Does $f\colon x\mapsto 2x+3$ mean the same thing as $f(x)=2x+3$? In my textbook there is a question like below:
If $$f:x \mapsto 2x-3,$$ then $$f^{-1}(7) = $$
As a multiple choice question, it allows for the answers:  
A. $11$
B. $5$
C. $\frac{1}{11}$
D. $9$
If what I think is correct and I read the equation as:
$$f(x)=2x-3$$
then,
$$y=2x-3$$
$$x=2y-3$$
$$x+3=2y$$
$$\frac {x+3} {2} = y$$
therefore:
$$f^{-1}(7)=\frac {7+3}{2}$$
$$=5$$
 A: (note: this answer was formulated in response to the original version of the question, which has since been edited by a third party)
You're parsing the expression wrong; it's not
$$ \color{red}{\mathbf{ f : x }} \mapsto 2x-3 $$
instead, it is 
$$ f : \color{red}{\mathbf{ x  \mapsto 2x-3 }}$$
That is, $x \mapsto 2x-3$ is one particular notation for "the function which, for all $x$, sends the value $x$ to the value $2x-3$". The colon notation is one particular way to attach a name to the mapping; to say "$f$ is the function which, for all $x$, ...". You might read it as "$f$ sends $x$ to $2x-3$".
The colon notation is more often used when you include a third term expressing the domain and codomain. e.g. if we are talking about a real-valued function on the reals, we would wrote
$$ f : \mathbb{R} \to \mathbb{R} : x \mapsto 2x-3 $$
Here:


*

*$\mathbb{R} \to \mathbb{R}$ denotes the type of mathematical object we call "real-valued functions of the reals"

*$f : \mathbb{R} \to \mathbb{R}$ says that $f$ is a variable of that type — i.e. that $f$ denotes a real-valued function of the reals.

*The whole expression additionally indicates which function $f$ specifically is.

A: Your mistake is here:
$$
x-3=2y
$$
It should be
$$
x+3=2y
$$
A: This is to just to elaborate on why someone would use the notation $$f: x \mapsto 2x-3$$ When we treat a function $f$ as an object, i.e. do more with it than just evaluate it at points, then we need to be able to distinguish between the object, the function $f$, and its  rule of assignment determining its values at points $x$, i.e. after evaluating at points $x$.
Since $f(x)$ denotes both the function $f$, as an object in its own right, as well as the value of that function evaluated at a point $x$, it is too ambiguous for such purposes, because it does not allow us to distinguish between the function and its rule of assignment.
Often we can identify the function $f$, as an object in its own right, with its rule of assignment taking $x$ to $f(x)$, since we are in such instances only considering the latter, so no ambiguity arises.
However, the notation $$f:x\mapsto 2x-3$$ is nice because it both presents the function $f$ as a distinct object while also specifying its rule of assignment. The colon : is meant to be read "such that", which implies that the expression $f:x\mapsto 2x-3$ reads "the function $f$ such that $x$ is mapped to $2x-3$".
The benefit of this notation is that it allows us to distinguish between the function $f$ and its rule of assignment when we need to distinguish between the two, while taking exactly as long to write as $f(x)=2x-3$, when we don't need to distinguish between the function and its rule of assignment, and thus economy of notation becomes a priority. 
Thus, the notation $f:x\mapsto2x-3$ is both exactly as efficient as the classical notation $f(x)=2x-3$ and less ambiguous.
A: Interpret
$$f:\quad x\mapsto y:=2x-3$$
as a flow diagram: The operation $f$ takes a variable value $x$ as input and produces a variable value $y$ as output, whereby the exact formula for computing $y$ from $x$ is given.
I'm writing here because in your argument you replace without hesitation $y=2x-3$ with $x=2y-3$. This seems to be a recommended procedure, but is a terrible mistake, because now both $x$ and $y$ have two different meanings in the same chain of reasoning. Keep the names of the variables as long as you are looking at $f$ and $f^{-1}$ simultaneously.
Argue as follows instead: The inverse function $f^{-1}$ should produce the required input value $x$ for a given output value $y$ of $f$. We therefore have to solve the equation $y=2x-3$ for $x$ "as a function of $y$". The result of course is $x={1\over2}(y+3)$, so that the flow diagram for $f^{-1}$ looks as follows:
$$f^{-1}:\quad y\mapsto x:={1\over2}(y+3)\ .$$
In particular $f^{-1}(7)=5$.
A: As a matter of not merely style but of writing in complete sentences you should write $x=f^{-1}(7)\iff f(x)=7\iff 2x-3=7\iff (2x-3)+3=7+3\iff 2x=10\iff x=5.$ This  reduces errors and shows the flow of the reasoning. In this case the sequence of "$\iff$" shows that the implications go in both directions. The importance of this is that we can conclude that $x=5\implies f(x)=7$ and that $x\ne 5\implies f(x)\ne 7.$ There are other equations that  imply $x=5$ but in fact have no solutions. For example $x=x+1\implies x=5,$ but $x=5$ does not imply $x=x+1.$ 
A: Answering your (main) question: Yes your interpretation of $f:x\rightarrow 2x-3$ being the same as $f(x)=2x-3$ is correct.
For your calculations: $\frac{7+3}{2}\neq10$.
