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