Question

What values of a and b make

f(x) = x^3 + ax^2 + bx

have a local max at x=-1 and a local min at x=3?

Answer 1

First of all, the derivative has to be zero at both x values. That requires
3x^2 + 2ax + b = 0
3 -2a +b = 0
27 + 6a + b = 0
8a = 24
a = 3
b = 2a -3 = 3
This already determines a and b, but we had better make sure the max and min requirements are correct. Otherwise there is no solution.

The second derivative is
y''(x) = 6x + 2a. This is negative at x = -1 (therefore a relative maximum) and postive when x = 3 (therefore a relative minimum).

We are OK with a = b = 3