You have $ f(x+1)=\frac{x+1}{x-1} (f(x)-f(1))$. Thus $ f(x+2)=\frac{x+2}{x} (\frac{x+1}{x-1}(f(x)-f(1))-f(1))$. But $ f(x+2)=\frac{x+2}{x} (f(x)-f(2))$. Eliminate $ f(x+2)$ among these relations and you will find a first degree equation in $f(x)$. From here it is easy.

the second equation Mr harazi is: $f(x+2)=\frac{x+2}{x-2}(f(x)-f(2))$

This is just a typo mistake. And the important is that he killed it in two lines. He proved that its Easy problem.

yeah it's a quit simple prob tell me AYMANE did you participe int the TST's this YEAR? :

Here is an other solution: 1-tend $y$ to $-x$ then$f(x)-f(-x)=2xf'(0)$ take $x=1$then$f'(0)=\frac{f(1)-f(-1)}{2}$ 2-take$x+y=1$ then $f(x)-f(1-x)=(2x-1)f(1)=f(x)-f(-(1-x))=f(x)-(f(x-1)-2a(x-1))$ ==>$f(x)-f(x-1)=(2x-1)f(1)-2a(x-1)$ 3-take $y=-1$ then$f(x-1)=\frac{x-1}{x+1}(f(x)-f(-1))$ ==>$f(x)-\frac{x-1}{x+1}(f(x)-f(-1))=(2x-1)f(1)-2a(x-1)$ ==>$f(x)=(f(1)-a)x^2+ax$ Hope did not make mistake on computing

i'm ok with it if the function is continous!

AYMANE wrote: Here is an other solution: 1-tend $y$ to $-x$ then$f(x)-f(-x)=2xf'(0)$ take $x=1$then$f'(0)=\frac{f(1)-f(-1)}{2}$ 2-take$x+y=1$ then $f(x)-f(1-x)=(2x-1)f(1)=f(x)-f(-(1-x))=f(x)-(f(x-1)-2a(x-1))$ ==>$f(x)-f(x-1)=(2x-1)f(1)-2a(x-1)$ 3-take $y=-1$ then$f(x-1)=\frac{x-1}{x+1}(f(x)-f(-1))$ ==>$f(x)-\frac{x-1}{x+1}(f(x)-f(-1))=(2x-1)f(1)-2a(x-1)$ ==>$f(x)=(f(1)-a)x^2+ax$ Hope did not make mistake on computing but there's NO condition about continous and differentiable

This is a tricky problem !! it seems easy , but under conditions exam , nobody knows how to deal with boring computations !! this problem needs just dare and^patience and time then it's really easy , however , only 2 students out of 11 could solve it . (i got caught with it) and i say to Ayman that he can't use derivatives , and it remains impossible proove continuity!!

Hi every body,Sorry for the delate i were on training Math-sipo,Chang hwan and Wisemaniac are right to tell me i am not allowed to use continuity, but some time like in this problem we can use it indirectly and this was what i want to transmite to you: i wrote:$f'(0)=\frac{f(x)-f(-x)}{2x}=a$ lets prove it: Put $g(x)=\frac{f(x)-f(-x)}{2x}$ for $x<>0$ (E) then $a=g(1)$ we have $g$ even. 1-$g(x+y)=\frac{x}{x-y}g(x)-\frac{y}{x-y}g(y)$ for $x-y<>0$ Take $y=-2x$ in (E)==>$g(-x)=g(x)=\frac{x}{3x}g(x)+\frac{2x}{3x}g(-2x)$==>$g(x)=g(2x)$ By induction easy to prove:$g(nx)=g(x)$ Take $x=1$==>$g(n)=g(1)=g(-n)$ Take $x=\frac{1}{n}$==>$g(\frac{1}{n})=g(1)$==>$g(\frac{m}{n})=g(1)$ So $g(x)=g(1)$ for all rationnals. 2-Take in (E) $x$ real non rational and$y=\frac{-1}{2^n}$==>$g(x-\frac{1}{2^n})=\frac{x}{x+\frac{1}{2^n}}g(x)+\frac{\frac{1}{2^n}}{x+\frac{1}{2^n}}g(\frac{-1}{2^n})=\frac{x}{x+\frac{1}{2^n}}g(x)+\frac{\frac{1}{2^n}}{x+\frac{1}{2^n}}g(1)$ Then $\lim_{n\to\infty}g(x-\frac{1}{2^n})=g(x)$ Do the same then $\lim_{n\to\infty}g(x+\frac{1}{2^n})=g(x)$ So $g$ is continious And thus $g(x)=g(1)$ for all reals. If you wont use continuity you can take$x=z-\frac{1}{2^n}$and$y=-z+\frac{2}{2^n}$ where $z$ is real non rational, use them in (E) and tend $n$ to infinity then $g(z)=g(1)$ And we done

good job Mr.AYMANE that's great!

