Question Number 173624 by mr W last updated on 15/Jul/22
$${find}\:{f}\left({x}\right)\:{such}\:{that}\:{f}'\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right). \\ $$
Commented by infinityaction last updated on 15/Jul/22
$$\:\:\:{f}'\left({x}\right)\:=\:{f}^{−\mathrm{1}} \left({x}\right) \\ $$$$\:\:\:{f}\left\{{f}'\left({x}\right)\right\}\:=\:{x}\:\:\:\:,\:\:{f}\left({x}\right)\:\:\:=\:\:{ax}^{{b}} \\ $$$$\:\:\:{f}\left({ab}\:{x}^{{b}−\mathrm{1}} \right)\:=\:\:{x} \\ $$$$\:\:\:\:{a}\centerdot{a}^{{b}} {b}^{{b}} \centerdot{x}^{{b}^{\mathrm{2}} −{b}} \:\:=\:\:{x} \\ $$$$\:\:\:\:{x}^{\underset{\mathrm{0}} {\underbrace{{b}^{\mathrm{2}} −{b}−\mathrm{1}}}\:} \:=\:\:\frac{\mathrm{1}}{\underset{\mathrm{1}} {\underbrace{{a}\centerdot{a}^{{b}} \:{b}^{{b}} }}} \\ $$$$\:\:{b}\:\:=\:\:\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\:=\:\phi\:\:\:{then}\:\:{a}\:\:=\:\:^{\phi} \sqrt{\frac{\mathrm{1}}{\phi}} \\ $$$$\:\:\:\:\:{f}\left({x}\right)\:\:\:=\:\:\:^{\phi} \sqrt{\frac{\mathrm{1}}{\phi}}\:{x}^{\phi} \\ $$
Commented by mr W last updated on 15/Jul/22
$${yes}.\:{any}\:{other}\:{possibilities}? \\ $$
Commented by infinityaction last updated on 15/Jul/22
$${i}\:{think}\:{no}\:{but}\:{i}\:{am}\:{not}\:{sure} \\ $$
Commented by infinityaction last updated on 15/Jul/22
$${sorry}\:{sir}\:{i}\:{don}'{t}\:{get}\:{it} \\ $$
Commented by infinityaction last updated on 15/Jul/22
$${okey} \\ $$
Commented by Tawa11 last updated on 15/Jul/22
$$\mathrm{Great}\:\mathrm{sirs} \\ $$