Question Number 187714 by 073 last updated on 20/Feb/23
Commented by 073 last updated on 20/Feb/23
$$\mathrm{fog}\left(\mathrm{x}\right)=\mathrm{5x}−\mathrm{1} \\ $$$$\mathrm{gof}^{−\mathrm{1}} \left(\mathrm{x}\right)=\mathrm{6x}+\mathrm{2} \\ $$$$\mathrm{g}\left(\mathrm{x}\right)=? \\ $$$$\mathrm{please}\:\mathrm{solution}?? \\ $$
Commented by 073 last updated on 20/Feb/23
$$\:\mathrm{i}\:\mathrm{need}\:\mathrm{it} \\ $$
Answered by aleks041103 last updated on 20/Feb/23
$${f}\left({g}\left({x}\right)\right)=\mathrm{5}{x}−\mathrm{1} \\ $$$${g}\left({f}^{−\mathrm{1}} \left({x}\right)\right)=\mathrm{6}{x}+\mathrm{2} \\ $$$$\Rightarrow{g}\left({g}\left({x}\right)\right)={g}\left({f}^{−\mathrm{1}} \left({f}\left({g}\left({x}\right)\right)\right)\right)={g}\left({f}^{−\mathrm{1}} \left(\mathrm{5}{x}−\mathrm{1}\right)\right)= \\ $$$$=\mathrm{6}\left(\mathrm{5}{x}−\mathrm{1}\right)+\mathrm{2}=\mathrm{30}{x}−\mathrm{4} \\ $$$${suppose}\:{g}\left({x}\right)={ax}+{b} \\ $$$$\Rightarrow{a}\left({ax}+{b}\right)+{b}=\mathrm{30}{x}−\mathrm{4} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\mathrm{30},\:\left({a}+\mathrm{1}\right){b}=−\mathrm{4} \\ $$$$\Rightarrow{a}=\pm\sqrt{\mathrm{30}},{b}=−\frac{\mathrm{4}}{\mathrm{1}\pm\sqrt{\mathrm{30}}}=−\frac{\mathrm{4}\left(\mathrm{1}\mp\sqrt{\mathrm{30}}\right)}{\mathrm{1}−\mathrm{30}}= \\ $$$$=\frac{\mathrm{4}}{\mathrm{29}}\mp\frac{\mathrm{4}\sqrt{\mathrm{30}}}{\mathrm{29}} \\ $$$$\Rightarrow{g}\left({x}\right)=\sqrt{\mathrm{30}}{x}+\left(\frac{\mathrm{4}}{\mathrm{29}}−\frac{\mathrm{4}\sqrt{\mathrm{30}}}{\mathrm{29}}\right)\:{or}\:{g}\left({x}\right)=−\sqrt{\mathrm{30}}{x}+\left(\frac{\mathrm{4}}{\mathrm{29}}+\frac{\mathrm{4}\sqrt{\mathrm{30}}}{\mathrm{29}}\right) \\ $$
Commented by 073 last updated on 20/Feb/23
$$\mathrm{thanks}\:\mathrm{alot} \\ $$