Answer :
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.
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