Consider $f: \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}$, where $f(x) = \frac{1}{x+1}$.
If you work out the inverse of $f$, you get $f^{-1} = \frac{1}{y} - 1$.
However, I'm confused about the role of the domain and codomain here. Clearly $f$ maps from all $\mathbb{R}_{\geq 0}$ to all $\mathbb{R}_{\gt 0}$. Note that $f$ does not map to $0$, even though it's in the definition of the function.
$f^{-1}$ maps from $\mathbb{R}_{\gt 0}$ to $\mathbb{R}_{\geq 0}$. Note that $f^{-1}$ does not map $0$.
My question is: under these conditions, do we consider $f$ invertible? The definition of $f$ states that its codomain is all the nonnegative reals. In practice, $f$ maps only to the positive reals. When we consider the domain of $f^{-1}$, is that the nonnegative reals or the positive reals? If the former, then $f$ is not invertible since $f^{-1}(0)$ doesn't work; if the latter, then $f$ is invertible.
Simply, everything depends on how you want to define $f$. If you consider $f$ as you wrote it, it is not invertible since the inverse at $x=0$ does not exist. However, if you restrict the codomain, it will be invertible.