Question

The inverse of the function $f\left(x\right)=\frac{9^x-9^{-x}}{9^x+9^{-x}}$ is:

• A. $f^{-1}\left(x\right)=$${\log }_9\left(\frac{1+x}{1-x}\right)$
• B. $f^{-1}\left(x\right)=$$\frac{1}{2}{\log }_9\left(\frac{1+x}{1-x}\right)$
• C. $f^{-1}\left(x\right)=$$\frac{1}{4}{\log }_9\left(\frac{1+x}{1-x}\right)$
• D. $f^{-1}\left(x\right)=$$\frac{1}{2}{\log }_9\left(\frac{2x}{2x-1}\right)$

Answer 1

Answer: (B)

$f^{-1}\left(x\right)=$$\frac{1}{2}{\log }_9\left(\frac{1+x}{1-x}\right)$

Let $y=f\left(x\right)$, then

$y=\frac{9^x-9^{-x}}{9^x+9^{-x}}$

or, $\frac{y}{1}=\frac{9^x-9^{-x}}{9^x+9^{-x}}$

or, $\frac{y+1}{1-y}=\frac{{2.9}^x}{{2.9}^{-x}}$

or, $\frac{1+y}{1-y}=9^{2x}$

or, $x=\frac{1}{2}{\log }_9\left(\frac{1+y}{1-y}\right)$

or, $f^{-1}\left(y\right)=$$\frac{1}{2}{\log }_9\left(\frac{1+y}{1-y}\right)$

or, $f^{-1}\left(x\right)=$$\frac{1}{2}{\log }_9\left(\frac{1+x}{1-x}\right)$