Question Number 64851 by naka3546 last updated on 22/Jul/19
$${Find}\:\:{all}\:\:{solutions}\:\:{of}\:\:{x}\:\:{real}\:\:{numbers}\: \\ $$$$\:\:\:\:\:\:\:\:{f}\:^{−\mathrm{1}} \left({x}\right)\:\:=\:\:{f}\left({x}\right) \\ $$
Commented by naka3546 last updated on 22/Jul/19
$${how}\:\:{many}\:\:{function}\:\:{which}\:\:{satisfy}\:\:{that}\:\:{condition}\:. \\ $$
Commented by mr W last updated on 22/Jul/19
$${i}\:{think}\:{only}\:{one}: \\ $$$${f}\left({x}\right)={x} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)={x} \\ $$
Commented by naka3546 last updated on 23/Jul/19
$${how}\:\:{about}\:\:{f}\left({x}\right)\:=\:−{x}\:+\:{b} \\ $$$${b}\:\in\:\mathbb{R}\:? \\ $$
Commented by mr W last updated on 23/Jul/19
$${that}\:{is}\:{correct}.\:{there}\:{two}\:{possibilities}: \\ $$$${the}\:{line}\:{y}={x}\:{and}\:{each}\:{line}\:{perpendicular} \\ $$$${to}\:{it},\:{i}.{e}.\:{y}=−{x}+{c}. \\ $$
Commented by ~ À ® @ 237 ~ last updated on 25/Jul/19
$$\:\:\:{how}\:{about}\:\:{f}\left({x}\right)=\frac{{a}}{{x}}\:\:\:{on}\:\mathbb{R}^{\ast\:\:\:} \:\:{with}\:\:\:{a}\in\mathbb{R} \\ $$
Commented by ~ À ® @ 237 ~ last updated on 25/Jul/19
$$ \\ $$$$ \\ $$$${If}\:\:{we}\:{considere}\:\:{G}\:\:{as}\:{the}\:{set}\:{of}\:{all}\:{the}\:{function} \\ $$$${we}\:{know}\:{that}\:{when}\:{G}\:{is}\:{equiped}\:{of}\:{the}\:{law}\:{of}\:{composition}\:{it}\:{becomes}\:{a}\:{group} \\ $$$${let}\:\:{H}=\left\{\:{f}\:\in{G}\:\:/\:\:\:\:{f}^{−\mathrm{1}} ={f}\:\:\right\}\:\:\:\:\:{H}\:{is}\:{sub}−{group}\:{not}\:{empty}\:{of}\:{G} \\ $$$${we}\:{just}\:{know}\:{that}\:{the}\:{order}\:{of}\:{H}\:{is}\:\mathrm{2}\:\:\:{but}\:\:{the}\:{question}\:{is}\:{about}\:{cardH}=? \\ $$$${to}\:{be}\:{continued} \\ $$