"Trade-off" between bound on the function and on the spectrum for functional calculus in spectral theory

Let $A$ be a self-adjoint (unbounded) operator on a separable Hilbert space $H$.
From the following form of spectral theorem, we may define a functional calculus by $f(A)=Q^{-1} M_{fcirc alpha} Q$. (See below for $Q,alpha$.)

There exists an at most countable collection of finite Borel measures $(mu_k)$ on $mathbb R$ and a isometric linear map $Q:L^2(X,mu) to H$, where $(X,mu)=bigsqcup_{k}(mathbb R , mu_k)$ is the disjoint union of copies of $mathbb R$, such that
$$
Q^{-1}AQ=M_alpha, text{Domain}(A)=D(A)=QD(M_alpha),
$$
where $alpha:X to mathbb R$ is given by $alpha(x)=x$ on each copy of $mathbb R$ and $M_alpha$ is the multiplication operator corresponding to $alpha$ on $L^2(X,mu)$.

This is a version of spectral theorem found in Spectral Theory –
Basic Concepts and Applications by David Borthwick. It is also proven in the book that this functional calculus is well-defined for all bounded borel functions $mathbb R to mathbb C$ (Theorem 5.9). However, this does not include many functions. For example, polynomial functions, like $f(x)=x$, are unbounded.

On the other hand, it is very easy to see that functional calculus for polynomials can easily be defined in this way, if $A$ is a bounded operator, because it has a bounded spectrum.

It appears that, to make the functional calculus work, we have to place a limit on either the "size" of spectrum of $A$ or the "growth rate" of $f$?

Can the above statement be made more precise? Are there any nice descriptions of this "trade-off"?

EDIT: Proof in the book that the functional calculus is unique (and hence well-defined) is based on the fact that $|f(A)|leq sup_{lambda in sigma(A)} |f(lambda)|$, which often does not make sense for unbounded functions $f$. What are some alternative ways to overcome this to establish that it is well-defined (i.e. independent of the choice of $Q$) for unbounded functions as well?