Question:

Let $y = f(x) = \frac{e^x - e^{-x}}{2}$. This is one-to-one since $f'(x) = \frac{e^x + e^{-x}}{2} > 0$ for all $x$. To find the inverse, set $y = \frac{e^x - e^{-x}}{2}$, multiply by $e^x$: $ye^x = \frac{e^{2x} - 1}{2}$, giving $e^{2x} - 2ye^x - 1 = 0$. Using the quadratic formula: $e^x = y \pm \sqrt{y^2 + 1}$. Since $e^x > 0$, we take $e^x = y + \sqrt{y^2 + 1}$, so $x = \ln(y + \sqrt{y^2 + 1})$ or $f^{-1}(x) = \ln(x + \sqrt{x^2 + 1})$. The range of $f(x)$ is $(-\infty, \infty)$.