Construct a function which is continuous in $[1,5]$ but not differentiable at $2, 3, 4$.

This question is just after the definition of differentiation and the theorem that if $f$ is finitely derivable at $c$, then $f$ is also continuous at $c$. Please help, my textbook does not have the answer.

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    $\begingroup$ Intuitively, a function is differentiable at a point if the graph of the function at that point is "smooth". $\endgroup$ – JavaMan Oct 20 '11 at 17:54
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    $\begingroup$ Just make some kind of saw-tooth with peeks in 2, 3, 4. $\endgroup$ – Jonas Teuwen Oct 20 '11 at 17:55
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    $\begingroup$ Intuitively, a function is continuous if you can "walk" on the graph and it is differentiable if you can see where you came from and where you are going. $\endgroup$ – AD. Oct 25 '11 at 18:05
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    $\begingroup$ But... a flight of stairs is discontinous and non-differentiable and yet you can walk up and down one step at a time. A better analogy would maybe that you could use a very tiny wheel to roll smoothly along it without any bumps. They see me rollin, they be differenti-ating. $\endgroup$ – mathreadler Dec 14 '15 at 22:15
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    $\begingroup$ I will refuse to upvote you question... but instead I gladly upvoted the infamous "W" answer below. $\endgroup$ – Matemáticos Chibchas Sep 24 '17 at 6:09

$|x|$ is continuous, and differentiable everywhere except at 0. Can you see why?

From this we can build up the functions you need: $|x-2| + |x-3| + |x-4|$ is continuous (why?) and differentiable everywhere except at 2, 3, and 4.

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    $\begingroup$ $\uparrow$ Change first + sign to a - sign for the infamous $\mathsf{W}$ solution...:) $\endgroup$ – Qmechanic Oct 22 '14 at 13:36

$$\ \ \ \ \mathsf{W}\ \ \ \ $$

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    $\begingroup$ @Shan Think about the $\mathsf{W}$ as the graph of a function. $\endgroup$ – Pedro Tamaroff May 4 '12 at 0:44
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    $\begingroup$ Avoid commenting. There are close to hundred deleted comments in this thread about: it not being the Lambert W function, it not being W for Weiertrass, the merits of this type of answer in general, and so on. If you do not understand what it is meant please read the comment above, if you still do not understand it read please read another explanation and the comment there. If you want to voice your opinion on the merits of this contribution, please, just make an effort to avoid it. Chances are what you want to say was said already multiple times. $\endgroup$ – quid Nov 21 '17 at 16:01

How about $f(x) = \max(\sin(n\pi x),0)$ or perhaps $g(x) = |\sin(n\pi x)|$?

  • $\begingroup$ You're right, I think, because we're considering only one-sided derivatives at $1$ and $5$. $\endgroup$ – Saaqib Mahmood Aug 16 '13 at 9:45
  • $\begingroup$ @SaaqibMahmuud Yes as for the W solution. However, there are more complex functions that lack derivatives everywhere like the Weierstrass function. $\endgroup$ – AD. Aug 22 '16 at 14:26

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