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Consider the implicit equation $ f^{-1} (x)=g(x)$. The function $g(x)$ is known and at least can be computed numerically. It may be piecewise continous or oscillating but it is always positive $ g(x) \ge 0 $. Here $ f(x) $ is not known.

Could it be that is there a function $ g(x) $ so it is never invertible and hence we cannot get $ f(x) $?

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What about $g(x)=0$? – Alex Becker Apr 29 '12 at 20:32
More generally: the inverse of a function is always one-to-one on its domain. – Robert Israel Apr 29 '12 at 20:36
$ g(x)=0 $ it can be viewed as the set of points $ (x,0) $ so its inverse would be the set of points $ (0,x) $ or more generally perhaps $ x=0 $ – Jose Garcia Apr 29 '12 at 20:41
@JoseGarcia You seem to be confusing "inverse" with "pre-image". The former doesn't necessarily always exist (as in the case g(x) = 0) but the latter does. An "inverse" by definition must be a function, and in particular cannot be "one-to-many" – Jonathan Apr 29 '12 at 21:45
@JoseGarcia Note that $x=0$ is not a function. How would you write it as $y=\text{something}$? – Pedro Tamaroff Apr 29 '12 at 22:31

This might be useful for you:

DEFINITION: Let $f:A\to B$ and $g:B \to A$ be given. The function $f$ is called the inverse of $g$ and the function $g$ is called the inverse of $f$ if $g(f(a))=a$ for each $a \in A$ and $f(g(b))=b$ for each $b \in B$. In this event we sall also say that $f$ and $g$ are inverse functions and that each of the is invertible.

It is a consequence of this definition that if $f$ and $g$ are inverses, then both of them are one-one and onto:

  1. $f$ is one-one if you have $x,y \in A$ then $f(x)=f(y)\Leftrightarrow x=y$.
  2. $f$ is onto if $f(A)=B$.

As a general result, it is necessary and sufficient that $f$ is onto and one-one for $f$ to be invertible.

An example is the definition of $\arcsin x$. To define it, we must first change

$$f:\mathbb R \to \mathbb R\text{ ; } x\mapsto\sin x$$


$$f:\left[-\frac {\pi} 2, \frac {\pi} 2\right] \to [-1,1]\text{ ; } x\mapsto\sin x$$

Since in such definition, $\sin x$ is onto and one-one, it follows we can define

$$g: [-1,1]\to \left[-\frac {\pi} 2, \frac {\pi} 2\right] \text{ ; } x\mapsto\arcsin x$$

Directly answering your question. Let

  1. $p$: $f$ is invertible.
  2. $q$: $f$ is onto.
  3. $r$: $f$ is one-one.


$$(q\wedge r) \equiv p $$ or

$$(-q\vee -r) \equiv -p $$ I reccomend you read Chapter 1 of Introduction to Topology by Bert Mendelson.

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Reading a topology book seems a little overkill! – The Chaz 2.0 Apr 29 '12 at 22:55
@TheChaz Not really! The first chapter is merely Theory of Sets and gives a great introduction to functions and relations in general. – Pedro Tamaroff Apr 29 '12 at 22:56
Ah. Key words - first chapter :D – The Chaz 2.0 Apr 29 '12 at 23:10
@TheChaz Indeed. Chapter 1. Are you still around mymathforum? I have stopped visiting it. – Pedro Tamaroff Apr 29 '12 at 23:10
I mostly just moderate and add snide comments (which might be counterproductive...). I've learned a WHOLE lot more from MSE in the past year! – The Chaz 2.0 Apr 29 '12 at 23:21

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