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.
 A: 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.
A: 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 $$\begin{align}f(g(x))&=\frac{1}{2}\left(1+\frac{1}{1-x}-\frac{1}{\frac{1}{(1-x)}\left(1-\frac{1}{1-x}\right)}\right)\\
&=\frac12\left(1+\frac1{1-x}-\frac{(1-x)^2}{(1-x)-1}\right)\\
&=\frac12\left(x-1+\frac1{x(1-x)}\right)
\end{align}$$
A: You didn't mention the functional equation holds for all $x\in\Bbb R_{\neq 1}$. I'm assuming it.
Let $y=\frac{1}{1-x}$. Then $y$ reaches all real values except $0$, so:$$f\left(\frac{y-1}{y}\right)+f(y)=\frac{y-1}{y},\, \forall y\in\Bbb R_{\neq0 }\tag{1}$$
Let $y=\frac{t-1}{t}$. Then $t=\frac{1}{1-y}$ reaches all real values except $0,1$ (because $y$ can't reach $0$):
$$f\left(\frac{1}{1-t}\right)+f\left(\frac{t-1}{t}\right)=\frac{1}{1-t},\,\forall t\in\Bbb R_{\neq 0,1}$$
$$\iff f\left(\frac{t-1}{t}\right)=\frac{1}{1-t}-f\left(\frac{1}{1-t}\right),\, \forall t\in\Bbb R_{\neq 0,1}$$
Substitute this into $(1)$:
$$\frac{1}{1-t}-f\left(\frac{1}{1-t}\right)+f(t)=\frac{t-1}{t},\, \forall t\in\Bbb R_{\neq 0,1}$$
$$\iff f\left(\frac{1}{1-t}\right)=\frac{1}{1-t}+f(t)-\frac{t-1}{t},\, \forall t\in\Bbb R_{\neq 0,1}\tag{2}$$
Your original equation says:
$$f(t)+f\left(\frac{1}{1-t}\right)=t,\, \forall t\in\Bbb R_{\neq 1}$$
Subsitute $(2)$ into it:
$$2f(t)+\frac{1}{1-t}-\frac{t-1}{t}=t,\,\forall t\in\Bbb R_{\neq 0,1}$$
$$\iff f(t)=\frac{t+\frac{t-1}{t}-\frac{1}{1-t}}{2}=\frac{t^3-t+1}{2(t-1) t},\,\forall t\in\Bbb R_{\neq 0,1}$$
(WolframAlpha agrees this is a solution).
Now to find $f(0), f(1)$, note that the original equation only cares about $f(0), f(1)$ when $x=0$, so $f(0),f(1)$ can be any real numbers such that $f(1)=-f(0)$.
