This is not an optimization problem, but rather a classic study of the variations of a function.
First, we define the function and plot it over a suitable interval. Then we check that the limits at plus and minus infinity and see that the function is always increasing (which was expected since it is a quartic with positive coefficient for the highest power). Therefore there exist no absolute maximum.
We can find the local maximum and absolute minimum by, first, computing the first derivative and equating to zero so we identify precisely the critical points, then we compute the second derivative and check the sign. We can also plot the first and second derivatives in one plot.
The following example should helps you started (note that you still have some work to do and fill in some blanks).
> f := proc (x) options operator, arrow; 3*x^4+2*x^3-15*x^2+13*x+3 end proc;
4 3 2
x -> 3 x + 2 x - 15 x + 13 x + 3
> plot(f, -4 .. 4);
> limit(f(x), x = infinity);
> limit(f(x), x = -infinity);
> diff(f(x), x);
12 x + 6 x - 30 x + 13
-2.011021846, 0.5763495115, 0.9346723349
> map(f, [-2.011021846, .5763495115, .9346723349]);
[-51.00552571, 6.223792575, 5.969233122]
> plot([12*x^3+6*x^2-30*x+13, diff(12*x^3+6*x^2-30*x+13, x)], x = -4 .. 4);