Well, this is going to happen when x is equal to 2 and x is equal to negative 2. Just to be clear, f prime of 2, you get 3 times 4 minus 12, which is equal to 0. And f prime of negative 2 is also, same exact reason, is also equal to 0. So we can say, and I'll switch colors here, that f has critical points at x equals 2 and x equals negative 2. Well, that's fair enough. But we still don't know whether the function takes on minimum values at those points, maximum values at those points, or neither. To figure that out, we have to figure out whether the derivative changes signs around these points.