I am given the function $$f(x) = |x| + |x - 1| + |x - 3| + |x - 6| + \cdots + |x - (1 + 2 + \cdots + 101)|$$ for all $x \in \mathbb R$
I am required to find the number of points at which the function is non differentiable, and number of integral points at which $f(x)$ is minimum. I have found the number of points where this function is non differentiable (which can also be thought of as the number of points where the graph has a sharp turn), namely at $x=0,1,3,..., (1+2+...+101)$ which turns out to be $102$.
But, I cannot come up for a logic on how to find the number integral points where this is a minimum. I did see this 1 year old thread on the exact question (Finding the no. of points at which $f$ is non-differentiable and the no. of integral points for which $f$ is minimum.) but I can't seem to understand quite properly the logic involved so I thought of reopening the question in a new thread. Any help would be appreciated.