Question Number 43353 by pieroo last updated on 10/Sep/18
$$\mathrm{Given}\:\mathrm{the}\:\mathrm{functions}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2x}−\mathrm{1}\:\mathrm{and}\:\mathrm{f}\bullet\mathrm{g}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{2}, \\ $$$$\mathrm{find}\:\mathrm{g}\left(\mathrm{x}\right) \\ $$
Commented by maxmathsup by imad last updated on 10/Sep/18
$${we}\:{have}\:{f}\left({x}\right)={y}\:\Rightarrow\mathrm{2}{x}−\mathrm{1}={y}\:\Rightarrow{x}\:=\frac{{y}+\mathrm{1}}{\mathrm{2}}\:\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)\:=\frac{{x}+\mathrm{1}}{\mathrm{2}} \\ $$$${but}\:{fog}\left({x}\right)={x}^{\mathrm{2}} −{x}+\mathrm{2}\:\Rightarrow{f}^{−\mathrm{1}} {of}\:{og}\left({x}\right)\:={f}^{−\mathrm{1}} \left({x}^{\mathrm{2}} −{x}+\mathrm{2}\right)\:\Rightarrow \\ $$$${g}\left({x}\right)\:={f}^{−\mathrm{1}} \left({x}^{\mathrm{2}} −{x}+\mathrm{2}\right)\:=\frac{{x}^{\mathrm{2}} −{x}+\mathrm{2}+\mathrm{1}}{\mathrm{2}}\:=\frac{{x}^{\mathrm{2}} −{x}+\mathrm{3}}{\mathrm{2}}\:. \\ $$
Answered by Joel578 last updated on 10/Sep/18
$${f}\left({x}\right)\:=\:\mathrm{2}{x}\:−\:\mathrm{1} \\ $$$$\:\:\:\:\:\:{f}\left({g}\left({x}\right)\right)\:=\:{x}^{\mathrm{2}} \:−\:{x}\:+\:\mathrm{2} \\ $$$$\mathrm{2}{g}\left({x}\right)\:−\:\mathrm{1}\:=\:{x}^{\mathrm{2}} \:−\:{x}\:+\:\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{g}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} \:−\:{x}\:+\:\mathrm{3}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{2}}{x}\:+\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18
$$\mathrm{2}{g}\left({x}\right)−\mathrm{1}={x}^{\mathrm{2}} −{x}+\mathrm{2} \\ $$$$\mathrm{2}{g}\left({x}\right)={x}^{\mathrm{2}} −{x}+\mathrm{3} \\ $$$${g}\left({x}\right)=\frac{{x}^{\mathrm{2}} −{x}+\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$