If f(x) = x³

consider the function f(x) = x³ + ax² + bx + c

Then f’(x) = 3x² + 2ax + b

It is given that f(x) is maximum at x = – 1

f’(– 1) = 3(– 1)2 + 2a(– 1) + b = 0

f’(– 1) = 3 – 3a + b = 0 ……(1)

it is given that f(x) is minimum at x = 3

f’(x) = 3(3)2 + 2a(3) + b = 0

f’(3) = 27 + 6a + b = 0 …… (2)

solving equation (1) and (2) we have

a = – 3 and b = – 9

since f’(x) is independent of constant c, it can be any real number