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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