TITLE: Domain and range of an inverse function
QUESTION [1 upvotes]: I was given the function $f(x) = \sqrt{2x+5}$ $x\in\Bbb R$ , $x \geq -2.5$
I was asked to find the inverse function which was $f^{-1}(x) = {x^2-5\over 2}$
Then I was asked to find the range of $f(x)$ and the domain of $f^{-1}(x)$ which is $x \geq 0$ and $f^{-1}(x) \geq 0$. This is because the range and domain swap for a function and its inverse. However if a value less than $0$ is in-putted for $x$ in the inverse function it obviously still works due to the fact that the inverse function is a quadratic.
Would someone be able to explain to me why the domain of the inverse function is not $x\in\Bbb R$?

REPLY [2 votes]: You wrote it yourself: it is because the domain of $f^{-1}$ is the range of $f$. Don't forget  defining a function consists in defining, not only a computational formula, but its domain of validity, and its target.
So, although $\dfrac{x^2-5}2$  is a perfectly valid function defining formula for all $x\in\mathbf R$, it does not define, per se, the inverse of $f$. If you consider it defines a function with domain $\mathbf R$, this function is a continuation of $f^{-1}$, which is defined on $\mathbf R^+$, to the whole of $\mathbf R$.

REPLY [0 votes]: Hint: By definition $f(f^{-1}(x))=f^{-1}(f(x))$  and use domain of composition of functions.

REPLY [0 votes]: Take care that $f^{-1}$ is not only the formal expression $$f^{-1}(x)=\frac{x^2-5}{2}$$
Since $f^{-1}$ is the function that take some element $y$ of $\text{Range }(f)$ and map it to one element $x$ of $\text{Dom }(f)$, such that $y=f(x)$. So, for $x<0$, the expression $f^{-1}$ has no sense since there is no element lees than $0$ in $\text{Range }(f)$.
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