<s>[INST] How to find $ f(x)$ if $f(1-f(x))=x$ for all $x$ $\in \mathbb{R}$ How can I determine $ f(x)$ if $f(1-f(x))=x$ for all real $x$? 
I have already recognized one problem caused from this: it 
follows that $ f(f(x))=1-x $, which is discontinuous. So how can I construct a function $f(x)$?
Best regards and thanks,
John
 [/INST] This is a partial answer.
We know that $f(x)$ is invertible, because $f^{-1}(x)=1-f(x),$ from the original; from here we get the very interesting relationship of $f(x)+f^{-1}(x)=1.$ Suppose we try to find out what $f(0)$ is (set it equal to $a$). By repeated alternating applications of $f$ and the equation $f^{-1}(x)=1-f(x),$ we wind up with the following interesting table:
$$
\begin{array}{c|c|c}
x &f(x) &f^{-1}(x) \\ \hline
0 &a &1-a \\ \hline
1-a &0 &1 \\ \hline
1 &1-a &a \\ \hline
a &1 &0
\end{array}
$$
One more step gets you where you started. In studying this table, we see that if $f$ and $f^{-1}$ are to be well-defined, we cannot have $a=0, 1/2,$ or $1$. We get a similar table if we start off with $x=-1:$
$$
\begin{array}{c|c|c}
x &f(x) &f^{-1}(x) \\ \hline
-1 &b &1-b \\ \hline
1-b &-1 &2 \\ \hline
2 &1-b &b \\ \hline
b &2 &-1
\end{array}
$$
From here we find that $b\not=-1, -3, 2, 1/2.$ Yet another table generates when we start with $x=-2:$
$$
\begin{array}{c|c|c}
x &f(x) &f^{-1}(x) \\ \hline
-2 &c &1-c \\ \hline
1-c &-2 &3 \\ \hline
3 &1-c &c \\ \hline
c &3 &-2
\end{array}
$$
From this we get that $c\not=3, 1/2, -2.$ This generalizes to the following table:
$$
\begin{array}{c|c|c}
x &f(x) &f^{-1}(x) \\ \hline
1-n &m &1-m \\ \hline
1-m &1-n &n \\ \hline
n &1-m &m \\ \hline
m &n &1-n
\end{array}
$$
From here, we can see that $n=1/2$ forces $m=1/2,$ which would be consistent in this table. So $f(1/2)=1/2.$
Moving on, we can see that the following are true:
\begin{align*}
f(1-x)&=y \\
f(1-y)&=1-x \\
f(x)&=1-y \\
f(y)&=x.
\end{align*}
Combining two of these equations yields $f(1-x)=1-f(x)$. Differentiating yields $f'(1-x)=f'(x).$ 
These are mostly negative results, obviously. My hope is that perhaps these ideas might spur someone else on to a solution.
</s>