Answer:
0 and 8/3
Step-by-step explanation:
y is continuous on [0,4]
It also also differentiable on (0,4)
Plugging in 0 and 4, we get -3 both times so y(0) = y(4).
Therefore there is at least one value c between 0 and 4 inclusive where y'(c) = 0
y' = -3x^2 + 8x
Set that to 0:
-3x^2 + 8x = 0
x(-3x+8) = 0
x = 0 or 8/3
And both values work.
