Let f(x) = x^3 - 3Lx^2 + 9x + 2 . Let L₀ be the maximum value of L for which f(x) is strictly increasing on the interval [1, 3] . If a function g(x) is defined as g(x) = 2x^3 - 3L₀^2 x^2 + 50 , then the local minimum value of g(x) is

Step-by-step solution

For f(x) to be strictly increasing on [1, 3] , we must have f'(x) 0 for all x [1, 3] . f'(x) = 3x^2 - 6Lx + 9 0 x^2 - 2Lx + 3 0 2Lx x^2 + 3 Since x [1, 3] , x > 0 , so we can divide by 2x : L 1 2 (x + 3 x ) Let h(x) = 1 2 (x + 3 x ) . For the inequality to hold for all x [1, 3] , L must be less than or equal to the minimum value of h(x) on [1, 3] . h'(x) = 1 2 (1 - 3 x^2 ) Setting h'(x) = 0 gives x^2 = 3 x = 3 (since x > 0 ). The critical point x = 3 lies in the interval [1, 3] . The minimum value is: h( 3 ) = 1 2