Hyperbolic Inverse: sinh(x)

[Trenters4325]

How can you find the inverse of the hyperbolic function sinh(x) by by switching the xs and ys.

 

[stapel]

Is the function "y = sinh(x)"? Then either use the inverse hyperbolic sine, or else convert the hyperbolic sine to its exponential form, and invert that.

Eliz.

 

[Trenters4325]

My question is how do you get the inverse after you convert it to exponential form and switch the variables?

 

[soroban]

Hello, Trenters4325!

This is not a pretty problem . . .

How can you find the inverse of \(\displaystyle \,\sinh(x)\,\) by by switching the \(\displaystyle x\)'s and \(\displaystyle y\)'s?

\(\displaystyle \text{We have: }\L\,y\:=\:\frac{e^x\,-\,e^{-x}}{2}\)

\(\displaystyle \text{Switch variables: }\L\,x\:=\:\frac{e^y\,-\,e^{-y}}{2}\)


\(\displaystyle \text{Now we solve for }y:\L\;\;\frac{e^y\,-\,e^{-y}}{2}\;=\;x\)

\(\displaystyle \text{Multiply both sides by }\L 2e^y:\;\;e^{2y}\,-\,1\;=\;2xe^y\)

\(\displaystyle \text{We have: }\L\,e^{2y}\,-\,2xe^y\,-\,1\;=\;0\)

\(\displaystyle \text{This is a }quadratic:\L\;(e^y)^2\,-\,2x(e^y)\,-\,1 \;= \;0\)

\(\displaystyle \text{Quadratic Formula: }\L\,e^y \;= \;\frac{-(-2x)\,\pm\,\sqrt{(-2x)^2\,-\,4(1)(-1)}}{2(1)} \;= \;\frac{2x\,\pm\,\sqrt{4x^2\,+\,4}}{2}\)

. . \(\displaystyle \L e^y\;=\;\frac{2x\,\pm\,\sqrt{4(x^2\,+\,1)}}{2}\;=\;\frac{2x\,\pm2\sqrt{x^2\,+\,1}}{2} \;= \;x\,\pm\,\sqrt{x^2\,+\,1}\)


\(\displaystyle \text{Hence: }\L\:y \;= \;\ln(x\,\pm\,\sqrt{x^2\,+\,1})\)

. . \(\displaystyle \text{But }\L\,x\,-\,\sqrt{x^2+1}\,\text{ is negative.}\)


\(\displaystyle \text{Therefore: }\L\:y \;=\;\ln(x\,+\,\sqrt{x^2\,+\,1})\)