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Question Number 66412 by aliesam last updated on 14/Aug/19

$$if$$$$$$$$f\left(x\right)=ln(x+\sqrt{x^2+1})$$$$$$$$find$$$$$$$$f^{-1}\left(x\right)=?$$

Commented by mathmax by abdo last updated on 14/Aug/19

$$wehavef\left(sh\left(x\right)\right)=ln(sh\left(x\right)+\sqrt{1+{sh}^{2}x})=ln(sh\left(x\right)+ch\left(x\right))$$$$=ln(\frac{e^x-e^{-x}}{2}+\frac{e^x+e^{-x}}{2})=ln\left(e^x\right)=x\Rightarrow f^{-1}\left(x\right)=sh\left(x\right)$$

Commented by mr W last updated on 14/Aug/19

$$f\left(x\right)=ln(x+\sqrt{x^2+1})={\sinh }^{-1}x$$$$\Rightarrow f^{-1}\left(x\right)=\sinh x=\frac{e^x-e^{-x}}{2}$$

Commented by kaivan.ahmadi last updated on 15/Aug/19

$$y=ln(x+\sqrt{x^2+1})\Rightarrow e^y-x=\sqrt{x^2+1}\Rightarrow e^{2y}-2{xe}^y+x^2=x^2+1\Rightarrow$$$$2{xe}^y=e^{2y}-1\Rightarrow x=\frac{e^{2y}-1}{2e^y}\Rightarrow f^{-1}\left(x\right)=\frac{e^{2x}-1}{2e^x}$$

Answered by MJS last updated on 14/Aug/19

$$x=\ln (y+\sqrt{y^2+1})$$$${\mathrm{e}}^x=y+\sqrt{y^2+1}$$$${({\mathrm{e}}^x-y)}^2=y^2+1$$$${\mathrm{e}}^{2x}-2y{\mathrm{e}}^x+y^2=y^2+1$$$$y=\frac{{\mathrm{e}}^x-{\mathrm{e}}^{-x}}{2}=\sinh x$$

Commented by Mikael last updated on 15/Aug/19

$$niceSir.$$

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