We have the next function
[tex]f(x)=\sqrt[]{x-2}[/tex]The domain is the set of all possible values that x can have in this case we need to remember that we can have a negative value for the radical therefore the domain is
[tex]\: x\ge\: 2[/tex]In interval notation
[tex]\: \lbrack2,\: \infty\: )[/tex]a) [2,inf)
Then for the range is the set of all the possible values that the function can have in this case the range is
[tex]\: \: f(x)\ge0[/tex]In interval notation
[tex]\: \lbrack0,\: \infty\: )[/tex]b) [0,inf)
Then we need to find the inverse of the function given, we make f(x)=y
[tex]y=\sqrt[]{x-2}[/tex]Then we make x=y and y=x
[tex]x=\sqrt[]{y-2}[/tex]Then we isolate the y
[tex]y=x^2+2[/tex]The domain of this function is
[tex]\: \mleft(-\infty\: ,\: \infty\: \mright)[/tex]c) (-inf,inf)
The range of this function is
[tex]\: \lbrack2,\: \infty\: )[/tex]d) [2,inf)
Then as we calculate before the inverse function of the function given is
[tex]f^{-1}(x)=x^2+2[/tex]e) f^-1(x)=x^2+2
ANSWER
a) domain of f: [2,inf)
b) range of f: [0,inf)
c) domain of f^-1: (-inf,inf)
d) range of f^-1: [2,inf)
e) f^-1(x)=x^2+2
[tex]f^{-1}(x)=x^2+2[/tex]
