I have this problem from my Calculus 1 homework and I don't know if I'm doing it correctly.

f (x) = x^3 - 12x

For stationary points: f ' (x) = 0

f ' (x) = 3x^2 - 12 (solve for 0)

3x^2 = 12

x^2= 4

x= +2 and -2

(2, -16) and (-2, 16)

f '' (x) = 6x

f '' (2)= 6(2)

= 12

f '' (-2)= 6(-2)

= -12

Also, i am not sure what is meant by the closed interval (-4, 4),