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Let $f(x)$ be a real-valued function with $c$ in its domain. Given
1. $f(c)=0, $
2. $f$ is not identically zero in any interval about $c$, and
3. $f$ is differentiable at $c$.

Claim:
a. If $f(x)$ changes sign at $c$, then $f’(x)$ does not change sign at $c$.
b. If $f(x)$ does not change sign at $c$, then $f’(x)$ changes sign at $c$.

How would I go about proving this claim?

Thoughts:
The main difficulty I have with this question is that I cannot express "changing sign" in a way involving derivatives that makes this result amenable to a proof I can come up with.

The one precise meaning of "changing sign" I thought of is:
$f(x)$ changes sign at $c$ if there exists an open set $S$ containing $c$ such that for all $y\in S$ and $y<c$, $f(y)$ is positive (negative) and for all $y\in S$ and $y>c$, $f(y)$ is negative (positive).

It's also clear that: If $f(x)$ changes sign at $c$, $f(c)=0$, but the converse is not true.

Suggestions would be appreciated.

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  • $\begingroup$ Hi all - thanks for the suggestion, I have some thoughts I have on this question so far. Suggestions to proceed, or a counterexample, would be appreciated! $\endgroup$
    – FreshAir
    Oct 31, 2015 at 1:13

1 Answer 1

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Here's a counterexample to (b) according to your definition.

$$f(x)=x^3sin(1/x), x \neq 0$$ $$f(0)=0$$

According to your definition, neither f nor f' changes sign at $x=0$.

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  • $\begingroup$ Your example clearly proves that $f$ differentiable at $c$ is not nearly good enough. And by replacing $x^3$ with $x^n$, it's also easy to show that $n$ times differentiable at $c$ isn't good enough either. That only leaves $C^\infty$ and analytic to check. I suspect all counter examples have bad oscillation like this. Can we prove that? $\endgroup$
    – Zach Stone
    Oct 31, 2015 at 2:44
  • $\begingroup$ @ZachStone, to eliminate the infinite oscillation case, you need to require that on each side of c there exists an open interval where f does not change sign. $\endgroup$
    – Paul
    Oct 31, 2015 at 2:52
  • $\begingroup$ For $C^{\infty}$ couldn't you do something like $e^{-1/x^2}\sin(1/x)$? I'm pretty sure that this is true for analytic though. $\endgroup$
    – Alex Zorn
    Oct 31, 2015 at 2:54
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    $\begingroup$ @Paul In the case where it has constant sign on either side of $c$, continuity plus mean value theorem combine to give a simple proof of (b). Even better, it doesn't require $f$ to even be continuously differentiable. $\endgroup$
    – Zach Stone
    Oct 31, 2015 at 2:56

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