Let $\mathcal{A}$ be a $k$-affinoid algebra, as in Berkovich (1990, Chapter 2), and let $X := \mathcal{M}(\mathcal{A})$.
In (loc. cit. 2.2.2) it is claimed that if $q > 0$, and $g \in \mathcal{A}$, then $X(q g^{-1}) := \{ x \in X | |g(x)| \geq q \}$ is an affinoid domain (actually something more general, but it can be reduced to this simpler statement easily), and that it is represented by the obvious homomorphism $$ \mathcal{A} \rightarrow \mathcal{A}\{q g^{-1}\} := \mathcal{A} \{qS\} / (gS -1) $$ Then Berkovich says that this follows easily from the following corollary 2.1.5:
Let $\phi: \mathcal{A} \rightarrow \mathcal{B}$ be a bounded homomorphism between affinoid $k$-algebras. Let $f \in \mathcal{B}$, $r >0$ such that $r \geq \rho(f)$. Then there exists a unique bounded homomorphism $\Phi:\mathcal{A}\{ r^{-1} T \} \rightarrow \mathcal{B}$ extending $\phi$ and sending $T$ to $f$.
Since this confused me a little bit, I add here a little explanation. I hope that you find it useful.
Recall that a morphism $\mathcal{A} \rightarrow \mathcal{A}'$ represents the affinoid domain $X':= \mathcal{M}(\mathcal{A'})$ if the following holds:
In our situation, $X' := X(qg^{-1})$ and $\mathcal{A}' = \mathcal{A}\{ qg^{-1}\}$, so we want to extend $\psi$ $\require{AMScd}$ \begin{CD} \mathcal{A}\{q g^{-1}\} \\ @AAA\\ \mathcal{A} @>>\psi> \mathcal{B} \end{CD} to a morphism $\mathcal{A}\{qg^{-1}\} \rightarrow \mathcal{B}$.
In the analytification, $\psi^*$ factors through $X(qg^{-1})$ if and only if $$ |g(\psi^*(y))| \geq q \quad \forall y \in \mathcal{M}(\mathcal{B}).$$
Now, note that $\psi(g)$ is a unit in $\mathcal{B}$, since $|g(\psi^*(y))| \geq q > 0$ for all $y \in X(qg^{-1})$ (c.f. for example Lemma 5.3.5 here). Hence we define $$ \mathcal{A} \{ qS \} \rightarrow \mathcal{B} $$ via $S \mapsto \psi(g)^{-1}$.
and we want to see that this is well defined. For this (and this is the part that was confusing me at first), instead of applying corollary 2.1.5, we follow it's proof and addapt it to our situation, since in principal we don't have a bound by the spectral radius $\rho(\psi(g))$ (or at least I don't see why, but it doesn't matter because we will not need it).
So let $h \in \mathcal{A}\{qS\}$, and write it as $h = \sum a_i S^i$. Then this is sent to $$ h \mapsto \sum \psi(a_i) \psi(g)^{-i}, $$ and we have that $$ | \psi(a_i) \psi(g)^{-i}| \leq C |a_i| q^{-i}, $$ where $C$ is the constant coming from the fact that $\psi$ is bounded. By definition of $\mathcal{A}\{qS\}$, the right hand side converges to zero as $i \rightarrow \infty$, so we conclude that the map is well defined and we are done, since it will also factor through the quotient $\mathcal{A} \{qS\} / (gS -1)$.