Looking for an example of differentiable function Does there exists a function $f(x)$, which is differentiable in $\mathbb{R}$, and the middle value point is unique, that is, the $\xi$ in formula
$$f'(\xi)=\frac{f(b)-f(a)}{b-a}$$
is unique, having following property:
Connect pairs of arbitrary point $x_0$ and $x_1$ (here $x_0 < x_1$), get $\xi_1$, with connecting $x_0$ and $\xi_1$, get $\xi_2$, with connecting $x_0$ and $\xi_2$, get $\xi_3$, $\dots\,$, satisfy: $\inf\{\xi_n\}>x_0$.
 A: Call $d(a,b) = \frac{f(a)-f(b)}{a-b}$ and $\delta(a,b) =$ the unique $\xi \in ]a;b[$ such that $f'(\xi) = d(a,b)$.
For any $a \in \mathbb{R}$, the function $d_a : b \mapsto d(a,b)$, extended on $b=a$ with $d_a(a) = f'(a)$ is continuous and injective.
To prove the injectiveness, if you have $d(a,b) = d(a,c)$, then $d(a,b) = d(a,c) = d(b,c)$, and if you look at the bigger of the three intervals, you will find one value for $\xi$ in each of the smaller intervalls, which contradicts your hypothesis.
Therefore, for every $a$, $d_a$ is either strictly increasing or strictly decreasing.
From this and from the continuity of $d$, you get that either they are all strictly increasing, either they are all strictly decreasing.
Without loss of generality, suppose they are increasing. Then you get that $f'$ is also strictly increasing : if $a \lt b$, then $f'(a) = d_a(a) \lt d_a(b) = d_b(a) \lt d_b(b) = f'(b)$.
From here, choose $x_0, x_1$ and build the sequence $\xi_n$ as you described.
$\xi_n$ is a strictly decreasing sequence, bounded below by $x_0$ so it converges to some limit $\xi$. If $\xi \gt x_0$ then $f'(\xi) = f'(\lim \xi_n) \le \lim f'(\xi_n) = \lim d(x_0, \xi_{n-1}) = d(x_0, \xi) = f'(\delta(x_0, \xi)) \lt f'(\xi)$, which is impossible.
Thus $\lim \xi_n = x_0$.
