Question

Find the inverse of the following: f(x) = (e^x − e^−x)/2

Answer 1

`\large y = (e^x -e^(-x))/(2)`
`\text{Find the inverse}`


`\text{Swap x and y, then solve for y}`


`\large x = (e^y -e^(-y))/(2) \\ \, \\ \large 2x = e^y -e^(-y) \\ \, \\ \large 2x = e^y - \frac 1 {e^(y)} \\ \, \\ \large 2x = e^y \cdot (e^y)/(e^y) - \frac 1 {e^(y)} \\ \, \\ \large 2x = \frac {(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 = (- (-2x) \pm √((-2x)^2-4(1)(-1) ) )/(2(1)) \\ \, \\ \large u = (2x \pm √(4x^2 + 4) )/(2(1)) \\ \, \\ \large u = (2x \pm √(4) √( x^2 + 1) )/(2(1)) \\ \, \\ \large u = x \pm √( x^2 + 1)√(4x^2 + 4) }{2(1)} \\ \, \\ \large u = (2x \pm √(4) √( x^2 + 1) )/(2(1)) \\ \, \\ \large u = x \pm √( x^2 + 1)`


`\text{Substitute back}`


`\large e^y = x \pm √( x^2 + 1) \\ \, \\ \large y = \ln \left( x \pm √( x^2 + 1) \right) \\ \, \\ \large y = \ln \left( x + √( x^2 + 1) \right)`


`\text{The negative part doesn't have any real solutions. so its just plus}`