Question Number 11238 by Joel576 last updated on 18/Mar/17
$${f}\left(\mathrm{1}\:−\:\mathrm{2}{x}\right)\:=\:{g}\left({x}\:+\:\mathrm{3}\right) \\ $$$${f}^{−\mathrm{1}} \left({x}\right)\:=\:? \\ $$$$ \\ $$$$\left(\mathrm{A}\right)\:\mathrm{7}\:−\:\mathrm{2}{g}^{−\mathrm{1}} \left({x}\right) \\ $$$$\left(\mathrm{B}\right)\:\mathrm{2}{g}^{−\mathrm{1}} \left({x}\right)\:+\:\mathrm{7} \\ $$$$\left(\mathrm{C}\right)\:\mathrm{2}{g}^{−\mathrm{1}} \left({x}\right)\:−\:\mathrm{5} \\ $$$$\left(\mathrm{D}\right)\:\mathrm{5}\:−\:\mathrm{2}{g}^{−\mathrm{1}} \left({x}\right) \\ $$
Answered by mrW1 last updated on 18/Mar/17
$${u}={f}\left(\mathrm{1}−\mathrm{2}{x}\right)={g}\left({x}+\mathrm{3}\right) \\ $$$$\Rightarrow\mathrm{1}−\mathrm{2}{x}={f}^{−\mathrm{1}} \left({u}\right)\:\:\:\:\:…\left({i}\right) \\ $$$$\Rightarrow{x}+\mathrm{3}={g}^{−\mathrm{1}} \left({u}\right)\:\:\:\:\:\:…\left({ii}\right) \\ $$$$\mathrm{2}×\left({ii}\right)+\left({i}\right)\:\Rightarrow\mathrm{7}={f}^{−\mathrm{1}} \left({u}\right)+\mathrm{2}{g}^{−\mathrm{1}} \left({u}\right) \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({u}\right)=\mathrm{7}−\mathrm{2}{g}^{−\mathrm{1}} \left({u}\right) \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\mathrm{7}−\mathrm{2}{g}^{−\mathrm{1}} \left({x}\right) \\ $$
Commented by Joel576 last updated on 18/Mar/17
$${thank}\:{you}\:{very}\:{much} \\ $$