Given $2xf(x)+(x-3)f(\frac{1}{1-x})=4x^2-10x-\frac{1}{2}$, find $f(x)$.

Given$$2xf(x)+(x-3)f\left(\frac{1}{1-x}\right)=4x^2-10x-\frac{1}{2}$$ Find $f(x)$.

This's the first time I see this kind of question, I have no idea. Please help. Thank you.

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Since on the RHS, degree is 2. I can not have $f(x)$ with degree more than 1. – Dilawar Nov 14 '12 at 12:48

Hint: In equatoin $$2xf(x)+(x-3)f\left(\frac{1}{1-x}\right)=4x^2-10x-\frac{1}{2}\tag{1}$$ make change of variables $x\to 1/(1-x)$ to get $$\frac{2}{1-x}f\left(\frac{1}{1-x}\right)+\frac{3x-2}{1-x}f\left(\frac{x-1}{x}\right)=-\frac{x^2-22x+13}{2(1-x)^2}\tag{2}$$ and make the same change one more time to get $$\frac{2 (x-1)}{x}f\left(\frac{x-1}{x}\right)-\frac{(2 x+1) f(x)}{x}=\frac{4}{x^2}+\frac{2}{x}-\frac{13}{2}\tag{2}$$ From $(1)$, $(2)$ and $(3)$ you can find $$f(x)\qquad f\left(\frac{1}{1-x}\right)\qquad f\left(\frac{x-1}{x}\right)$$
+1 upvote! Though solution is easy to devise (as it involves the famous Moebius transform $x \mapsto (1-x)^{-1}$ with order 3), the answer seems somewhat messy: $$f(x) = \frac{49 x^4-273 x^3+240 x^2+22 x-48}{6 x \left(2 x^3-9 x^2+3 x+2\right)}.$$ (This solution is obtained by using Solve method in Mathematica 8.) – Sangchul Lee Nov 14 '12 at 12:44
You're welcome! Anyway, sorry for a wrong answer above. I must have made a typo. Here is a correct answer: $$f(x) = \frac{3 x^4-31 x^3+60 x^2-6 x-16}{4 x^4-18 x^3+6 x^2+4 x}.$$ – Sangchul Lee Nov 14 '12 at 12:54