# 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

• Literally divide $\frac{f(x+2)}{f(x)}$ and you will get what you want. – Lucas Henrique Dec 10 '18 at 5:45
• 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. – platty Dec 10 '18 at 5:45
• 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. – Lucas Henrique Dec 10 '18 at 5:47
• Something like $f(x+2) = f(f^{-1}(f(x))+2) = \frac{f(x)-2}{2f(x)-3}$??? – 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

$$g(f(x))=f(x+2).$$

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

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

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

• Oh...I was too slow typing the answer :) – 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.

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