Question Detail

To find the minimum point, we first find the critical points by setting the first derivative to zero.y = x3 + 6x2 - 15x + 5dy/dx = 3x2 + 12x - 15.Set dy/dx = 0:3x2 + 12x - 15 = 0Divide by 3: x2 + 4x - 5 = 0Factor the quadratic: (x + 5)(x - 1) = 0.The critical points are x = -5 and x = 1.Now, use the second derivative test to determine if these points are maxima or minima.d2y/dx2 = d/dx (3x2 + 12x - 15) = 6x + 12.At x = -5: d2y/dx2 = 6(-5) + 12 = -30 + 12 = -18. Since -18 < 0, there is a local maximum at x = -5.At x = 1: d2y/dx2 = 6(1) + 12 = 6 + 12 = 18. Since 18 > 0, there is a local minimum at x = 1.Therefore, the function has a minimum point at x = 1.