How prove or disprove $f$ Non-differentiable points countably infinite Today,my frend ask this follow question:and I consider sometime,and I can't solve it.
I hope see someone can help me
Question:

let $f$ is  continuous strictly increasing function,
prove or disprove :the $f$  Non-differentiable points  countably infinite？

since  the contably infinite define:
http://en.wikipedia.org/wiki/Countable_set
and I find somebook,and I found this 
This follow  famous example(Pringsheim) is $f$ is   continuous strictly increasing function,and there exsit a point not differentiable
$$\begin{cases}
x(1+\dfrac{1}{3}\sin{(\ln{x^2})}&x\neq 0\\
0&x=0
\end{cases}$$
we can show $f$ is strictly increaing and continuous on $R$,but $x=0$,$f$ is not differentable.
But My question:
there exsit $f$ such strictly increaing and continuou,but have  infinite point not differentable?
I can't have coumexample,Thank you very much!
 A: Take a look at this Essay on non-differentiability points of monotone functions, by Dave L Renfro, for a wealth of excellent information on the topic. 
In particular, Renfro mentions that if $f$ is monotone, then the set of points where $f'(x)$ does not exist has Lebesgue measure zero, and this is sharp even if $f$ is continuous, in the sense that for any set $E\subseteq[a,b]$ having measure zero there is a strictly increasing continuous function $f:[a,b]\to\mathbb R$ with $f'(x)=+\infty$ at all $x$ in $E$.
The example he refers to, from A. M. Bruckner,  J. B. Bruckner, and B. S. Thomson, Real analysis (downloadable here), appears as Theorem 7.9 in page 460 in their current edition. Here is a brief sketch: Let $E$ be a set of measure zero, and (for $n=1,2,\dots$) let $G_n$ be open sets covering $E$ with measure $\lambda(G_n)<1/2^n$. If $g_n(x)=\lambda(G_n\cap[a,x])$, then $g_n$ is nondecreasing, continuous, and $0\le g_n\le 1/2^n$. Now let 
 $$ g(x) =\sum_n g_n(x), $$
so $g$ is increasing, $0\le g\le 1$, and $g$ is continuous (by the $M$-test). If $f(x)=g(x)+x$, then $f$ is strictly increasing. Also, at any point of $E$, $f'(x)=\infty$ since $g'(x)=\infty$, because $g_n'(x)=1$ for all $n$.  
A: The Cantor function has uncountably many points of nondifferentiability. By adding $x$, you get it to be strictly increasing.
