Find the global maximization

find the global maximization of g(x)=(x^3)-12x on interval -3 less/equal to x less than/equal to 1

y = x^3 - 12x
y' = 3x^2 - 12 = 0
3x^2 = 12
x^2 = 4
x = -2 , 2
y = 16 , -16
y" = 6x
y"(-2) = -12
y"(2) = 12
hence the critical points are local extrema
on the interval: -3 ≤ x ≤ 1 , the global maimma is simply the largest value of g(x):
g(-3) = 9
g(-2) = 16
g(-1) = 11
which is at(-2 , 16)

We want to find the global maximum of g(x) = x^3 - 12x on [-3, 1].

Since the interval [-3, 1] is closed, the global maximum (and minimum) of g(x) on [-3, 1] will occur at either the critical points on [-3, 1], or at [-3, 1]'s endpoints (x = -3 and x = 1).

By differentiating g(x), we have:
g'(x) = 3x^2 - 12x.

Then, the critical points occur when:
g'(x) = 3x^2 - 12x = 3x(x - 4) = 0 ==> x = 0 and x = 4.

x = 0 is the only solution to g'(x) that lies on [-3, 1], so the only critical point on [-3, 1] is x = 0.

Evaluating g(x) at x = -3, x = 0, and x = 1 gives:
g(-3) = 9, g(0) = 0, and g(1) = -11.

Since 9 > 0 > -11, we see that the global maximum of g(x) on [-3, 1] is 9 at x = -3.

I hope this helps!