Mathematics - Functions Question with Solution | TestHub

Let $y=f(x)$. We want to solve for $x$ in terms of $y$. So, $y=\frac{e^x-e^{-x}}{e^x+e^{-x}}$. Multiplying both sides by the denominator, we get $y(e^x+e^{-x})=e^x-e^{-x}$. This can be rewritten as $y{e}^x+y{e}^{-x}=e^x-e^{-x}$. Rearranging terms to isolate $e^x$, we have $y{e}^x-e^x=-e^{-x}-y{e}^{-x}$, which simplifies to $e^x(y-1)=-e^{-x}(1+y)$. Multiplying by $e^x$ on both sides, we get $e^{2x}(y-1)=-(1+y)$. Therefore, $e^{2x}=\frac{-(1+y)}{y-1}=\frac{1+y}{1-y}$. Taking the natural logarithm of both sides, $2x=\ln \left(\frac{1+y}{1-y}\right)$. Finally, $x=\frac{1}{2}\ln \left(\frac{1+y}{1-y}\right)$. Replacing $y$ with $x$ for the inverse function, $f^{-1}(x)=\frac{1}{2}\ln \left(\frac{1+x}{1-x}\right)$. Thus, option A is correct.