consider the function f(x)=(x-1)^2+2
1. Determine the inverse of f(x)
2. Determine the domain and range of f(x) and its inverse f^-1(x)
$\displaystyle y=(x-1)^2+2$
for finding inverse, interchange x and y and solve for y,
$\displaystyle x=(y-1)^2+2$
$\displaystyle x-2 = (y-1)^2$
$\displaystyle \pm \sqrt{x-2}=y-1$
$\displaystyle y=1\pm \sqrt{x-2}$
$\displaystyle f^{-1}(x)=1\pm \sqrt{x-2}$
For f(x)
Domain of f(x) $\displaystyle = \{x\in \mathbb{R}\}$
Range of f(x) $\displaystyle = \{y\in \mathbb{R}\;| \;y\ge2\}$
Now, For $\displaystyle f^{-1}(x)$
Domain of $\displaystyle f^{-1}(x) =\{x\in \mathbb{R}\;|\;x\ge 2\}$
Range of $\displaystyle f^{-1}(x) = \{y\in \mathbb{R}\}$
Strictly speaking $\displaystyle f(x)= (x-1)^2+ 2$ does NOT have an inverse because it is not a "one to one" function.
Yes, you can recognize that the graph of y= f(x) has a vertex at (1, 2) and then break it into TWO functions:
$\displaystyle g(x)= (x-1)^2+ 2$ with domain $\displaystyle 1\le x$ which then has range $\displaystyle 2\le y$ and write $\displaystyle g^{-1}(x)= 1+ \sqrt{x-2}$ or
$\displaystyle h(x)= (x-1)^2+ 2$ with domain $\displaystyle x\le 1$ which also has range $\displaystyle 2\le y$ and write $\displaystyle h^{-1}(x)= 1- \sqt{x- 2}$.
But neither of those functions is f(x) so, strictly speaking, neither of those is the inverse of f.