Let f(x) be a function defined as f(x) = cases x^3 - 3x + 5, & x 1 3(x-2)^2, & 1 3 cases The sum of all local maximum and local minimum values of f(x) is

Step-by-step solution

First, we check the continuity of f(x) at the boundary points x=1 and x=3 . At x=1 : f(1) = 1^3 - 3(1) + 5 = 3 . Also, _ x 1^+ 3(x-2)^2 = 3(-1)^2 = 3 . Thus, f(x) is continuous at x=1 . At x=3 : f(3) = 3(3-2)^2 = 3 . Also, _ x 3^+ (-x^2 + 8x - 12) = -9 + 24 - 12 = 3 . Thus, f(x) is continuous at x=3 . Next, we find the derivative f'(x) for non-boundary points: f'(x) = cases 3x^2 - 3, & x 3 cases Critical points within the intervals: For x For 1 For x > 3 : -2x + 8 = 0 x = 4 Now we check the sign changes of f'(x) at