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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 $g(x)=0$?2012-04-29
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    More generally: the inverse of a function is always one-to-one on its domain.2012-04-29
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    $ 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 $2012-04-29
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    @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"2012-04-29
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    @JoseGarcia Note that $x=0$ is not a function. How would you write it as $y=\text{something}$?2012-04-29

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