# Preserving the extrema of one function after using an additional

Suppose we have some function $f(x)$ with neighborhood extrema at $x_1, x_2, \dots$, and also a 2nd function $g(x)$ which is continual, purely raising and also non-zero almost everywhere over the series of the $x_i$. Will $g(f(x))$ have its neighborhood extrema at the very same $x_i$ and also nothing else?

If so, exist any kind of noticeable helpings to loosen of the restraints on $g$ for which this will hold?

(I'm actually considering this in the context of signal processing, considering makeovers that maintain the aesthetic framework of a photo, yet it feels like a basic inquiry that has to have been trivially confirmed by a person 250 years ago ... )

Since g is continual and also purely raising, its inverted $g^{-1}$ is a function and also purely raising. Given that both are purely raising, $a<b\Leftrightarrow g(a)<g(b)\Leftrightarrow g^{-1}(a)<g^{-1}(b)$. From this, it adheres to that $x_i$ is a neighborhood max (minutes) of g (f (x)) iff it is a neighborhood max (minutes) of f (x).

If g were continual and also purely lowering, it would certainly trade neighborhood optimums and also minimums (due to the fact that $a<b\Leftrightarrow g(a)>g(b)$ and also $a<b\Leftrightarrow g^{-1}(a)>g^{-1}(b)$).

You do not require to think that $g$ is non - absolutely no, and also maybe purely lowering too. In addition, the problems on $g$ just require to hang on the photo of $f$ (which does not require to the be entire of $\mathbb{R}$, as an example).

On the various other hand, if $g$ has a neighborhood extremum at $y=f(z)$ and also $f$ is purely raising around $z$, after that you're clearly in problem, due to the fact that $g\circ f$ will certainly have a neighborhood extremum at $z$. Yet having no neighborhood extrema amounts being purely monotonic.

The only that could loosen up the conition on $g$ is that it has neighborhood extrema specifically where $f$ does and also they "terminate each various other out" or "intensify each various other".

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