$$f : R^+ \rightarrow R^+$$

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

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.


  • $\begingroup$ Literally divide $\frac{f(x+2)}{f(x)}$ and you will get what you want. $\endgroup$ – Lucas Henrique Dec 10 '18 at 5:45
  • 3
    $\begingroup$ One way to do it is $f(x+2) = \frac{x+2}{x+3} \times \frac{x+1}{x} \times f(x)$. It's not really clear what you're looking for. $\endgroup$ – platty Dec 10 '18 at 5:45
  • $\begingroup$ If I get you right, you want a product/sum independent of x as a factor but only as an argument, because, as in my first comment, it would be obvious. $\endgroup$ – Lucas Henrique Dec 10 '18 at 5:47
  • $\begingroup$ Something like $f(x+2) = f(f^{-1}(f(x))+2) = \frac{f(x)-2}{2f(x)-3}$??? $\endgroup$ – achille hui Dec 10 '18 at 5:50

I can't quite tell, but I think you're looking for a function $g$ so that


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


So, your answer is


  • $\begingroup$ Oh...I was too slow typing the answer :) $\endgroup$ – tonychow0929 Dec 10 '18 at 5:53

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 $f(1+2)=\frac{2-f(1)}{3-2f(1)}=\frac{2-\frac{1}{2}}{3-2(\frac{1}{2})}=\frac{\frac{3}{2}}{2}=\frac{3}{4}$.


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.

  • 2
    $\begingroup$ Your sign of $x$ is wrong. That should be $1-f(x)$ instead of $f(x)-1$. $\endgroup$ – tonychow0929 Dec 10 '18 at 5:56
  • 1
    $\begingroup$ @tonychow0929 I am sorry, thank you for that :) $\endgroup$ – PradyumanDixit Dec 10 '18 at 5:56

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