Answer:
[tex]f^{-1}(x)= \dfrac{1}{2}\ln \left(-\dfrac{x^2}{x^2-1}\right), \quad \textsf{for}\:\{x:0 < x < 1\}[/tex]
Step-by-step explanation:
Given function:
[tex]f(x)=\dfrac{e^x}{\sqrt{e^{2x}+1}}[/tex]
The domain of the given function is unrestricted: {x : x ∈ R}
The range of the given function is restricted: {f(x) : 0 < f(x) < 1}
To find the inverse of a function, swap x and y:
[tex]\implies x=\dfrac{e^y}{\sqrt{e^{2y}+1}}[/tex]
Rearrange the equation to make y the subject:
[tex]\implies x\sqrt{e^{2y}+1}=e^y[/tex]
[tex]\implies x^2(e^{2y}+1)=e^{2y}[/tex]
[tex]\implies x^2e^{2y}+x^2=e^{2y}[/tex]
[tex]\implies x^2e^{2y}-e^{2y}=-x^2[/tex]
[tex]\implies e^{2y}(x^2-1)=-x^2[/tex]
[tex]\implies e^{2y}=-\dfrac{x^2}{x^2-1}[/tex]
[tex]\implies \ln e^{2y}= \ln \left(-\dfrac{x^2}{x^2-1}\right)[/tex]
[tex]\implies 2y \ln e= \ln \left(-\dfrac{x^2}{x^2-1}\right)[/tex]
[tex]\implies 2y(1)= \ln \left(-\dfrac{x^2}{x^2-1}\right)[/tex]
[tex]\implies 2y= \ln \left(-\dfrac{x^2}{x^2-1}\right)[/tex]
[tex]\implies y= \dfrac{1}{2}\ln \left(-\dfrac{x^2}{x^2-1}\right)[/tex]
Replace y with f⁻¹(x):
[tex]\implies f^{-1}(x)= \dfrac{1}{2}\ln \left(-\dfrac{x^2}{x^2-1}\right)[/tex]
The domain of the inverse of a function is the same as the range of the original function. Therefore, the domain of the inverse function is restricted to {x : 0 < x < 1}.
Therefore, the inverse of the given function is:
[tex]f^{-1}(x)= \dfrac{1}{2}\ln \left(-\dfrac{x^2}{x^2-1}\right), \quad \textsf{for}\:\{x:0 < x < 1\}[/tex]
