Find the ab

It is given that f (x) = x³, x ∈ [–2, 2]

⇒ f’(x) = 3x²

Now, f’(x) = 0

⇒ x = 0

Now, we evaluate the value of f at critical point x = 0 and at end points of the interval [-2, 2].

f(0) = 0

f(-2) = (-2)³ = -8

f(2) = (2)³ = 8

Therefore, we have the absolute maximum value of f on [-2, 2] is 8 occurring at x = 2.

And, the absolute minimum value of f on [-2, 2] is -8 occurring at x =-2.