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

Question

Given function $y=f(x)$ such that $$f\left(\frac{x-3}{x+1}\right)+f\left(\frac{x+3}{x+1}\right)=x \quad\forall x\neq -1$$ find $f(x)$ and $f(2007)$.

Answer 1

With $$f(x) = \frac{x+3}{x-1}-\frac{3}{2}$$ we have $$a := f\left(\frac{x-3}{x+1}\right) = -x-\frac{3}{2}$$ and $$b := f\left(\frac{x+3}{x+1}\right)=2x+\frac{3}{2}$$ Therefore the sum is $a+b=x$ and $f(2007)=-\frac{999}{2006}.$

Edit regarding the comments: My function $f(x)\,$ is a solution of the functional equation (FE). The road to it was as follows: If the FE is valid for all $x\ne -1\;$ and you plug in large $x$ into the FE, you see that there should be a singularity at $x=1.\;$ Making the ansatz $f(x) = \frac{g(x)}{x-1}+\cdots$ and playing arround with Maple arrives at the given function.

If the OP has a different function solving the FE, this only shows that there are at least two solutions.

Answer 2

I get an answer for your question but I am not sure that my answer is unique. I think the exclusivity of solution is still to be proved.

Firstly I make some basic transformation to you origin equation: $$f(1-\frac{4}{1+x})+f(1+\frac{2}{1+x})=x$$ Let $y=\frac{2}{1+x}$, the equation can be transformed into: $$f(1-2y)+f(1+y)=\frac{2}{y}-1 \;where \;y\neq 0$$ Again let $m=y+1$: $$f(3-2m)+f(m)=\frac{3-m}{m-1}$$

Here I assume that $f(m)$ has the form of $\frac{cm+d}{am+b}$ after the investigation of above equation

This assumption is mostly based on experience rather than strict proof. So it's possible that there exists some solutions where $f(m)$ is not in this form.

We can reorganize and apply the form to the equation $$(m-1)((c(3-2m)+d)(am+b)+(cm+d)(a(3-2m))+b))=(3-m)(am+b)(a(3-2m)+b)$$ This is a very long equation. But we can just specific the value of m since this equation holds for all of m.

When $m=1$ $$LHS=0=-(a+b)^2=RHS$$ We can conclude that $a=-b$

When $m=3$, with the knowledge that $a=-b$ $$b(d+9c)=0$$ Because $b \ne 0$(if $b=0$, denominator of $f(x)$ would be $0$ which is impossible) ,so $d=-9c$

We apply $a=-b$ and $d=-9c$ to the original equation. $$f(3-2m)+f(m)=\frac{2cm-6c}{a(m-1)}=\frac{2m-3}{m-1}$$ So $a=1$, $b=-1$, $c=-0.5$, $d=4.5$ And one solution for your question is $$f(x)=\frac{-0.5x+4.5}{x-1}$$

Answer 3

Below we show the following

Claim. Let $y_0\in\mathbb R$ be arbitrary. Then there exists a continuous function $f\colon\mathbb R\setminus\{1\}\to \mathbb R$ such that $$f\left(\frac{x-3}{x+1}\right) +f\left(\frac{x+3}{x+1}\right)=x\qquad\text{for all $x\ne -1$}$$ and $f(2007)=y_0$.

