Question Number 187027 by yaslm last updated on 12/Feb/23
Answered by Farhadazizi last updated on 12/Feb/23
$${fog}\left({x}\right)={f}\left({g}\left({x}\right)\right)=\frac{\mathrm{2}{g}\left({x}\right)−\mathrm{4}}{{g}\left({x}\right)}=\frac{\mathrm{2}\sqrt{{x}−\mathrm{1}}−\mathrm{4}}{\:\sqrt{{x}−\mathrm{1}}} \\ $$$${gof}\left({x}\right)={g}\left({f}\left({x}\right)\right)=\sqrt{{f}\left({x}\right)−\mathrm{1}}=\sqrt{\frac{{x}−\mathrm{4}}{{x}}} \\ $$$${f}\left({x}\right)=\frac{\mathrm{2}{x}−\mathrm{4}}{{x}}=\mathrm{2}−\frac{\mathrm{4}}{{x}}\Rrightarrow{f}\left({x}\right)−\mathrm{2}=−\frac{\mathrm{4}}{{x}} \\ $$$$−\frac{\mathrm{1}}{{f}\left({x}\right)−\mathrm{2}}=\frac{{x}}{\mathrm{4}}\Rrightarrow{x}=\frac{−\mathrm{4}}{{f}\left({x}\right)−\mathrm{2}}\Rrightarrow{f}^{−\mathrm{1}} \left({x}\right)=\frac{−\mathrm{4}}{{x}−\mathrm{2}} \\ $$$$ \\ $$$${g}\left({x}\right)=\sqrt{{x}−\mathrm{1}}\Rightarrow\left({g}\left({x}\right)\right)^{\mathrm{2}} +\mathrm{1}={x} \\ $$$$\:\:\:\:\Rightarrow{g}^{−\mathrm{1}} \left({x}\right)={x}^{\mathrm{2}} +\mathrm{1} \\ $$
Commented by yaslm last updated on 12/Feb/23
thanks sir,