Question Number 1280 by Rasheed Soomro last updated on 19/Jul/15
$$\:{f}\:\left(\:\frac{\mathrm{1}}{{f}\left({x}\right)}\right)={f}\left({x}\right)\: \\ $$$${f}\left({x}\right)=? \\ $$
Commented by 123456 last updated on 19/Jul/15
$${f}\left({x}\right)=\frac{\mathrm{1}}{{x}} \\ $$
Commented by prakash jain last updated on 19/Jul/15
]=[(c^2 +ad)x+d(b+c)](ax+b) ac(b+c)=a(c^2 +ad)⇒bc=ad b^2 c+acd+abd+acd=ad(b+c)+(c^2 +ad)b ⇒b^2 c+acd=c^2 b+abd⇒bc=ad d(b^2 +ad)=bd(b+c) ⇒db^2 +ad^2 =b^2 d+bcd⇒bc=ad f(x) = ((ax+b)/(cx+((bc)/a))) = ((a^2 x+ab)/(acx+bc)) This functional equation will also have multiple solutions.](https://www.tinkutara.com/question/Q1288.png)
$${f}\left({x}\right)=\frac{{ax}+{b}}{{cx}+{d}} \\ $$$${f}\left(\frac{\mathrm{1}}{{f}\left({x}\right)}\right)=\frac{{a}\frac{{cx}+{d}}{{ax}+{b}}+{b}}{{c}\frac{{cx}+{d}}{{ax}+{b}}+{d}}=\frac{{ax}+{b}}{{cx}+{d}} \\ $$$$\left.\left[{a}\left({b}+{c}\right){x}+\left({b}^{\mathrm{2}} +{ad}\right)\right]\left({cx}+{d}\right)\right]=\left[\left({c}^{\mathrm{2}} +{ad}\right){x}+{d}\left({b}+{c}\right)\right]\left({ax}+{b}\right) \\ $$$${ac}\left({b}+{c}\right)={a}\left({c}^{\mathrm{2}} +{ad}\right)\Rightarrow{bc}={ad} \\ $$$${b}^{\mathrm{2}} {c}+{acd}+{abd}+{acd}={ad}\left({b}+{c}\right)+\left({c}^{\mathrm{2}} +{ad}\right){b} \\ $$$$\:\:\:\:\Rightarrow{b}^{\mathrm{2}} {c}+{acd}={c}^{\mathrm{2}} {b}+{abd}\Rightarrow{bc}={ad} \\ $$$${d}\left({b}^{\mathrm{2}} +{ad}\right)={bd}\left({b}+{c}\right) \\ $$$$\:\:\:\:\Rightarrow{db}^{\mathrm{2}} +{ad}^{\mathrm{2}} ={b}^{\mathrm{2}} {d}+{bcd}\Rightarrow{bc}={ad} \\ $$$${f}\left({x}\right)\:=\:\frac{{ax}+{b}}{{cx}+\frac{{bc}}{{a}}}\:=\:\frac{{a}^{\mathrm{2}} {x}+{ab}}{{acx}+{bc}} \\ $$$$\mathrm{This}\:\mathrm{functional}\:\mathrm{equation}\:\mathrm{will}\:\mathrm{also} \\ $$$$\mathrm{have}\:\mathrm{multiple}\:\mathrm{solutions}. \\ $$