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?

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  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

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