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Question Number 173624 by mr W last updated on 15/Jul/22
find f(x) such that f′(x)=f^(−1) (x).
$${find}\:{f}\left({x}\right)\:{such}\:{that}\:{f}'\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right). \\ $$
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
yes. any other possibilities?
$${yes}.\:{any}\:{other}\:{possibilities}? \\ $$
Commented by infinityaction last updated on 15/Jul/22
i think no but i am not sure
$${i}\:{think}\:{no}\:{but}\:{i}\:{am}\:{not}\:{sure} \\ $$
Commented by infinityaction last updated on 15/Jul/22
sorry sir i don′t get it
$${sorry}\:{sir}\:{i}\:{don}'{t}\:{get}\:{it} \\ $$
Commented by infinityaction last updated on 15/Jul/22
okey
$${okey} \\ $$
Commented by Tawa11 last updated on 15/Jul/22
Great sirs
$$\mathrm{Great}\:\mathrm{sirs} \\ $$

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