Question

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

• A. $\frac{1}{2}\ell n\frac{1+x}{1-x}$
• B. $\frac{1}{2}\ell n\frac{2+x}{2-x}$
• C. $\frac{1}{2}\ell n\frac{1-x}{1+x}$
• D. $2\ell n(1+x)$

Answer 1

Answer: (A)

$\frac{1}{2}\ell n\frac{1+x}{1-x}$

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

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

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

for universe replace;$x$ with $y$

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

using componendo divodendo

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

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

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

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

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