Question Number 165376 by MikeH last updated on 31/Jan/22
$$\mathrm{Verify}\:\mathrm{wether}\:{f}\:\mathrm{is}\:\mathrm{invertible}\: \\ $$$${f}\:\left({x}\right)\:=\:\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{3}} \\ $$
Answered by Rasheed.Sindhi last updated on 31/Jan/22
$${f}\:\left({x}\right)\:=\:\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{3}} \\ $$$${y}=\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{3}} \\ $$$$\mathrm{1}+\mathrm{2}{x}=\sqrt[{\mathrm{3}}]{{y}}\: \\ $$$${x}=\frac{\sqrt[{\mathrm{3}}]{{y}}\:−\mathrm{1}}{\mathrm{2}} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=\frac{\sqrt[{\mathrm{3}}]{{x}}\:−\mathrm{1}}{\mathrm{2}} \\ $$
Commented by aleks041103 last updated on 01/Feb/22
$${If}\:{x}\in\mathbb{R},\:{g}\left({x}\right)={x}^{\mathrm{3}} \:{is}\:{bijective}. \\ $$$${For}\:{all}\:{x},\:{l}\left({x}\right)=\mathrm{1}+\mathrm{2}{x}\:{is}\:{bijective}. \\ $$$${Therefore}\:{f}={g}\circ{l}\:{is}\:{a}\:{composition} \\ $$$${of}\:\mathrm{2}\:{bijective}\:{functions}\Rightarrow{f}\:{is}\:{bijective}. \\ $$$$\Rightarrow{f}\:{is}\:{invertible},\:{but}\:{only}\:{on}\:\mathbb{R}. \\ $$