TITLE: Find $f$ if $ f(x)+f\left(\frac{1}{1-x}\right)=x $
QUESTION [7 upvotes]: 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.

REPLY [3 votes]: In case you're a bit miffed by the "trickiness" of the solution, here's how I came upon the answer by playing around with the definition of $f$. The key, as in both answers above, is to realize there's some cyclicality induced by $\frac{1}{1-x}$.
My first instinct was to play around with $x=0$, from which we get $f(0)+f(1)=0$. In order to get something out of this, though, we look for another way to express $f(0)$ or $f(1)$. However, to get the left-hand term to be $f(1)$ requires plugging an undefined point ($x=1$) in the right-hand term, and getting $f(0)$ out of the right-hand term requires taking an infinite limit of the left-hand term. So that seems like a dead end.
I moved on to try $x=-1$ since this is also simple to play with. We get
$$f(-1)+f(\frac{1}{2})=-1$$
Just for kicks, I tried plugging in $x=\frac{1}{2}$ since the numbers in the first equation were pretty palatable:
$$f(\frac{1}{2}) + f(2)=\frac{1}{2}$$
Again, not bad -- nice, pretty numbers. So I tried once more:
$$f(2)+f(-1) = 2$$
And now we're on to something. We note that we got $f(-1)$ to show up again, and that in the three equations we've written, there are only three unknowns -- namely, $f(-1), f(\frac{1}{2})$, and $f(2)$. This means we can solve for these three values (and luckily it's a simple linear form).
Now we begin to wonder whether we happened upon this by accident or whether we've found a more general phenomenon -- it appears that churning $-1$ through $\frac{1}{1-x}$ recursively brought us back to $-1$; is this true for other starting values?
Indeed it is -- $\frac{1}{1-\frac{1}{1-x}}=-\frac{1-x}{x}$ and $\frac{1}{1-(-\frac{1-x}{x})}=x$, so we'll get the system of three unknowns we saw in our concrete example for any (valid) starting value of $x$.
The rest is details, which are sufficiently covered in the other answers.