Find the local minimum and maximum values of the function below. f(x) = 3x4 - 413 - 180x2 + 2 f'(x) = 12x3 - 12x2 - 360x = 12x(x - 6)(x + 5) From the chart Interval 12x 6 x + 5 f'(x) I 6 + + increasing on (6, co) we see that f' (x) changes from negative to positive at - 5, so f( - 5) = is a local minimum value by the First Derivative Test. Similarly, f' (x) changes from negative to positive at 6, so f (6) = is a local minimum value. And, f(0) = is a local maximum value because f'(x) changes from positive to negative at 0. Question Help: Message instructor Submit