What is the value of $x$, given that $f\left(x\right)+f\left(\frac{1}{1-x}\right)=x$ and $f^{-1}\left(x\right)=2$? I tried to approach the problem by using the equation $f\left(2\right)=x$ but I always get stuck in the middle of the process.
 A: I think a function $f$ doesn't have an inverse in $\mathbb{R}$, This is my attempt : 
We have $$f(x)+f\left(\frac{1}{1-x}\right)=x $$
If we change variable $x \to\ 1-\frac{1}{x}$, we have $$f\left(1-\frac{1}{x}\right)+f(x)=1-\frac{1}{x}$$
Also, $x\to\frac{1}{1-x}$ from original one, we have $$f\left(\frac{1}{1-x}\right)+f\left(1-\frac{1}{x}\right)=\frac{1}{1-x}$$

From these, we can get 
\begin{eqnarray}
f\left(\frac{1}{1-x}\right)-f(x)&=&\frac{1}{1-x}+\frac{1}{x}-1 \\
f\left(\frac{1}{1-x}\right)+f(x)&=&x \,   \text{(This is just the original one.)}
\end{eqnarray}
And you can get original function $f$ from subtracting these.
A: Try sub in $x=x$,$x=\frac{1}{1-x}$,$x=\frac{x-1}{x}$.
you should get  $3$ equations with $3$ variables, namely $f(x),f(\frac{1}{1-x}), f(\frac{x-1}{x})$.
From which you can find $f(x)$ (left as an exercise for you), then sub $x=2$.
A: $$f(2)+f(-1)=2,$$ $$f(-1)+f\left(\frac{1}{2}\right)=-1$$ and $$f\left(\frac{1}{2}\right)+f(2)=\frac{1}{2},$$ which gives $$f(2)+f(-1)+f\left(\frac{1}{2}\right)=\frac{3}{4}$$ and $$\frac{3}{4}-f(2)=-1$$ or $$f(2)=\frac{7}{4}.$$
A: We have
$$x=2\Rightarrow f(2)+f(-1)=2$$
$$x=1/2\Rightarrow f(1/2)+f(2)=1/2$$
$$x=-1\Rightarrow f(-1)+f(1/2)=-1$$
This is a system of three equations with three unknowns. If it helps, we can rewrite it as
$$s+y=2$$
$$s+z=1/2$$
$$y+z=-1$$
where $f(2)=s$, $f(-1)=y$, and $f(1/2)=z$. We can easily solve this by noting that
$$2+1/2=(s+y)+(s+z)=2s+(y+z)=2s-1$$
$$7/2=2s$$
$$s=7/4$$
We conclude $f(2)=7/4$ and therefore $x=7/4$.
