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

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.

• Intuitively, a function is differentiable at a point if the graph of the function at that point is "smooth". Commented Oct 20, 2011 at 17:54
• Just make some kind of saw-tooth with peeks in 2, 3, 4. Commented Oct 20, 2011 at 17:55
• 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. Commented Oct 25, 2011 at 18:05
• 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. Commented Dec 14, 2015 at 22:15
• I will refuse to upvote you question... but instead I gladly upvoted the infamous "W" answer below. Commented Sep 24, 2017 at 6:09

## 3 Answers

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

• @Shan Think about the $\mathsf{W}$ as the graph of a function.
– Pedro
Commented May 4, 2012 at 0:44
• 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.
– quid
Commented Nov 21, 2017 at 16:01

$|x|$ 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.

• $\uparrow$ Change first + sign to a - sign for the infamous $\mathsf{W}$ solution...:) Commented Oct 22, 2014 at 13:36

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

• You're right, I think, because we're considering only one-sided derivatives at $1$ and $5$. Commented Aug 16, 2013 at 9:45
• @SaaqibMahmuud Yes as for the W solution. However, there are more complex functions that lack derivatives everywhere like the Weierstrass function. Commented Aug 22, 2016 at 14:26