In my textbook, the following example is given. I am confused as to why the domain for $f^-1(x) = x^2+5$ is "all real $x$ greater than $0$". Doesn't this function, take all real x-values? Or is it something to do with this inverse function ($x^2+5$) not passing the horizontal one-to-one test when plotted? Any explanation would be appreciated.
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The inverse of $f:A\to B$ is a function $f^{-1}:B\to A$, where $B$ is the range of $f$.
The range of $f(x)=\sqrt{x-5}$ is $[0,\infty)$ hence the domain of its inverse is $[0,\infty)$.
The domain of $f(x)$ is $[5,\infty)$ hence the range of its inverse is $[5,\infty)$.
It just so happens that we can extend $f^{-1}$ over $\Bbb R^-$. But that will mean it is no longer the inverse of $f$ since for no value of $x$ does $f(x)\in\Bbb R^-$.