When we substitute variables do we compose functions? Suppose that we have the function $f(u)=2u+3$ and we define $u=2x$. What of the below expressions are correct?
$$f(u(x))=4x+3 \ \text{or} \ f(x)=4x+3 $$
Does the "definition" serves only as a substitution (shorthand) i.e. is "$f$" still the same function or the composition of $f \circ u$ irrespective of the fact that we didn't write $u(x)=2x$ at first (abuse of notation)?
 A: The second of your suggestions is the correct interpretation. We often abuse notation and write $f(x)$, although one actually defines a new function by composition $\hat f(x)=f\circ u(x)$. The most common application is the substitution, when calculating integrals. The error cancels in the end of those calculations by resubstitution (for indefinite integrals) or because you also substitute the boundaries (for definite integrals).
A: The first of your suggestions is correct:
$f(u(x)) = 2u(x) + 3 = 2(2x) + 3 = 4x + 3$.
The second is incorrect, since $f(x) = 2x + 3$.
However, it is common, especially in physics, to abuse notation and write $f(x) = 4x+3$.
A: In my personal opinion, what notation you use is irrelevant as long as you know what you are doing. If you are just a beginner it's best to go with the standard notation of $$f(u(x))=4x+3 $$
To me the expression above reads: In $f(x)$ if I feed in the function $u(x)$'s output as an input then I have the algebraic expression $4x+3$ denoted as $ f(u(x))$ but perhaps you could understand the same thing if it was written as $f(x)$. Being attentive to notation is only important when you're talking to someone else haha
A: We consider two functions $f$ and $g$:
\begin{align*}
&f:\mathbb{R}\to\mathbb{R}&\qquad &g:\mathbb{R}\to\mathbb{R}\\
&f(u)=2u+3&\qquad &g(u)=4u+3\tag{1}
\end{align*}
Here we have two linear real-valued functions $f$ and $g$ which are different, since e.g. $f(1)\ne g(1)$.
A substitution $u=2x$ in $f$ and $g$ is a composition of functions with a function $u$:
\begin{align*}
&u:\mathbb{R}\to\mathbb{R}\\
&u(x)=2x
\end{align*}
We obtain
\begin{align*}
&f\circ u:\mathbb{R}\to\mathbb{R}&\qquad &g\circ u:\mathbb{R}\to\mathbb{R}\\
&f(u(x))=f(2x)=4x+3&\qquad &g(u(x))=g(2x)=8x+3\tag{2}\\
\end{align*}


*

*We see from (1) and (2) that $f(2x)=g(x)$ is an identity.

*But setting $f(x)=g(x)$ is just an equation which is true iff $2x+3=4x+3$, i.e. $x=0$.


