1
$\begingroup$

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.

  • 0
    Since on the RHS, degree is 2. I can not have $f(x)$ with degree more than 1.2012-11-14

1 Answers 1

3

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) $

  • 0
    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}. $2012-11-14