Answer
critical points are (0,0) and (4,32)
Work Step by Step
Given $y=6x^2-x^3$
The critical point occurs where the first derivative is zero:
so $\frac{dy}{dx}=\frac{d{(6x^2-x^3)}}{dx}$
${\frac{dy}{dx}}=12x-3x^2$
so the critical point is ${\frac{dy}{dx}}=0$
$f'(x)=12x-3x^2=0$
$x=0, x=4$
the critical value of y can be obtained by putting in the value of x:
$y=6\times0^2-0=0$
$y=6\times4^2-4^3=32$
thus critical points are (0,0) and (4,32)