Question Number 191579 by 073 last updated on 26/Apr/23
$$\mathrm{f}\left(\mathrm{x}\right)=\mid\mathrm{x}\mid \\ $$$$\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)=? \\ $$$$\mathrm{solution}? \\ $$
Answered by mr W last updated on 26/Apr/23
$${y}={f}\left({x}\right)=\mid{x}\mid \\ $$$$\Rightarrow{x}=\pm{y}\:{with}\:{y}\geqslant\mathrm{0} \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\pm{x}\:{with}\:{x}\geqslant\mathrm{0} \\ $$
Answered by mehdee42 last updated on 26/Apr/23
$${f}\:,\:{in}\:\mathbb{R}\:,{is}\:{not}\:,{one}\:{to}\:{one}.{therefore}\:{it}\:{is}\:{not}\:{invertable}. \\ $$