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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.

Techie5879
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  • "I don't understand the answer" without pinpointing where exactly is the problem? It probably would get closed very quickly as a duplicate. – user10354138 Sep 05 '21 at 14:30
  • I suggest that you analyze a simpler function first , then slowly work your way up to your original function (perhaps $g(x):=|x|+|x-1|+|x-3|$, then $h(x):=g(x)+|x-6|$, then $j(x):=h(x)+|x-10|$, etc.). – Andrew Chin Sep 05 '21 at 14:33
  • https://math.stackexchange.com/a/3013178/668747 This link to a related answer points to a answer that I tried to look at, but I didnt understand the logic why the median of the numbers in the question would be the number of integral points. Also, in the link given in the post, https://math.stackexchange.com/a/3552944/668747, in this answer, I'm not able to understand why we have "$r+2$ regions when drawing stumps at every '$a_k$' " if that makes sense. So I would like a more elaborate answer @user10354138 – Techie5879 Sep 05 '21 at 14:33
  • @AndrewChin Would that be particularly helpful here? I did try that approach and I can see that the graph is of a bit different nature depending upon if the number of terms is even or odd. For example, if the number of terms is even (for example, $|x|+|x-1|$ which is two terms, the graph is constant between $[0,1]$, if the number of terms is odd, example, $|x|+|x-1|+|x-3|$, the nature of the graph changes and is no longer constant for any value and is a bit "tapered" at $x=1$), – Techie5879 Sep 05 '21 at 14:39
  • Also, if I introduce another term, now there being 4 terms, $|x|+|x-1|+|x-3|+|x-6|$, the graph is constant but still only between $[1,3]$ – Techie5879 Sep 05 '21 at 14:41
  • I know that I only need to find the interval where $f(x)$ is constant, but I'm struggling to exactly get to that point – Techie5879 Sep 05 '21 at 14:47
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    @Techie5879 Do you realize that it will constant exactly in the "middle"? – Azlif Sep 05 '21 at 14:49
  • @Azlif I did realize that after trying out some graphs in desmos but how do I arrive at that mathematically is my question.... – Techie5879 Sep 05 '21 at 14:51
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    It is constant because the $x$ terms cancel out. Note that, for example, $|x - 1|$ could be $x - 1$ or $-(x -1)$, depends on the interval. The same goes for the other.

    Anther example is that $$|x| + |x - 1| + |x - 3| + |x - 6| = -x -(x - 1) + (x - 3) + (x -6)$$ exactly in $[1,3]$.

     – Azlif Sep 05 '21 at 14:52
  • @Azlif Yes and I now see that the middle terms are the 51st and 52nd terms here which mean $|x-(1+2+....+50)|$ and $x-(1+2+...+51)$ It just clicked right now, thanks. – Techie5879 Sep 05 '21 at 14:54
  • I would highly suggest a graphical approach. See the graphs of easier ones like $|x|$ , $|x| + |x-1|$ and you will see a pattern which will quickly solve the problem. – An_Elephant Nov 08 '22 at 18:08

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