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Question Number 118960 by bramlexs22 last updated on 21/Oct/20
$$\:{Given}\:{f}:\:{R}\rightarrow{R}\:{and}\:{g}:\:{R}\rightarrow{R} \\ $$$${where}\:{f}\left({x}\right)={x}^{\mathrm{3}} +\mathrm{3}\:{and}\:{g}\left({x}\right)=\mathrm{2}{x}+\mathrm{1}.\:{Find} \\ $$$${the}\:{value}\:{of}\:{f}^{−\mathrm{1}} \left({g}^{−\mathrm{1}} \left(\mathrm{23}\right)\right). \\ $$
Answered by benjo_mathlover last updated on 21/Oct/20
$${let}\:{f}^{−\mathrm{1}} \left({g}^{−\mathrm{1}} \left(\mathrm{23}\right)\right)=\:{q} \\ $$$${then}\:{g}^{−\mathrm{1}} \left(\mathrm{23}\right)\:=\:{f}\left({q}\right)\:{or}\:{g}\left({f}\left({q}\right)\right)=\mathrm{23} \\ $$$${g}\left({q}^{\mathrm{3}} +\mathrm{3}\right)=\mathrm{23}\:\Rightarrow\mathrm{2}\left({q}^{\mathrm{3}} +\mathrm{3}\right)+\mathrm{1}=\:\mathrm{23} \\ $$$$\mathrm{2}{q}^{\mathrm{3}} \:+\:\mathrm{7}\:=\:\mathrm{23}\:,\:{q}\:=\:\sqrt[{\mathrm{3}\:}]{\mathrm{8}}\:=\:\mathrm{2}.\: \\ $$$${Hence}\:{f}^{−\mathrm{1}} \left({g}^{−\mathrm{1}} \left(\mathrm{23}\right)\right)=\:\mathrm{2} \\ $$