Find the inverse of the following: f(x) = (e^x − e^−x)/2
\(\large y = \dfrac{e^x -e^{-x}}{2}\) \(\text{Find the inverse}\)
\(\text{Swap x and y, then solve for y}\)
\(\large x = \dfrac{e^y -e^{-y}}{2} \\ \, \\ \large 2x = e^y -e^{-y} \\ \, \\ \large 2x = e^y - \dfrac 1 {e^{y}} \\ \, \\ \large 2x = e^y \cdot \dfrac{e^y}{e^y} - \dfrac 1 {e^{y}} \\ \, \\ \large 2x = \dfrac {(e^{y})^2 -1 } {e^{y}} \\ \, \\ \large 2xe^y = (e^{y})^2 -1 \\ \, \\ \large 0 = (e^{y})^2 - 2xe^y - 1 \\ \, \\ \large 0 = (e^{y})^2 - 2x(e^y) - 1\)
\(\text{Let} \ u=e^y\)
\(\large 0 = u^2 -2xu - 1 \\ \, \\ \large u = \dfrac{- (-2x) \pm \sqrt{(-2x)^2-4(1)(-1) } }{2(1)} \\ \, \\ \large u = \dfrac{2x \pm \sqrt{4x^2 + 4} }{2(1)} \\ \, \\ \large u = \dfrac{2x \pm \sqrt{4} \sqrt{ x^2 + 1} }{2(1)} \\ \, \\ \large u = x \pm \sqrt{ x^2 + 1}\sqrt{4x^2 + 4} }{2(1)} \\ \, \\ \large u = \dfrac{2x \pm \sqrt{4} \sqrt{ x^2 + 1} }{2(1)} \\ \, \\ \large u = x \pm \sqrt{ x^2 + 1}\)
\(\text{Substitute back}\)
\(\large e^y = x \pm \sqrt{ x^2 + 1} \\ \, \\ \large y = \ln \left( x \pm \sqrt{ x^2 + 1} \right) \\ \, \\ \large y = \ln \left( x + \sqrt{ x^2 + 1} \right)\)
\(\text{The negative part doesn't have any real solutions. so its just plus}\)