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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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Does crescent mean increasing? –  Gordon Craig May 19 '10 at 15:06
I think it's a translation from something that means "waxing" (like the moon, or "waning" I can never remember which is which). –  Michael Hoffman May 19 '10 at 15:38
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). –  Andrey Rekalo May 19 '10 at 15:52
"Croissant" has the obvious dual sense in French (AFAIK) so I wouldn't be surprised to see this in other, related languages –  Yemon Choi May 19 '10 at 23:05
Crescendo... $ $ –  Pete L. Clark May 20 '10 at 7:55
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1 Answer

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 $f'(0)=1$.

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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. –  Julián Aguirre May 19 '10 at 15:26
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... –  Pete L. Clark May 20 '10 at 7:53
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