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,
$$y=2x-3$$ $$x=2y-3$$ $$x+3=2y$$ $$\frac {x+3} {2} = y$$


$$f^{-1}(7)=\frac {7+3}{2}$$ $$=5$$

  • $\begingroup$ For mathjax, you need to put things into dollar signs, i.e. $\frac{1}{11}$. Also, welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? $\endgroup$ – 5xum Nov 8 '16 at 10:06
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    $\begingroup$ Hint: $2\cdot 5 -3 =7$. $\endgroup$ – Niklas Nov 8 '16 at 10:06
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    $\begingroup$ Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote. $\endgroup$ – 5xum Nov 8 '16 at 10:06
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    $\begingroup$ if you replace $\rightarrow$ with $\mapsto$ I'd be okay with that. Although as Mauro ALLEGRANZA points out, $f(x)$ is more usual. $\endgroup$ – user259242 Nov 8 '16 at 10:10
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    $\begingroup$ @SooKyungAhn Indeed it is. However, how you get the equality $\frac{7+3}{2}=10$ is not clear :) $\endgroup$ – 5xum Nov 8 '16 at 10:43

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

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    $\begingroup$ Prejudiced against base 5 much? $\endgroup$ – Yakk Nov 8 '16 at 19:28
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    $\begingroup$ @Yakk 7 is not a symbol in base 5. $\endgroup$ – jpmc26 Nov 8 '16 at 19:51
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    $\begingroup$ 5 is not a symbol in base 5 either. $\endgroup$ – fluffy Nov 8 '16 at 22:35
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    $\begingroup$ Every base is base 10. $\endgroup$ – Joel Cohen Nov 8 '16 at 23:00
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    $\begingroup$ @JoelCohen except base 1. $\endgroup$ – hobbs Nov 9 '16 at 0:03

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.

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    $\begingroup$ IMO, $f(x)$ should never be used to denote the function as an object in its own right, but indeed many people (most?) do this all the time. $\endgroup$ – leftaroundabout Nov 8 '16 at 18:30
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    $\begingroup$ @leftaroundabout See here $\endgroup$ – Daniel Fischer Nov 8 '16 at 18:45

(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

$$ f : \mathbb{R} \to \mathbb{R} : x \mapsto 2x-3 $$


  • $\mathbb{R} \to \mathbb{R}$ 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.

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


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


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 $x=x+1.$


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