Are all continuous one one functions differentiable? I was reading about one one functions and found out that they cannot have maxima or minima except at endpoints of domain. So their derivative , if it exists,  must not change it sign , i.e. , the function should be either strictly increasing or strictly decreasing. From this I've a feeling that all continuous one one functions must be differentiable . Is this true?  
 A: Well, you could do something like:
Let $a_i$ be an enumeration of the rationals.
Define $f_i(x)$ as a continuous, nondecreasing function, strictly between $0$ and $1$ which is differentiable everywhere but $a_i$.
Define $g(x) := x + \sum_{i=0}^{\infty}  f_i(x)\times  2^{-i}$
That should give a function that is continuous, but not differentiable at any rational number.  (It's differentiable at every irrational number, though.)
A: $\newcommand{\Reals}{\mathbf{R}}$There exists a strictly increasing continuous function $F:\Reals \to \Reals$ that fails to be differentiable at each rational number.
Let $(a_{k})_{k=1}^{\infty}$ be an enumeration of your favorite dense countable set $A$, such as the set of rational numbers, and $H:\Reals \to \Reals$ the unit step function
$$
H(x) = \begin{cases}
  0 & \text{if $x \leq 0$,} \\
  1 & \text{if $0 < x$.}
\end{cases}
$$
Form the sum of scaled translates
$$
f(x) = \sum_{k=1}^{\infty} 2^{-k} H(x - a_{k}),
$$
and its definite integral
$$
F(x) = \int_{0}^{x} f(t)\, dt.
$$
The following are easy "honors calculus"/elementary analysis exercises:


*

*The function $f$ is strictly positive (in fact, $0 < f(x) < 1$ for all real $x$), strictly increasing (hence Riemann integrable over an arbitrary compact interval), has a jump discontinuity at each point of $A$, and is continuous elsewhere.

*The function $F$ is continuous (as a definite integral), strictly increasing (positive integrand), strictly convex (increasing integrand), and differentiable at $x$ if and only if $f$ is continuous at $x$ (fundamental theorem of calculus, since $f$ has only jump discontinuities), hence non-differentiable at each point of $A$.
A: Not by a long shot. Take, for example, the function
$$f(x) = \begin{cases}x & x\leq 0\\ 2x & x\geq 0\end{cases}$$
Which is continuous and one-to-one on $\mathbb R$, but is not differentiable at $0$.
This is of course just one example, but in general, any time you "stick" two functions together at a point where their derivatives are not equal, like in my example, you can cause the resulting function to have a point at which it is not differentiable.
A: 
$x^{1/3}$ is not differentiable at $0$. See its graph above. It's qualitatively different from the example given by 5xum.
The Cantor function $ +\, x$ is an example of a function that's continuous and one-to-one, but non-differentiable at uncountably many points.
There's a limit to how bad an example can get. The set of points where a continuous one-to-one functions is non-differentiable always has Lebesgue measure $0$.
