We'll need to find the 1st and 2nd derivatives of F(x) to answer that question.
F '(x) = -4x^3 - 27x^2 - 48x - 16 You must set this = to 0 and solve for the roots (which we call "critical values).
F "(x) = -12x^2 - 54x - 48
Now suppose you've found the 3 critical values. We use the 2nd derivative to determine which of these is associated with a max or min of the function F(x).
Just supposing that 4 were a critical value, we ask whether or not we have a max or min of F(x) there:
F "(x) = -12x^2 - 54x - 48 becomes F "(4) = -12(4)^2 - 54(4) = -192 - 216 Because F "(4) is negative, the graph of the given function opens down at x=4, and so we have a relative max there. (Remember that "4" is only an example, and that you must find all three critical values and then test each one in F "(x).
