Does the usual procedure for finding the inverse function also prove that the function is invertible? Students are usually taught to find inverse functions using a procedure such as this from Stewart (Calculus, 2016, p. 58):

Question. Does the above procedure also prove that $f$ is invertible?

Let's take Stewart's example:

In the above example, have we also shown that $f$ is invertible? If so, at which step(s)?
 A: Definition. A function $f$ is invertible if for any $a,b\in \text{Domain}\,f$, $$f(a)=f(b) \implies a=b.$$

Claim. Let $f$ be a function. Suppose we can find a function $g:\text{Range}\,f\rightarrow\text{Domain}\,f$ such that for every $x\in\text{Range}\,f$,
$$x=f\left(t\right) \implies t=g\left(x\right).$$
Then $f$ is invertible.
(Moreover, the function $g$ is the inverse of $f$.)
Proof. Let $a,b\in  \text{Domain}\,f$. Suppose $f(a)=f(b)$.
Suppose we can find the function $g$. Then $g(f(a))=a$ and $g(f(b))$=b.
But since $g$ is a function and $f(a)=f(b)$, we have $a=b$. (Proof of moreover bit omitted.) ∎

Example. Define $f: \mathbb R \rightarrow \mathbb R$ by $f(x)=x^3+2$.
For each $x\in\text{Range}\,f$, there exists some $t\in\text{Domain}\,f$ such that
$$x=f(t)=t^3+2 \text{ or } t=\sqrt[3]{x-2}.$$
We've just shown there exists the function $g:\text{Range}\,f\rightarrow\text{Domain}\,f$ defined by $g(x)=\sqrt[3]{x-2}$ such that for every $x\in\text{Range}\,f$,
$$x=f\left(t\right) \implies t=g\left(x\right).$$
Hence, by the above Claim, $f$ is invertible. (Moreover, $g$ is the inverse of $f$.)
A: When he solved for $x$, he showed there is a unique $x$ corresponding to any $y$ in the range.  That is, $f$ is one-to-one, hence invertible.
A: Maybe the following way to determine the inverse also proves that the original function is invertible. 
Suppose function $g$ is the inverse of $f(x)= x^3+2$. 
In that case : 
$f(g(x))=x$
$\iff [g(x)]^3+2 = x $
$\iff [g(x)]^3 = x -2$
$\iff \sqrt[3]{[g(x)]^3} = \sqrt[3] {x-2}$
$\iff g(x) = \sqrt[3] {x-2}$
So if function $f$ has an inverse, it must be $g(x) = \sqrt[3] {x-2}$. 
Now, to prove that $f$ actually has an inverse, we can produce the inverse composition, an check whether it is equal to the identity function $Id(x)=x$. 
$g(f(x)) $
$= \sqrt[3] {(x^3+2) -2}$
$=\sqrt[3] {(x^3)}$
$ =x$. 
So we have a function $g$ such that $f(g(x)) = g(f(x)) = x $
which means that 
(1) $g$ is the inverse of $f$ 
(2) and , a fortiori , that $f$ is invertible. 
