Let $\;f : \; \stackrel{\circ}{D}\; \subset \mathbb{R} \to \mathbb{R}$ differentiable in $x_0 \in \; \stackrel{\circ}{D}\;$ and $f\;'(x_0) > 0$.
Does exists a neighborhood $A \subset \; \stackrel{\circ}{D}\;$ of $x_0$ where $f$ is crescent?
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3$\begingroup$ Does crescent mean increasing? $\endgroup$– Gordon CraigCommented May 19, 2010 at 15:06
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$\begingroup$ I think it's a translation from something that means "waxing" (like the moon, or "waning" I can never remember which is which). $\endgroup$– Michael HoffmanCommented May 19, 2010 at 15:38
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$\begingroup$ According to Wikipedia, the word "crescent" itself, derived from the Latin verb crescere "to grow", literally means "waxing" or "increasing", and was originally applied to the form of the waxing moon (luna crescens). $\endgroup$– Andrey RekaloCommented May 19, 2010 at 15:52
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$\begingroup$ "Croissant" has the obvious dual sense in French (AFAIK) so I wouldn't be surprised to see this in other, related languages $\endgroup$– Yemon ChoiCommented May 19, 2010 at 23:05
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$\begingroup$ Crescendo... $ $ $\endgroup$– Pete L. ClarkCommented May 20, 2010 at 7:55
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No. The function $$f(x)= x+2x^2\sin\frac{1}{x},\quad x\in\mathbb R,$$ is not monotonic in any neighborhood of $x=0$ yet $f'(0)=1$.
answered May 19, 2010 at 15:15
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4$\begingroup$ You just beat me. I would like to comment that f' is not continuous at 0. If we add the condition that f' be continuous at x_0, then f is increasing in an interval around x_0. $\endgroup$ Commented May 19, 2010 at 15:26
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$\begingroup$ Right, because then $f'$ is necessarily positive on an entire interval, not just at the point. Even in freshman calculus I try to make the distinction between "pointwise positivity" of $f'$ implies $f$ is increasing "through" the point -- i.e., smaller to the left, bigger to the right -- whereas "intervalwise positivity" implies increasing on an interval. I assumed this was part of the standard spiel. Now I wonder whether I am being too ambitious... $\endgroup$ Commented May 20, 2010 at 7:53