Question Number 142674 by Snail last updated on 03/Jun/21
$${If}\:{f}\left({x}\right)=\frac{{x}−\mathrm{3}}{{x}+\mathrm{1}}\:{and}\:{h}\left({x}\right)={f}\left({f}\left({f}\left({f}\left(..\mathrm{2022}\:{times}\left({f}\left({x}\right)\right)\right)\right)\right)\right) \\ $$$${then}\:\:\:{h}\left(\mathrm{1}\right)\:=? \\ $$
Answered by iloveisrael last updated on 04/Jun/21
$${f}\left({f}\left({x}\right)\right)=\frac{\frac{{x}−\mathrm{3}}{{x}+\mathrm{1}}−\mathrm{3}}{\frac{{x}−\mathrm{3}}{{x}+\mathrm{1}}+\mathrm{1}}\:=\:\frac{{x}−\mathrm{3}−\mathrm{3}{x}−\mathrm{3}}{{x}−\mathrm{3}+{x}+\mathrm{1}} \\ $$$${f}\left({f}\left({x}\right)\right)=\:\frac{−\mathrm{2}{x}−\mathrm{6}}{\mathrm{2}{x}−\mathrm{2}}\:=\:\frac{−{x}−\mathrm{3}}{{x}−\mathrm{1}}={f}^{−\mathrm{1}} \left({x}\right) \\ $$$$\underset{{h}\left({x}\right)={x}} {\underbrace{{f}\left({f}\left({f}\left(....\left({f}\left({f}\left({x}\right)\right)\right)\right)}={x}}\right.\right. \\ $$$${h}\left({x}\right)={x}\Rightarrow{h}\left(\mathrm{1}\right)=\mathrm{1} \\ $$