Let:
[tex]\begin{gathered} f(x)=2x+1 \\ g(x)=e^x \end{gathered}[/tex]The inverses of each function are:
[tex]\begin{gathered} f^{-1}(x)=\frac{x-1}{2} \\ g(x)^{-1}=\ln (x) \end{gathered}[/tex]Both:
[tex]\begin{gathered} f^{-1}(x) \\ and \\ g^{-1}(x) \end{gathered}[/tex]Are functions since for every x:
[tex]\begin{gathered} D\colon\mleft\lbrace(x1,y1\mright),(x2,y2),\ldots,(xn,yn)\} \\ x1\ne x2\ne xn \end{gathered}[/tex]all the inverse of the composite function made by the original function also be a function?.
Yes, since they satisfy the following property:
[tex]\begin{gathered} (f_{\text{ }}o_{\text{ }}f^{-1})(x)=x \\ (f^{-1}_{\text{ }}o_{\text{ }}f^{})(x)=x \end{gathered}[/tex]the inverse of the sum or difference of the original function also be a function?
Let's check it out:
[tex]\begin{gathered} f(x)+f^{-1}(x)=2x+1+\frac{x-1}{2}=\frac{4x+2+x-1}{2}=\frac{5x+1}{2} \\ g(x)+g^{-1}(x)=e^x+\ln (x) \end{gathered}[/tex]The result for both cases is a function. Therefore, the sum or difference of the original function also will be a function.
From the pic:
The green graph is g(x)
The purple graph is g^-1(x)
The red graph is f(x)
The blue graph is f^-1(x)
The black graph is f(x)+f^-1(x)
The orange graph is g(x)+g^-1(x)
We can verify from the pic that all of them are functions
