(x^{\frac{3}{2}})-32)}{(2\sqrt{x})}$
so the critical point is ${\frac{dy}{dx}}=0$
$f'(x)=\frac{(4(x^{\frac{3}{2}})-32)}{(2\sqrt{x})}=0$
$x=4$
since denominator is undefined at ${\sqrt{x}}=0$
$x=0,x=4$
the critical value of y can be obtained by putting in the value of x:
$y={(x)^2}-32\sqrt{x}=0$
$y={(x)^2}-32\sqrt{x}={(4)^2}-32\sqrt{4}=-48$
thus the critical points are: (0,0), (4,-48)
