One-to-one differentiable function with zero derivative in every neighborhood Is it possible to have a one-to-one, differentiable function $f: (a,b) \to \mathbb{R}$ such that, at some point $c \in (a,b)$, we have $f'(c) = 0$, and every neighborhood of $c$ has a point $d \neq c$ such that $f'(d) = 0$ as well?
 A: Let $F(x)=x^{2}\sin^{2}(1/x)$ for $x\ne 0$ and $F(0)=0$.
$F$ is integrable, now we let $f(x)=\int_{0}^{x}F(t)dt$.
Note that $f(x)-f(y)=\int_{y}^{x}F(t)dt>0$ for $x>y$.
A: Yes. As the example by user284331 suggests, just consider any continuous function $F: (a,b) \to \mathbb{R}$ such that:
(1) $F(c) = 0$ for some $c \in (a,b)$.
(2) For every $\delta > 0$, there is $d \in (c-\delta) \cup (c+\delta)$ such that $F(d) = 0$.
(3) $\int_x^y F(t)\,dt > 0$ for every $x, y \in (a,b)$ with $x < y$.
By the fundamental theorem of calculus, $f(x) = \int_c^x F(t)\,dt$ satisfies $f’ = F$, and $f$ is one-to-one because $x < y$ implies $f(y) - f(x) = \int_x^y F(t)\,dt > 0$.
Basically, $F$ is a continuous function that is zero at $c$ and at infinitely many points close to $c$, and $F$ has positive integral between any two points. There are many possible examples of such a function $F$.
(E.g. consider $F(x) = |x|g(x)$ at $c = 0$, where $g$ is the "zig-zag" function defined by: $g(0) = 0$; for each $k \in \mathbb{Z}$, $g(1/k) = 0$ if $k$ is even and $g(1/k) = 1$ if $k$ is odd; and in between $1/k, 1/(k+1)$, connect the two points on the graph by a straight line.)
