How do we write $f(x+2)$ in terms of $f(x)$? $$f : R^+ \rightarrow R^+$$
$$f(x)  = \dfrac{x}{x+1} $$
How do we write $f(x+2)$ in terms of $f(x)$? This is a general question that I wondered how to algebraically write it. 
Regards
 A: I can't quite tell, but I think you're looking for a function $g$ so that
$$g(f(x))=f(x+2).$$
To do this, we find $f^{-1}(y)$ first. If $f(x)=y$, then
$$\frac{x}{x+1}=y\implies x=y(x+1)\implies y=x(1-y)\implies x=\frac{y}{1-y}.$$
So, if $f(x)=y$, then
$$f(x+2)=\frac{x+2}{x+3}=\frac{\frac{y}{1-y}+2}{\frac{y}{1-y}+3}=\frac{y+2(1-y)}{y+3(1-y)}=\frac{2-y}{3-2y}.$$
So, your answer is
$$f(x+2)=\frac{2-f(x)}{3-2f(x)}.$$
A: A more general method:
$$f(x) = \frac{x}{x+1}$$
$$f(x)(x+1) = x$$
$$(f(x)-1)x=-f(x)$$
$$x=\frac{f(x)}{1-f(x)}$$
Hence,
$$f(x+2)=\frac{x+2}{x+3}=\frac{\frac{f(x)}{1-f(x)}+2}{\frac{f(x)}{1-f(x)}+3}=\frac{f(x)+2(1-f(x))}{f(x)+3(1-f(x))}=\frac{2-f(x)}{3-2f(x)}$$
Note: I usually think that this can easily go wrong, so you may want to do some checking: $f(1+2)=\frac{3}{4}$ by the original expression, and indeed $f(1+2)=\frac{2-f(1)}{3-2f(1)}=\frac{2-\frac{1}{2}}{3-2(\frac{1}{2})}=\frac{\frac{3}{2}}{2}=\frac{3}{4}$.
A: If I get your question correctly, then you can try this:
$$f(x+2) = \frac{x+2} {x+3}$$
$$f(x) = \frac{x} {x+1}$$
$$ x = x.f(x) + f(x)$$
$$\implies x(1-f(x)) = f(x)$$
$$\implies x = \frac {f(x)}{1-f(x)}$$
Now substitute this $x$ in $f(x+2)$ and get the desired answer.
