9

There is a function given $f\left(\dfrac{x-1}{x}\right)+ f\left(\dfrac{1}{1-x}\right)= 2- 2x ,f\colon \Bbb R\setminus\{0,1\}\to \Bbb R$ How many fuction exist? I have no idea how to start

Git Gud
  • 31,706
Veritas
  • 963
  • Is $f$ defferentiable? – Haha Nov 09 '13 at 13:04
  • 2
    The problem comes from the KöMaL magazine: http://www.komal.hu/verseny/feladat.cgi?a=feladat&f=B4567&l=en It seems we have a cheater who asks illegal help for solving the contest problems of KöMaL. – G.Kós Mar 04 '14 at 10:43

2 Answers2

16

$f\left(\dfrac{x-1}{x}\right)+ f\left(\dfrac{1}{1-x}\right)= 2- 2x\tag{1}$

Replace $\displaystyle x\rightarrow \frac{x-1}{x} = 1-\frac{1}{x}$ in $(1)$

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

Similarly Replace $\displaystyle x\rightarrow \frac{x-1}{x} = 1-\frac{1}{x}$ in $(2)$

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

Now $(1)-(2)-(3),$ we get

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

So $\displaystyle f(x) = \frac{x^3+x-1}{x^2-x}$ for all $x\in \mathbb{R}\setminus\{0,1\}$

Norbert
  • 58,398
juantheron
  • 56,203
  • Nice. So the thing to notice here was that for $g(x)=\dfrac{x-1}x$ we have $g^2(x)=\dfrac 1{x-1}$ and $g^3=\mathrm{id}$. – Carsten S Nov 30 '13 at 12:56
  • At the end one would have to check whether the found "solution candidate" is actually a solution of the original problem. – Christian Blatter Nov 30 '13 at 12:57
-2

Hint: If you let $g(x)=\dfrac{x-1}x$, then $g^{-1}(x)=\dfrac 1{x-1}$.

So $f(g(x))+f(g^{-1}(x))=2-2x$.

Git Gud
  • 31,706
Khosrotash
  • 25,772