Let f(x) and g(x) be twice differentiable functions on R satisfying f''(x) - g''(x) = 24x . If the function f(x) - g(x) has a local minimum at x = 1 with a minimum value of -3 , then the local maximum value of f(x) - g(x) is equal to :

Step-by-step solution

Let h(x) = f(x) - g(x) . We are given that h''(x) = 24x . Integrating with respect to x , we get h'(x) = 12x^2 + C . Since h(x) has a local minimum at x = 1 , we must have h'(1) = 0 . Substituting x = 1 , 12(1)^2 + C = 0 C = -12 . Thus, h'(x) = 12x^2 - 12 . Integrating again with respect to x , we obtain h(x) = 4x^3 - 12x + D . The minimum value at x = 1 is given as -3 , so h(1) = -3 . Substituting x = 1 , 4(1)^3 - 12(1) + D = -3 -8 + D = -3 D = 5 . So, h(x) = 4x^3 - 12x + 5 . To find the local maximum, we set h'(x