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Let $g_0\colon[1,4]\to\mathbb R$ be a continuous function with $$\tag0g_0(4)=12+g_0(1).$$ Given a continuous function $g_k\colon[4^{-k},4^{k+1}]\to\mathbb R$, $k\ge 0$, with $$\tag1 g_k(4x)=12x+g_k(x)\qquad\text{whenever $x,4x\in[4^{-k},4^{k+1}]$},$$ we can define $g_{k+1}\colon[4^{-k-1},4^{k+2}]\to\mathbb R$ by $$ g_{k+1}(x)=\begin{cases}g_k(x)&\text{if $4^{-k}\le x\le4^{k+1}$}\\ g_k(4x)-12x&\text{if $4^{-k-1}\le x\le4^{k}$}\\ g_k(x/4)+3x&\text{if $4^{-k+1}\le x \le 4^{k+2}$} \end{cases}$$ Indeed, the cases cover all of the desired domain, i.e. $$[4^{-k},4^{k+1}]\cup [4^{-k-1},4^{k}] \cup [4^{-k+1},4^{k+2}]= [4^{-k-1},4^{k+2}]$$ if $k\in\mathbb N_0$ and $(1)$ guarantees that the overlapping between the cases is not harmful. Moreover, whenever $x,4x\in[4^{-k-1},4^{k+2}]$, one of the three cases covers both $x$ and $4x$ so that $(1)$ with $k\leftarrow k+1$ holds. Since $g_{k}$ is the restriction of $g_{k+1}$ to its domain, we can define $g\colon(0,\infty)\to\mathbb R$ $$ g(x)=g_k(x)\qquad \text{for any $k\in\mathbb N_0$ with $x\in[4^{-k},4^{k+1}]$}.$$ Then $g$ is continuous and $$\tag2g(4x)=12x+g(x)\qquad \text{for all $x>0$}.$$ Define $h\colon\mathbb R\setminus\{0\}\to\mathbb R$ as $$ h(x)=\begin{cases}g(x)&\text{if $x>0$}\\2x-1-g(-x/2)&\text{if $x<0$}\end{cases}$$ Then we have $$\tag3h(2x)=4x-1-h(-x)\qquad\text{for all $x\ne0$}. $$ Indeed, this follows for $x>0$ from $$h(2x)+h(-x)=g(2x)+2(-x)-1-g(x/2)\stackrel{(2)}=6x+g(x/2)-2x-1-g(x/2) =4x-1$$ and for $x<0$ from $$ h(2x)+h(-x)=4x-1-g(-x)+g(-x)=4x-1.$$

Finally define $f\colon\mathbb R\setminus\{1\}\to \mathbb R$ as $$ f(x)=h\left(\frac1{x-1}\right).$$ Then for all $x\ne-1$, we have $$\tag4\begin{align}f\left(\frac{x-3}{x+1}\right) +f\left(\frac{x+3}{x+1}\right)&=h\left(-\frac{x+1}{4}\right)+h\left(\frac{x+1}{2}\right)=x. \end{align}$$ Vice versa, any $f$ like this allows us to define $h(x)=f(1+\tfrac 1x)$ for $x\ne0$ such that $(4)$ implies $(3)$; then define $g(x)=h(x)$ for $x>0$ by which $(3)$ implies $(2)$; finally define $g_0(x)=g(x)$ for $1\le x\le 4$, by which $2)$ implies $0)$.

So this gives us a large family of solutions, one for each choice of continuous $g_0$ with $(0)$. Specifically, for any $x_0\in[1,4]$ and $y_0\in\mathbb R$, there exists $g_0$ with $g_0(x_0)=y_0$. From $f(2007)=h(\frac1{2006})=g(\frac1{2006})$ we find using $(2)$ that $f(2007)$ is a constant plus $g_0(\frac{4^6}{2006})$ and therefore $f(2007)$ can assume any arbitrary value.

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Remark: We cannot expect the solutions to be "more unique" by adding assumptions such as that $f$ is continuous on all of $\mathbb R$: That would require that $\lim_{x\to\pm\infty}h(x)$ exists, which is not possible according to $(3)$.

Answer 4

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

$f\left(1-\dfrac{4}{x+1}\right)+f\left(1+\dfrac{2}{x+1}\right)=x$

$f\left(1-\dfrac{4}{x}\right)+f\left(1+\dfrac{2}{x}\right)=x-1$

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

$f(3-2x)+f(x)=\dfrac{2}{x-1}-1~......(1)$

$f(3-2(3-2x))+f(3-2x)=\dfrac{2}{3-2x-1}-1$

$f(4x-3)+f(3-2x)=-\dfrac{1}{x-1}-1~......(2)$

$(2)-(1)$ :

$f(4x-3)-f(x)=-\dfrac{3}{x-1}$

$f(4(4^x+1)-3)-f(4^x+1)=-\dfrac{3}{4^x+1-1}$

$f(4^{x+1}+1)-f(4^x+1)=-3\times4^{-x}$

$f(4^x+1)=\Theta(x)+4^{1-x}$ , where $\Theta(x)$ is an arbitrary periodic function with unit period

$f(x)=\Theta(\log_4(x-1))+\dfrac{4}{x-1}$ , where $\Theta(x)$ is an arbitrary periodic function with unit period