# 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". Oct 20, 2011 at 17:54
• Just make some kind of saw-tooth with peeks in 2, 3, 4. 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. 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. Dec 14, 2015 at 22:15
• I will refuse to upvote you question... but instead I gladly upvoted the infamous "W" answer below. Sep 24, 2017 at 6:09

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

• @Shan Think about the $\mathsf{W}$ as the graph of a function.
– Pedro
May 4, 2012 at 0:44
$|x|$ 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...:) 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 $5$. Aug 16, 2013 at 9:45