find-f-x-such-that-f-x-f-1-x-

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′(x) = f^(−1) (x) f{f′(x)} = x , f(x) = ax^b f(ab x^(b−1) ) = x a∙a^b b^b ∙x^(b^2 −b) = x x^(b^2 −b−1_(0) ) = (1/(a∙a^b b^b _(1) )) b = ((1±(√5))/2) = φ then a = ^φ (√(1/φ)) f(x) = ^φ (√(1/φ)) x^φ

$$\:\:\:{f}'\left({x}\right)\:=\:{f}^{−\mathrm{1}} \left({x}\right) \\ $$$$\:\:\:{f}\left\{{f}'\left({x}\right)\right\}\:=\:{x}\:\:\:\:,\:\:{f}\left({x}\right)\:\:\:=\:\:{ax}^{{b}} \\ $$$$\:\:\:{f}\left({ab}\:{x}^{{b}−\mathrm{1}} \right)\:=\:\:{x} \\ $$$$\:\:\:\:{a}\centerdot{a}^{{b}} {b}^{{b}} \centerdot{x}^{{b}^{\mathrm{2}} −{b}} \:\:=\:\:{x} \\ $$$$\:\:\:\:{x}^{\underset{\mathrm{0}} {\underbrace{{b}^{\mathrm{2}} −{b}−\mathrm{1}}}\:} \:=\:\:\frac{\mathrm{1}}{\underset{\mathrm{1}} {\underbrace{{a}\centerdot{a}^{{b}} \:{b}^{{b}} }}} \\ $$$$\:\:{b}\:\:=\:\:\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\:=\:\phi\:\:\:{then}\:\:{a}\:\:=\:\:^{\phi} \sqrt{\frac{\mathrm{1}}{\phi}} \\ $$$$\:\:\:\:\:{f}\left({x}\right)\:\:\:=\:\:\:^{\phi} \sqrt{\frac{\mathrm{1}}{\phi}}\:{x}^{\phi} \\ $$

Commented by mr W last updated on 15/Jul/22