Find $f(x)$ if for every $x$: $f(x) + f(\frac {2x-3}{x-1}) = x$

Question

I want to find $f(x)$ if for every $x$ (except one and two):

$$f(x) + f\left(\frac {2x-3}{x-1}\right) = x$$

I know that the answer goes something like $g(x)= \frac {2x-3}{x-1} $ and in conclusion $g(g(g(x)))=x$ But i don't know what to do from there on...

Just a clarification to those looking to help, are you trying to write $2x-\frac{3}{x}-1$ or $\frac{2x-3}{x-1}$? — Eleven-Eleven, Feb 21 '16 at 14:50
@A.nishaburi Then you should have written it as $(2x-3)/(x-1)$ or \frac{2x-3}{x-1}. Strictly speaking $2x-3/x-1=2x-\frac{3}{x}-1$. — user236182, Feb 21 '16 at 14:58
@stella biderman That's what the question declares I'm not sure of the reason... — A.nishaburi, Feb 21 '16 at 15:07
@StellaBiderman That's because $\frac{2x-3}{x-1}$ can't equal $2$ (but it can equal anything else). — user236182, Feb 21 '16 at 15:10
@StellaBiderman, If you write out $g(g(x))$, you find that $g(g(x))=\frac{x-3}{x-2}$, so we need to assert that $x\neq 2$ so that the denominator is non-zero. — Ali, Feb 21 '16 at 15:14

score 6 · Accepted Answer · answered Feb 21 '16 at 15:07

Since $g(g(g(x))) = x$, we have $$\begin{array}{crcll} (1) & f(x) &+& f(g(x)) &= x\\ (2) & f(g(x)) &+& f(g(g(x)) &= g(x)\\ (3) & f(g(g(x)) &+& f(x) &= g(g(x)) \end{array} \quad\xrightarrow{(1)-(2)+(3)}\quad 2 f(x) = x - g(x) + g(g(x)) $$ This leads to $\displaystyle\;f(x) = \frac{x^3-4x^2+5x-3}{2x^2-6x+4}$

Find $f(x)$ if for every $x$: $f(x) + f(\frac {2x-3}{x-1}) = x$

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