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Question Number 93786 by mr W last updated on 14/May/20

find f(x) such that f ′(x)=f^(−1) (x)

$${find}\:{f}\left({x}\right)\:{such}\:{that} \\ $$$${f}\:'\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right) \\ $$

Answered by john santu last updated on 15/May/20

f(x)= ((1/ϕ))^(1/ϕ) x^ϕ

$$\mathrm{f}\left(\mathrm{x}\right)=\:\sqrt[{\varphi}]{\frac{\mathrm{1}}{\varphi}}\:\mathrm{x}^{\varphi} \: \\ $$

Answered by  M±th+et+s last updated on 15/May/20

f(x)=kx^r / k,r∈R f′(x)=krx^(r−1) , f^(−1) (x)=((x/k))^(1/r) f′(x)=f^(−1) (x)⇔krx^(r−1) =((x/k))^(1/r) krx^(r−1) =x^(1/r) ((1/k))^(1/r) x^(r−1−(1/r)) =((1/k))^(1/r) ((1/(kr))) f ′(x)=f^(−1) (x)⇔(r−1−(1/r)=0&((1/k))^((1/r)+1) ((1/r))=1) f ′(x)=f^(−1) (x)⇔(r^2 −r−1=0&((1/k))^((1/r)+1) ((1/r))=1p) f ′(x)=f^(−1) (x)⇔(r=((1±(√5))/2)&((1/k))^((1/r)+1) ((1/r))=1) notice/((1+(√5))/2)=ϕ,((1−(√5))/2)=ϕ^� ;1+(1/r)=r f ′(x)=f^(−1) (x)⇔(r=ϕ,ϕ^� &((1/k))^r =r) f ′(x)=f^(−1) (x)⇔(r=ϕ,ϕ^� &k=(1/(r)^(1/r) )) f_ϕ (x)=(1/(ϕ)^(1/ϕ) )x^ϕ ,f_ϕ^� (x)=(1/(ϕ^� )^(1/ϕ^� ) )x^ϕ notice/′′ϕ′′ is golden ratio

$${f}\left({x}\right)={kx}^{{r}} /\:\:{k},{r}\in{R} \\ $$$${f}'\left({x}\right)={krx}^{{r}−\mathrm{1}} \:,\:{f}^{−\mathrm{1}} \left({x}\right)=\left(\frac{{x}}{{k}}\right)^{\frac{\mathrm{1}}{{r}}} \\ $$$${f}'\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right)\Leftrightarrow{krx}^{{r}−\mathrm{1}} =\left(\frac{{x}}{{k}}\right)^{\frac{\mathrm{1}}{{r}}} \\ $$$${krx}^{{r}−\mathrm{1}} ={x}^{\frac{\mathrm{1}}{{r}}} \left(\frac{\mathrm{1}}{{k}}\right)^{\frac{\mathrm{1}}{{r}}} \\ $$$${x}^{{r}−\mathrm{1}−\frac{\mathrm{1}}{{r}}} =\left(\frac{\mathrm{1}}{{k}}\right)^{\frac{\mathrm{1}}{{r}}} \left(\frac{\mathrm{1}}{{kr}}\right) \\ $$$$ \\ $$$${f}\:'\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right)\Leftrightarrow\left({r}−\mathrm{1}−\frac{\mathrm{1}}{{r}}=\mathrm{0\&}\left(\frac{\mathrm{1}}{{k}}\right)^{\frac{\mathrm{1}}{{r}}+\mathrm{1}} \left(\frac{\mathrm{1}}{{r}}\right)=\mathrm{1}\right) \\ $$$${f}\:'\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right)\Leftrightarrow\left({r}^{\mathrm{2}} −{r}−\mathrm{1}=\mathrm{0\&}\left(\frac{\mathrm{1}}{{k}}\right)^{\frac{\mathrm{1}}{{r}}+\mathrm{1}} \left(\frac{\mathrm{1}}{{r}}\right)=\mathrm{1}{p}\right) \\ $$$${f}\:'\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right)\Leftrightarrow\left({r}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\&\left(\frac{\mathrm{1}}{{k}}\right)^{\frac{\mathrm{1}}{{r}}+\mathrm{1}} \left(\frac{\mathrm{1}}{{r}}\right)=\mathrm{1}\right) \\ $$$${notice}/\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}=\varphi,\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}=\bar {\varphi};\mathrm{1}+\frac{\mathrm{1}}{{r}}={r} \\ $$$${f}\:'\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right)\Leftrightarrow\left({r}=\varphi,\bar {\varphi}\&\left(\frac{\mathrm{1}}{{k}}\right)^{{r}} ={r}\right) \\ $$$${f}\:'\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right)\Leftrightarrow\left({r}=\varphi,\bar {\varphi}\&{k}=\frac{\mathrm{1}}{\sqrt[{{r}}]{{r}}}\right) \\ $$$${f}_{\varphi} \left({x}\right)=\frac{\mathrm{1}}{\sqrt[{\varphi}]{\varphi}}{x}^{\varphi} ,{f}_{\bar {\varphi}} \left({x}\right)=\frac{\mathrm{1}}{\sqrt[{\bar {\varphi}}]{\bar {\varphi}}}{x}^{\varphi} \\ $$$$ \\ $$$${notice}/''\varphi''\:{is}\:{golden}\:{ratio} \\ $$

Commented by mr W last updated on 15/May/20

thank you both! you assumed y=kx^r . but are there any other solutions?

$${thank}\:{you}\:{both}! \\ $$$${you}\:{assumed}\:{y}={kx}^{{r}} .\:{but}\:{are}\:{there} \\ $$$${any}\:{other}\:{solutions}? \\ $$

Commented by  M±th+et+s last updated on 15/May/20

mybe there is, but i couldn′t find just this

$${mybe}\:{there}\:{is},\:{but}\:{i}\:{couldn}'{t}\:{find}\:{just}\:{this} \\ $$

Answered by hovero clinton last updated on 15/May/20

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