Question Detail

To find the maximum value, first find the critical points by taking the first derivative and setting it to zero:y' = d/dx (x3 + 6x2 - 15x + 5) = 3x2 + 12x - 15Set y' = 0:3x2 + 12x - 15 = 0Divide by 3: x2 + 4x - 5 = 0Factor the quadratic: (x+5)(x-1) = 0Critical points are x = -5 and x = 1.Now, use the second derivative test to determine if these are maximums or minimums:y'' = d/dx (3x2 + 12x - 15) = 6x + 12Evaluate y'' at each critical point:At x = -5: y''(-5) = 6(-5) + 12 = -30 + 12 = -18. Since y''(-5) < 0, x = -5 corresponds to a local maximum.At x = 1: y'' (1) = 6(1) + 12 = 18. Since y''(1) > 0, x = 1 corresponds to a local minimum.To find the maximum value of y, substitute x = -5 into the original function:y(-5) = (-5)3 + 6(-5)2 - 15(-5) + 5y(-5) = -125 + 6(25) + 75 + 5y(-5) = -125 + 150 + 75 + 5y(-5) = 25 + 75 + 5 = 105.The maximum value of y is 105.