Question Number 143595 by ZiYangLee last updated on 16/Jun/21
$$\mathrm{If}\:{f}\left({x}\right)=\frac{\mathrm{10}^{{x}} −\mathrm{10}^{−{x}} }{\mathrm{10}^{{x}} +\mathrm{10}^{−{x}} },\:\mathrm{find}\:{f}^{\:−\mathrm{1}} \left({x}\right). \\ $$
Answered by bobhans last updated on 16/Jun/21
$${let}\:\mathrm{10}^{{x}} ={t} \\ $$$$\Rightarrow{y}={f}\left({t}\right)=\frac{{t}−\frac{\mathrm{1}}{{t}}}{{t}+\frac{\mathrm{1}}{{t}}}=\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow\left({y}−\mathrm{1}\right){t}^{\mathrm{2}} =−{y}−\mathrm{1} \\ $$$$\Rightarrow{t}=\pm\:\sqrt{\frac{−{y}−\mathrm{1}}{{y}−\mathrm{1}}}=\pm\:\sqrt{\frac{{y}+\mathrm{1}}{\mathrm{1}−{y}}} \\ $$$$\Rightarrow\mathrm{10}^{{x}} \:=\:\sqrt{\frac{{y}+\mathrm{1}}{\mathrm{1}−{y}}}\: \\ $$$$\Rightarrow{x}\:=\:\mathrm{log}\:_{\mathrm{10}} \left(\sqrt{\frac{{y}+\mathrm{1}}{\mathrm{1}−{y}}}\:\right) \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\mathrm{log}\:_{\mathrm{10}} \left(\sqrt{\frac{{x}+\mathrm{1}}{\mathrm{1}−{x}}}\:\right) \\ $$
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
$$\mathrm{y}=\frac{\mathrm{10}^{\mathrm{x}} −\mathrm{10}^{−\mathrm{x}} }{\mathrm{10}^{\mathrm{x}} +\mathrm{10}^{−\mathrm{x}} } \\ $$$$\Rightarrow\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lg}\frac{\mathrm{1}+\mathrm{y}}{\mathrm{1}−\mathrm{y}} \\ $$$$\Rightarrow\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lg}\frac{\mathrm{1}+\mathrm{x}}{\mathrm{1}−\mathrm{x}}.....\left(\mid\mathrm{x}\mid<\mathrm{1}\right) \\ $$