Does $f\colon x\mapsto 2x+3$ mean the same thing as $f(x)=2x+3$?

Solution 1:

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.

Solution 2:

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: $\frac{7+3}{2}\neq10$.

Solution 3:

(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.

Solution 4:

Your mistake is here: $$ x-3=2y $$ It should be $$ x+3=2y $$

Solution 5:

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 $f^{-1}(7)=5$.