Question Number 203305 by Ari last updated on 15/Jan/24
Answered by mr W last updated on 15/Jan/24
$${y}={mx}+{c} \\ $$$$\Rightarrow{x}=\frac{{y}}{{m}}−\frac{{c}}{{m}}\:\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\frac{{x}}{{m}}−\frac{{c}}{{m}} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)={f}\left({x}\right) \\ $$$$\Rightarrow{m}=\frac{\mathrm{1}}{{m}}\:\Rightarrow{m}=\pm\mathrm{1} \\ $$$$\Rightarrow{c}=−\frac{{c}}{{m}}\:\Rightarrow{c}=\mathrm{0}\:{or}\:{c}={any},\:{m}=−\mathrm{1} \\ $$$$\Rightarrow{f}\left({x}\right)={x}\:{or}\:{f}\left({x}\right)=−{x}+{c} \\ $$
Commented by Ari last updated on 15/Jan/24
$${thanks}\:{Mr}! \\ $$