Let f (x) = ax^3 + 6x^2 + bx + 4. determine the constants a and b so that f has a relative minimum at x = - 1 and a relative maximum at x = 2. show workings

F' (x) = 3ax^2 + 12x + b ... (1)

Applying the formula for solving any quadratic leads to two equations:

(-12 + sq rt (144 - 12ab)) / 6a = 2 ... (2)

(-12 - sq rt (144 - 12ab)) / 6a = - 1 ... (3)

Adding equations (2) and (3), we get:

-24/6a = 1

a = - 4

Plugging the value a = - 4 into equation (2) and solving for b, we get:

b = 24.

The answer is: a = - 4, b = 24.