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Question Number 143595 by ZiYangLee last updated on 16/Jun/21

$$\mathrm{If}f\left(x\right)=\frac{{10}^x-{10}^{-x}}{{10}^x+{10}^{-x}},\mathrm{find}{f}^{-1}\left(x\right).$$

Answered by bobhans last updated on 16/Jun/21

$$let{10}^x=t$$$$\Rightarrow y=f\left(t\right)=\frac{t-\frac{1}{t}}{t+\frac{1}{t}}=\frac{t^2-1}{t^2+1}$$$$\Rightarrow (y-1)t^2=-y-1$$$$\Rightarrow t=\pm \sqrt{\frac{-y-1}{y-1}}=\pm \sqrt{\frac{y+1}{1-y}}$$$$\Rightarrow {10}^x=\sqrt{\frac{y+1}{1-y}}$$$$\Rightarrow x=\log {}_{10}\left(\sqrt{\frac{y+1}{1-y}}\right)$$$$\Rightarrow f^{-1}\left(x\right)=\log {}_{10}\left(\sqrt{\frac{x+1}{1-x}}\right)$$

Commented by ZiYangLee last updated on 16/Jun/21

$$\mathrm{thank}\mathrm{you}\mathrm{very}\mathrm{much}!$$

Answered by qaz last updated on 16/Jun/21

$$\mathrm{y}=\frac{{10}^{\mathrm{x}}-{10}^{-\mathrm{x}}}{{10}^{\mathrm{x}}+{10}^{-\mathrm{x}}}$$$$\Rightarrow \mathrm{x}=\frac{1}{2}\lg \frac{1+\mathrm{y}}{1-\mathrm{y}}$$$$\Rightarrow {\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\frac{1}{2}\lg \frac{1+\mathrm{x}}{1-\mathrm{x}}.....(\mid \mathrm{x}\mid <1)$$

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