Given the function g(x) 4x3 42.x2 + 120x, find the first derivative, g (x). g'(x) = Notice that g'(x) = 0 when x = 5, that is, g'(5) = 0. Now, we want to know whether there is a local minimum or local maximum at æ 5, so we will use the second derivative test. Find the second derivative, g''(a). = (x), ,6 Evaluate g''(5). g'"(5)