0
$\begingroup$

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?

$\endgroup$
5
  • 3
    $\begingroup$ Does crescent mean increasing? $\endgroup$ May 19, 2010 at 15:06
  • $\begingroup$ I think it's a translation from something that means "waxing" (like the moon, or "waning" I can never remember which is which). $\endgroup$ May 19, 2010 at 15:38
  • $\begingroup$ 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). $\endgroup$ May 19, 2010 at 15:52
  • $\begingroup$ "Croissant" has the obvious dual sense in French (AFAIK) so I wouldn't be surprised to see this in other, related languages $\endgroup$
    – Yemon Choi
    May 19, 2010 at 23:05
  • $\begingroup$ Crescendo... $ $ $\endgroup$ May 20, 2010 at 7:55

1 Answer 1

5
$\begingroup$

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 $f'(0)=1$.

$\endgroup$
2
  • 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. $\endgroup$ May 19, 2010 at 15:26
  • $\begingroup$ 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... $\endgroup$ May 20, 2010 at 7:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.