Solving a functional equation $f(x)+f\left(\frac{1}{1-x}\right)=x$

Question

I was given the following homework: list all functions $f:\mathbb{R}\setminus\{0,1\}\rightarrow\mathbb{R}$ such that $f(x)+f\left(\frac{1}{1-x}\right)=x$. And obviously have I no idea what should I do here. A step-by-step explanation welcome.

Answer 1

Let $g(x)=\frac{1}{1-x}$. Then, we have the system \begin{align*} f(x)+f[g(x)]&=x;\\ f[g(x)]+f[g(g(x))]&=g(x);\\ f[g(g(x))]+f[g(g(g(x)))]&=g(g(x)). \end{align*} But $g(g(g(x)))=x$ so the above is a system of 3 equations in 3 unknowns $\{f(x),f[g(x)],f[g(g(x))]\}$. Solve for $f(x)$.

Answer 2

The system matrix is $\left( {\begin{array}{*{20}{c}} 1&1&0&& x\\ 0&1&1&& {\frac{1}{{1 - x}}}\\ 1&0&1&& {\frac{{x - 1}}{x}} \end{array}} \right)$

The solution of the system is $\left\{ {\frac{{1 - x + {x^3}}}{{2\left( { - 1 + x} \right)x}},\frac{{ - 1 + x - 2{x^2} + {x^3}}}{{2\left( { - 1 + x} \right)x}},\frac{1}{2}\left( {1 + \frac{1}{{1 - x}} - \frac{1}{x} - x} \right)} \right\}$ so $f(x)={\frac{{1 - x + {x^3}}}{{2\left( { - 1 + x} \right)x}}}$