<s>[INST] Find $f$ if $ f(x)+f\left(\frac{1}{1-x}\right)=x $ Find $f$ if
$$
f(x)+f\left(\frac{1}{1-x}\right)=x
$$
I think, that I have to find x that $f(x) = f\left(\frac{1}{1-x}\right)$
I've tried to put x which make $x = \frac{1}{1 - x}$, but this equation has no roots in real numbers.
 [/INST] If $g(x)=\frac{1}{1-x}$ then verify that $g(g(g(x)))=x$.
Then:
$$\begin{align}
f(x)+f(g(x))&=x\\
f(g(x)) + f(g(g(x))) &= g(x)=\frac{1}{1-x}\\
f(g(g(x))) + f(x) &= g(g(x))=\frac{x-1}{x}
\end{align}$$
So this gives three linear equations, so solve for $f(x)$.
Subtract the second row from the first:
$$f(x)-f(g(g(x)))=x-\frac{1}{1-x}$$
Then add this to the third:
$$2f(x)=x-\frac{1}{1-x}+\frac{x-1}{x}=x+1-\frac{1}{x(1-x)}$$
Then $$\begin{align}f(g(x))&=\frac{1}{2}\left(1+\frac{1}{1-x}-\frac{1}{\frac{1}{(1-x)}\left(1-\frac{1}{1-x}\right)}\right)\\
&=\frac12\left(1+\frac1{1-x}-\frac{(1-x)^2}{(1-x)-1}\right)\\
&=\frac12\left(x-1+\frac1{x(1-x)}\right)
\end{align}$$
</s>