Respuesta :
[tex]\large y = \dfrac{e^x -e^{-x}}{2}[/tex] [tex]\text{Find the inverse}[/tex]
[tex]\text{Swap x and y, then solve for y}[/tex]
[tex]\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[/tex]
[tex]\text{Let} \ u=e^y[/tex]
[tex]\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}[/tex]
[tex]\text{Substitute back}[/tex]
[tex]\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)[/tex]
[tex]\text{The negative part doesn't have any real solutions. so its just plus}[/tex]
[tex]\text{Swap x and y, then solve for y}[/tex]
[tex]\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[/tex]
[tex]\text{Let} \ u=e^y[/tex]
[tex]\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}[/tex]
[tex]\text{Substitute back}[/tex]
[tex]\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)[/tex]
[tex]\text{The negative part doesn't have any real solutions. so its just plus}[/tex]
