The sum of all local maximum and local minimum values of the function f(x) = ₀^x (t-2)(t-4)|t-3| dt is

Step-by-step solution

By the Fundamental Theorem of Calculus, the derivative of the function is f'(x) = (x-2)(x-4)|x-3| The critical points are x=2 , x=3 , and x=4 . Analysing the sign of f'(x) around these points: For x 0 For 2 For 3 For x > 4 , f'(x) > 0 Since f'(x) changes sign from positive to negative at x=2 , there is a local maximum at x=2 . Since f'(x) does not change sign at x=3 , it is a point of inflection. Since f'(x) changes sign from negative to positive at x=4 , there is a local minimum at x=4 . Now, we evaluate the funct