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Is there any function whose derivative at a point is positive but it is not ascending or whose

derivative is negative but is not descending?

I have thought about this a lot, but I cannot find anything

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    $\begingroup$ What do you mean by ascending/descending? $\endgroup$ – apnorton Oct 25 '13 at 21:35
  • $\begingroup$ rowdy.msudenver.edu/~talmanl/PDFs/Quiz.pdf $\endgroup$ – Andrés E. Caicedo Oct 25 '13 at 21:47
  • $\begingroup$ Yes, you need to clarify your question. Do you mean ascending or descending on an interval? Ascending or descending at a point doesn't really mean anything. $\endgroup$ – Stefan Smith Oct 26 '13 at 0:02
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This question relates to a somewhat subtle issue, so rather than just leaving a link in the comments above, let me add an explanation:

Suppose first that $I$ is an interval on $\mathbb R$, that $f:I\to\mathbb R$, that $f'(a)>0$, and that $a$ is an interior point of $I$. We could just consider one-sided derivatives if $a$ is an end-point of $I$, but that seems an unnecessary distraction.

That $f'(a)>0$ means that $$\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}>0,$$ so if $h$ is sufficiently small, we can assume both that $a+h\in I$, and that $$\frac{f(a+h)-f(a)}{h}>0.$$ Considering $h>0$, this means that if $b>a$ is sufficiently close to $a$, then $f(b)>f(a)$. Considering $h<0$, this means that if $b<a$ is sufficiently close to $a$, then $f(b)<f(a)$. If this is all we mean by "$f$ is increasing at $a$" then indeed $f'(a)>0$ implies that. Similarly, $f'(a)<0$ would imply that $f$ is decreasing at $a$.

The definition here would be that $h$ is increasing at $a$ iff there is an interval $I$ about $a$ such that $h(b)<h(a)$ if $b<a$ and $b\in I$, and $h(a)<h(b)$ if $b>a$ and $b\in I$. But, really, this is a silly notion: The function below given by $$ h(x)=\left\{\begin{array}{cl}0&\mbox{ if }x=0,\\-x-3&\mbox{ if }-2\le x<0,\mbox{ and }\\-x+1&\mbox{ if }0<x\le 2,\end{array}\right. $$ is increasing at $0$. alt text

However, what is not true is that if $f'(a)>0$, then $f$ is increasing on an interval around $a$ (even if we assume that $f$ is differentiable everywhere). It is common to define the notions of increasing and decreasing so that they apply to functions defined on intervals, rather than to individual points of the domain: We say that $f$ is increasing on a set $A$ iff whenever $x<y$ are in $A$ then $f(x)\le f(y)$ (with $\le$ replaced by $<$ if we insist that "increasing" be interpreted in the strict sense.)

To see that indeed $f$ needs not be increasing on any interval containing $a$, no matter how small, consider the example suggested in these slides (that I mentioned in a comment above) by Louis A. Talman, on The Mother of All Calculus Quizzes: Let $g(x)$ be the function given by $g(0)=0$ and, if $x\ne0$, then $$ g(x)=\frac x2+x^2\sin\left(\frac1x\right). $$

This function is differentiable everywhere, with $\displaystyle g'(0)=\frac 12+\lim_{h\to 0}\frac{h^2\sin(1/h)}h=\frac12$, and $$g'(x)=\frac12+2x\sin\left(\frac1x\right)-\cos\left(\frac1x\right)$$ for $x\ne0$. Note that there are (both positive and negative) values of $x$ arbitrarily close to $0$ where $g'(x)<0$, so "$g$ is decreasing at $x$". This is because we can find arbitrarily small $x$ with $\cos(1/x)=-1$, so $\sin(1/x)=0$ and $g'(x)=-1/2<0$. (Note that the $x/2$ in the definition of $g$ could be replaced with any function with small enough derivative to ensure the same behavior.) alt text

Finally, note that $g'$ is not continuous at $0$. This is an essential feature of the example. For suppose that $f'$ exists on a neighborhood of $a$ and is continuous at $a$. If $f'(a)>0$, then for $b$ sufficiently close to $a$ we have $f'(b)>0$ as well. This means that $f$ is indeed increasing on a neighborhood of $a$. For example, if $x<y$ and they are close enough to $a$ to ensure that $f'(t)>0$ at all $t\in[x,y]$, then use the mean value theorem to see that there is a $z\in(x,y)$ with $$f(y)-f(x)=f'(z)(y-x)>0,$$ that is, $f(x)<f(y)$.


In a comment to another question, Dave L. Renfro suggested the following paper as "one literature entry point" to work on this notion and its variants: Jack B. Brown, Udayan B. Darji, and Eric P. Larsen. Nowhere monotone functions and functions of nonmonotonic type, Proceedings of the American Mathematical Society, 127 (1), (1999), 173-182. MR1469402 (99b:26015).

From the abstract:

We investigate the relationships between the notions of a continuous function being monotone on no interval, monotone at no point, of monotonic type on no interval, and of monotonic type at no point. In particular, we characterize the set of all points at which a function that has one of the weaker properties fails to have one of the stronger properties.

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    $\begingroup$ Short is such a relative term... :-) $\endgroup$ – Asaf Karagila Oct 25 '13 at 22:50
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One needs to be more precise what "is not ascending" means. We say that $f\colon I\to\mathbb R$ is ascending if for $a,b\in I$ with $a<b$ always $f(a)<f(b)$. Note that there is no notion of being ascending at a single point. While $f'(a)>0$ implies that there exists some $\delta>0$ with $f(b)>f(a)$ whenever $a<b<a+\delta$ (and similarly to the left of $a$), we cannot conclude from the derivative at a single point that there is in interval around $a$ where $f$ is ascending.

Take for example $$ f(x) =\begin{cases} x +x^2&\text{if }x\in \mathbb Q\\x&\text{if }x\notin \mathbb Q\end{cases}$$ Then $f$ is continuous and differentiable at $0$ and we find $f'(0)=1$. But there is no interval where $f$ would be increasing.

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