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How many extremum points (local maximums and minimums) can a continuous and differentiable function have on the bounded interval.

My guess it is still countable many.

If so what extra condition can we impose on the function that the number of extremum points is at most finite? Basically, if we show that $A=\{x: f'(x)=0 \}$ is finite should be enough.

I would like to clarify that we look for extremum points and not critical points. That is points at which the derivative is zero but around which the derivative is not zero. That is we exclude constant functions and functions constant on some intervals.

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  • $\begingroup$ do you have an example? $\endgroup$ – Dr. Sonnhard Graubner Oct 5 '16 at 13:11
  • $\begingroup$ If you asking if I have an example of a function with countable many extreme points. I do not have such an example. I would really like to see though. $\endgroup$ – Boby Oct 5 '16 at 13:27
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    $\begingroup$ I deleted my answer, as it was no longer relevant. To ensure only finitely many, there are some criteria that would be sufficient. If the function is $C^1$, then you could check that there are only finitely many zeroes in $f^{\prime}$, that would of course work. Otherwise, if there were some partition of $[a,b]$, $(x_1-a, x_2-x_1,....)$ with $x_i-x_{i-1}>\epsilon$ for some $\epsilon>0$ with $f$ monotone on every $[x_i-x_{i-1})$, this would also be good enough. What kind of criteria are you looking for? $\endgroup$ – Andres Mejia Oct 5 '16 at 13:57
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    $\begingroup$ Well that will be true as well, but I mean continuous $\endgroup$ – Andres Mejia Oct 5 '16 at 14:49
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    $\begingroup$ No. That is actually incorrect. You can in fact define a smooth function that has uncountably many zeroes, so long as the uncountable set is closed. I believe this is known as Whitney's theorem. My observation was much more trivial: check the zeroes of the derivative, if the derivative is continuous. $\endgroup$ – Andres Mejia Oct 5 '16 at 15:52

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