F(x)=-x^(4)-9x^(3)-24x^(2)-16x what is the relative maxium and minimum?

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).