Construct a function $f_1:\mathbb{R}\to\mathbb{R}$ with the following properties or show that no such function exists:
$1.$ $f_1$ is differentiable everywhere except one point $x_1.$
$2.$ Define $f_2 : \mathbb{R}\setminus\{x_1\} \to \mathbb{R}$ as $f_2(x) := $ derivative of $f_1$ at $x.$ This $f_2$ must be differentiable everywhere in its domain except one point $x_2.$
$3.$ Define $f_3 : \mathbb{R}\setminus\{x_1,\;x_2\} \to \mathbb{R}$ as $f_3(x) := $ derivative of $f_2$ at $x.$ This $f_3$ must be differentiable everywhere in its domain except one point $x_3.$
$\vdots$
$n.$ Define $f_n : \mathbb{R}\setminus\{x_1, \cdots, x_{n-1}\} \to \mathbb{R}$ as $f_n(x) := $ derivative of $f_{n-1}$ at $x.$ This $f_n$ must be differentiable everywhere in its domain except one point $x_n.$
$\vdots$
(Note that we do not stop at any $n.$)
I found this question in a collection of extra questions for my Calculus course.
I started out by trying something along the lines of $f(x) = \lim_{n\to\infty}\sum_{i=1}^n (x-i)^i|x-i|$, but the function itself isn't defined anywhere and I couldn't figure out how to fix it with minimum effort.
So next, I tried out something that might actually be defined somewhere such as $$f(x) = \lim_{n\to\infty}\sum_{i=1}^n \frac{\left(\frac{2}{\pi}\arctan(x-i)\right)^i\left(\frac{2}{\pi}\arctan|x-i|\right)}{(i+1)!}$$
which is defined for $x \in \mathbb{R}$, but I wasn't able to prove continuity or differentiability. Intuitively, I feel that since it's a sum of continuous functions, it should be continuous, but I'm not sure whether this intuition is correct because it's an infinite sum.
I asked the person whose website I found the question on (another student), and he said that he wasn't sure whether such a function was even possible.
Any help would be appreciated!