Question

The inverse of the function $y=\frac{e^x-e^{-x}}{e^x+e^{-x}}$ 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. $2\log (1+x)$

Answer 1

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

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

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

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

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

So the inverse is $\frac{1}{2}\log \frac{1+x}{1-x}$.