Hi math-sipo. this exercice is quite easy I solved it in the exam in 2 mn. I'll be in Mexico this year:MAR2.

ok! mon ami so i will say good luck for u all! moroccan team! u can do it! allez Maroc!!!!

harazi wrote: You have $ f(x+1)=\frac{x+1}{x-1} (f(x)-f(1))$. Thus $ f(x+2)=\frac{x+2}{x} (\frac{x+1}{x-1}(f(x)-f(1))-f(1))$. But $ f(x+2)=\frac{x+2}{x} (f(x)-f(2))$. Eliminate $ f(x+2)$ among these relations and you will find a first degree equation in $f(x)$. From here it is easy. Is the answer $f(n) = \frac{n}{2}f(2)-\frac{n(n-2)}{n-1}f(1)$?

Probably, I did not make the computations since the two equations show our function is linear and then it's easy.

I was just wondering...

I believe that the answer is $f(x)=2ax-ax^2$ for any $a$ (derived from harazi's solution).

Bojan Basic wrote: I believe that the answer is $f(x)=2ax-ax^2$ for any $a$ (derived from harazi's solution). I don't understand. Is my result wrong? Can you show your work?

Sorry, I made a mistake, but here are the calculations. From \[\frac{x+2}x\left(\frac{x+1}{x-1}\bigl(f(x)-f(1)\bigr)-f(1)\right)=\frac{x+2}{x-2}\bigl(f(x)-f(y)\bigr)\] you get \[(x^2-x-2)\bigl(f(x)-f(1)\bigr)-(x^2-3x+2)f(1)=(x^2-x)\bigl(f(x)-f(2)\bigr)\] from where is easily to find \[f(x)=2xf(1)-x^2f(1)+\frac{x^2f(2)-xf(2)}2\]Now put $x=1$ and $x=2$ to obtain $f(1)=a$ and $f(2)=b$ for any $a$ and $b$, and hence follows \[f(x)=\left(\frac b 2-a\right)x^2+\left(2a-\frac b 2\right)x\] or, if $\frac b 2-a=k$, \[f(x)=kx^2+(a-k)x\]

Assuming $f(x)$ is continuous and differentiable over the reals, let us differentiate the functional equation with respect to $x$ and $y$ to obtain: $f(x) - f(y) + (x+y)f'(x) = f(x+y) + (x-y)f'(x+y)$ (i), $f(x) - f(y) - (x+y)f'(y) = -f(x+y) + (x-y)f'(x+y)$ (ii) and subtracting (ii) from (i) gives: $(x+y)[f'(x) + f'(y)] = 2f(x+y)$ (iii). Next, we differentiate (iii) with to respect to $x$ and $y$ to get: $f'(x) + f'(y) + (x+y)f''(x) = 2f'(x+y)$ (iv), $f'(x) + f'(y) + (x+y)f''(y) = 2f'(x+y)$ (v) and equating (iv) with (v) ultimately yields $f''(x) = f''(y) =$ constant, or $f(x) = Ax^2 + Bx + C$ (vi) for real constants $A, B, C$. Substituting (iv) back into the original function equation now produces: $(x+y)[Ax^2 + Bx + C - Ay^2 - By - C] = (x-y)[A(x+y)^2 + B(x+y) + C]$; or $Ax^3 + Bx^2 -Axy^2 - Bxy + Ax^2y + Bxy - Ay^3 - By^2 = Ax^3 + 2Ax^2y + Axy^2 + Bx^2 + Bxy + Cx - Ax^2y - 2Axy^2 - Ay^3 - Bxy - By^2 - Cy$; which entirely simplifies into $0 = C(x - y) \Rightarrow C = 0$ for all real $x$ and $y$. Therefore, the solution is the set $f(x) = Ax^2 + Bx$ for all $x \in \mathbb{R}$.