Use the 2nd derivative test to find any local extrema for f

Solution

The first derivative is 3x^2 - 12x - 15 = 3(x^2 - 4x - 5) = 3(x - 5)(x+1), so the critical values are at x = -1 and 5 The second derivative is 6x - 12 At x = -1, this equals 6(-1) - 12 = -18 < 0 (the function is strictly concave there), so we have a relative maximum there. At x = 5, this equals 6(5) - 12 = 18 > 0 (the function is strictly convex there), so we have a relative minimum here. We can verify this in two ways. One is that, as the cubic coefficient is positive, the function goes to positive infinity as x goes to infinity and negative infinity as x goes to negative implies the nature of these critical points. Another way is to calculate f(-1) and f(5). f(-1) = -1 - 6 + 15 + 12 = 20 (this is the value of the relative maximum) f(5) = 125 - 150 - 75 + 12 = -88 (this is the value of the relative minimum)