Question

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

• A. $\frac{1}{2}\log \frac{1+x}{1-x}$
• B. $\frac{1}{2}\log \frac{2+x}{2-x}$
• C. $\frac{1}{2}\log \frac{1-x}{1+x}$
• D. none of these

Answer 1

The correct option is A $\frac{1}{2}\log \frac{1+x}{1-x}$

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

$y=\frac{e^{2x}-1}{e^{2x}+1}$

$y(e^{2x}+1)=e^{2x}-1$

$\Rightarrow e^{2x}\left(y\right)+y=e^{2x}-1$

$\Rightarrow y+1=e^{2x}(1-y)$

$e^{2x}=\left(\frac{1+y}{1-y}\right)$

Taking logarithm on both the sides, we get

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

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

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

Alternatively, $f\left(x\right)$ is $\tan h\left(x\right)$

So, it's inverse will be $\frac{1}{2}\log \left(\frac{1+x}{1-x}\right)$.