Let f : R R be the function given by f(x) = |x - 2| + 3|x - 1| + ||x - 2| - 1| . What is the number of points where f is NOT differentiable?

Step-by-step solution

The given function is f(x) = |x - 2| + 3|x - 1| + ||x - 2| - 1| . The possible points of non-differentiability are the critical points where the expressions inside the absolute values become zero: x - 2 = 0 x = 2 x - 1 = 0 x = 1 |x - 2| - 1 = 0 |x - 2| = 1 x = 1 or x = 3 Thus, the critical points to check are x = 1, 2, 3 . Let us define f(x) in the intervals around these points. For x f(x) = (2 - x) - 3(x - 1) + (2 - x - 1) f(x) = 2 - x - 3x + 3 + 1 - x = 6 - 5x The derivative is f'(x) = -5 . For 1 f(x) = (2 - x) +