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

$$findf\left(x\right)suchthat$$$$f{}^{\prime }\left(x\right)=f^{-1}\left(x\right)$$

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

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

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

$$f\left(x\right)={kx}^r/k,r\in R$$$$f^{\prime }\left(x\right)={krx}^{r-1},f^{-1}\left(x\right)={\left(\frac{x}{k}\right)}^{\frac{1}{r}}$$$$f^{\prime }\left(x\right)=f^{-1}\left(x\right)\iff {krx}^{r-1}={\left(\frac{x}{k}\right)}^{\frac{1}{r}}$$$${krx}^{r-1}=x^{\frac{1}{r}}{\left(\frac{1}{k}\right)}^{\frac{1}{r}}$$$$x^{r-1-\frac{1}{r}}={\left(\frac{1}{k}\right)}^{\frac{1}{r}}\left(\frac{1}{kr}\right)$$$$$$$$f{}^{\prime }\left(x\right)=f^{-1}\left(x\right)\iff (r-1-\frac{1}{r}=0\mathrm{\&}{\left(\frac{1}{k}\right)}^{\frac{1}{r}+1}\left(\frac{1}{r}\right)=1)$$$$f{}^{\prime }\left(x\right)=f^{-1}\left(x\right)\iff (r^2-r-1=0\mathrm{\&}{\left(\frac{1}{k}\right)}^{\frac{1}{r}+1}\left(\frac{1}{r}\right)=1p)$$$$f{}^{\prime }\left(x\right)=f^{-1}\left(x\right)\iff (r=\frac{1\pm \sqrt{5}}{2}\mathrm{\&}{\left(\frac{1}{k}\right)}^{\frac{1}{r}+1}\left(\frac{1}{r}\right)=1)$$$$notice/\frac{1+\sqrt{5}}{2}=\phi ,\frac{1-\sqrt{5}}{2}=\overline{\phi };1+\frac{1}{r}=r$$$$f{}^{\prime }\left(x\right)=f^{-1}\left(x\right)\iff (r=\phi ,\overline{\phi }\mathrm{\&}{\left(\frac{1}{k}\right)}^r=r)$$$$f{}^{\prime }\left(x\right)=f^{-1}\left(x\right)\iff (r=\phi ,\overline{\phi }\mathrm{\&}k=\frac{1}{\sqrt[r]{r}})$$$$f_{\phi }\left(x\right)=\frac{1}{\sqrt[\phi ]{\phi }}{x}^{\phi },f_{\overline{\phi }}\left(x\right)=\frac{1}{\sqrt[\overline{\phi }]{\overline{\phi }}}{x}^{\phi }$$$$$$$$notice{/}^{″}{\phi }^{″}isgoldenratio$$

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

$$thankyouboth!$$$$youassumedy={kx}^r.butarethere$$$$anyothersolutions?$$

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

$$mybethereis,buti{couldn}^{\prime }tfindjustthis$$

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

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