Question Number 10694 by ABD last updated on 23/Feb/17
Answered by mrW1 last updated on 23/Feb/17
$${f}\left(\mathrm{3}{x}^{\mathrm{4}} +{x}^{\mathrm{3}} \right)=\mathrm{6}{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} −\mathrm{3}=\mathrm{2}\left(\mathrm{3}{x}^{\mathrm{4}} +{x}^{\mathrm{3}} \right)−\mathrm{3} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{2}{x}−\mathrm{3} \\ $$$${y}=\mathrm{2}{x}−\mathrm{3} \\ $$$${x}=\frac{{y}+\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\frac{{x}+\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$$${g}^{−\mathrm{1}} \left({x}\right)=\frac{{x}+\mathrm{2}}{\mathrm{3}} \\ $$$${y}=\frac{{x}+\mathrm{2}}{\mathrm{3}} \\ $$$${x}=\mathrm{3}{y}−\mathrm{2} \\ $$$$\Rightarrow{g}\left({x}\right)=\mathrm{3}{x}−\mathrm{2} \\ $$$$ \\ $$$${g}\left({f}^{−\mathrm{1}} \left({x}\right)\right)=\mathrm{3}×\left(\frac{{x}+\mathrm{3}}{\mathrm{2}}\right)−\mathrm{2}=\frac{\mathrm{3}{x}+\mathrm{5}}{\mathrm{2}} \\ $$$${f}^{−\mathrm{1}} \left({g}^{−\mathrm{1}} \left({x}\right)\right)=\frac{\frac{{x}+\mathrm{2}}{\mathrm{3}}+\mathrm{3}}{\mathrm{2}}=\frac{{x}+\mathrm{11}}{\mathrm{6}} \\ $$
Commented by ABD last updated on 23/Feb/17
Commented by mrW1 last updated on 23/Feb/17
$${thank}\:{you}! \\ $$