Find the absolute maximum and absolute minimum values of the function

f(x) = x3 + 6x2 – 63x + 11

f'(x) = 3x2 + 12x – 63

Putting f' (x) = 0, we get

3x2 + 12x – 63 = 0

x2 + 4x – 21 = 0

(x + 7) (x – 3) = 0

x = – 7, 3 – critical points of this function

(A) Interval = [-8, 0]. The critical point in this interval is x = -7

(B) Interval = [-5, 4]. The critical point in this interval is x = 3

(C) Interval = [-8, 4]. Critical points in this interval are: x = -7, 3