Question Number 219404 by golsendro last updated on 24/Apr/25
$$\:\:\mathrm{given}\:\mathrm{g}\left(\mathrm{x}\right)=\:\frac{\mathrm{x}−\mathrm{2023}}{\mathrm{x}−\mathrm{1}} \\ $$$$\:\:\mathrm{find}\:\left(\mathrm{gogogogogog}\right)\left(\mathrm{2024}\right) \\ $$
Commented by kapoorshah last updated on 25/Apr/25
$${g}^{−\mathrm{1}} \left({x}\right)=\frac{{x}−\mathrm{2023}}{{x}−\mathrm{1}} \\ $$$$\:\:\:\:\:\left({g}\:{o}\:{g}\:{o}\:{g}\:{o}\:{g}\:{o}\:{g}\:{o}\:{g}\right)\left(\mathrm{2024}\right) \\ $$$$=\:\left({g}\:{o}\:{g}^{−\mathrm{1}} \:{o}\:{g}\:{o}\:{g}^{−\mathrm{1}} \:{o}\:{g}\:{o}\:{g}^{−\mathrm{1}} \right)\left(\mathrm{2024}\right) \\ $$$$=\:\mathrm{2024} \\ $$$$ \\ $$$$ \\ $$
Commented by kapoorshah last updated on 25/Apr/25
$$\left({g}\:{o}\:{g}^{−\mathrm{1}} \right)\left({a}\right)\:=\:{a} \\ $$
Answered by y0o0o last updated on 24/Apr/25
Answered by mehdee7396 last updated on 26/Apr/25
$$ \\ $$$${f}\left({x}\right)=\frac{{x}−{a}}{{x}−\mathrm{1}}\Rightarrow{f}\left({f}\left({x}\right)\right)=\frac{\frac{{x}−{a}}{{x}−\mathrm{1}}−{a}}{\frac{{x}−{a}}{{x}−\mathrm{1}}−\mathrm{1}} \\ $$$$=\frac{{x}−{a}−{ax}+{a}}{{x}−{a}−{x}+\mathrm{1}}=\frac{\left(\mathrm{1}−{a}\right){x}}{\mathrm{1}−{a}}={x}\:\: \\ $$$$\Rightarrow{f}^{{n}} \left({x}\right)={f}\left({x}\right)\:\:\:;\:{n}=\mathrm{2}{k} \\ $$$$\:\&\:\:{f}^{{n}} \left({x}\right)={x}\:\:\:;\:{n}=\mathrm{2}{k}+\mathrm{1} \\ $$$$ \\ $$$$ \\ $$