What Mark M. meant (I think) was, use calculus to look for possible global minima and maxima just from the function itself. So differentiate once: f'(x) = 3x^2 - 12x . This will have zeros (== possible global maxima or minima, or maybe just an inflection point) at x=0 and x=4 (substitute into the f'(x) equation to check this!). Since these two values are both outside the stated interval, all you have to do is look at the original function values at the two interval endpoints; one will be the local (i.e. over the interval) minimum and the other the local maximum (do you see why? b/c the function MUST be monotonically increasing or decreasing over the interval, since it can only change that attribute at the global minimum or maximum, and these points are OUTSIDE the interval).
You have to calculate those two (local maximum and minimum) values yourself. But make sure you understand the logic in the above paragraph, thoroughly. Sketch the function if you need to, to do this.