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Question Number 1280 by Rasheed Soomro last updated on 19/Jul/15

 f ( (1/(f(x))))=f(x)   f(x)=?

$$\:{f}\:\left(\:\frac{\mathrm{1}}{{f}\left({x}\right)}\right)={f}\left({x}\right)\: \\ $$$${f}\left({x}\right)=? \\ $$

Commented by 123456 last updated on 19/Jul/15

f(x)=(1/x)

$${f}\left({x}\right)=\frac{\mathrm{1}}{{x}} \\ $$

Commented by prakash jain last updated on 19/Jul/15

f(x)=((ax+b)/(cx+d))  f((1/(f(x))))=((a((cx+d)/(ax+b))+b)/(c((cx+d)/(ax+b))+d))=((ax+b)/(cx+d))  [a(b+c)x+(b^2 +ad)](cx+d)]=[(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.

$${f}\left({x}\right)=\frac{{ax}+{b}}{{cx}+{d}} \\ $$$${f}\left(\frac{\mathrm{1}}{{f}\left({x}\right)}\right)=\frac{{a}\frac{{cx}+{d}}{{ax}+{b}}+{b}}{{c}\frac{{cx}+{d}}{{ax}+{b}}+{d}}=\frac{{ax}+{b}}{{cx}+{d}} \\ $$$$\left.\left[{a}\left({b}+{c}\right){x}+\left({b}^{\mathrm{2}} +{ad}\right)\right]\left({cx}+{d}\right)\right]=\left[\left({c}^{\mathrm{2}} +{ad}\right){x}+{d}\left({b}+{c}\right)\right]\left({ax}+{b}\right) \\ $$$${ac}\left({b}+{c}\right)={a}\left({c}^{\mathrm{2}} +{ad}\right)\Rightarrow{bc}={ad} \\ $$$${b}^{\mathrm{2}} {c}+{acd}+{abd}+{acd}={ad}\left({b}+{c}\right)+\left({c}^{\mathrm{2}} +{ad}\right){b} \\ $$$$\:\:\:\:\Rightarrow{b}^{\mathrm{2}} {c}+{acd}={c}^{\mathrm{2}} {b}+{abd}\Rightarrow{bc}={ad} \\ $$$${d}\left({b}^{\mathrm{2}} +{ad}\right)={bd}\left({b}+{c}\right) \\ $$$$\:\:\:\:\Rightarrow{db}^{\mathrm{2}} +{ad}^{\mathrm{2}} ={b}^{\mathrm{2}} {d}+{bcd}\Rightarrow{bc}={ad} \\ $$$${f}\left({x}\right)\:=\:\frac{{ax}+{b}}{{cx}+\frac{{bc}}{{a}}}\:=\:\frac{{a}^{\mathrm{2}} {x}+{ab}}{{acx}+{bc}} \\ $$$$\mathrm{This}\:\mathrm{functional}\:\mathrm{equation}\:\mathrm{will}\:\mathrm{also} \\ $$$$\mathrm{have}\:\mathrm{multiple}\:\mathrm{solutions}. \\ $$

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