Here , the given function to us is [tex]{\bf{f(x)=e^{x}+1}}[/tex] which is clearly an exponential function , so it's inverse will be a logarithmic function . So , now let's solve for it's inverse
We have ;
[tex]{:\implies \quad \sf f(x)=e^{x}+1}[/tex]
Now , let y = f(x) . So :
[tex]{:\implies \quad \sf y=e^{x}+1}[/tex]
Now , solve for x
[tex]{:\implies \quad \sf e^{x}=y-1}[/tex]
Take natural log on both sides :
[tex]{:\implies \quad \sf ln(e^{x})=ln(y-1)}[/tex]
[tex]{:\implies \quad \sf x\: ln(e)=ln(y-1)\quad \qquad \{\because ln(a^{b})=b\: ln(a)\}}[/tex]
[tex]{:\implies \quad \sf x=ln(y-1)\quad \qquad \{\because ln(e)=1\}}[/tex]
Now , replace x by [tex]{\bf{f^{-1}(x)}}[/tex] and y by x
[tex]{:\implies \quad \bf \therefore \quad \underline{\underline{f^{-1}(x)=ln(x-1)}}}[/tex]
Now , we got the inverse of f(x) which is a logarithmic function with domain [tex]{\bf{(-\infty , +\infty)}}[/tex] and Range [tex]{\bf{(1,\infty)}}[/tex].So , the domain is the set of all real numbers and range being y > 1
Hence , The required answers are a logarithmic, all real numbers and y > 1