Answer
critical points are (1,27) and (4,0)
Work Step by Step
Given $y=x(4-x)^3$
The critical points occur where the first derivative is zero:
so$\frac{dy}{dx}=\frac{d{(x(4-x)^3)}}{dx}$
${\frac{dy}{dx}}=(4-x)^3-x(4-x)^2$
so the critical point is ${\frac{dy}{dx}}=0$
$f'(x)=(4-x)^3-x(4-x)^2=0$
$x=1,x=4$
the critical value of y can be obtained by putting in the value of x:
$y=1(4-1)^3=27$
$y=4(4-4)^3=0$
thus the critical points are (1,27) and (4,0)