Can a function have a positive derivative at a point but not be increasing/decreasing around that point?

Question

TL:DR, if you have a function centered at "a" with the derivative at "a" being positive, can the function not be increasing around a? What does that mean about points around "a"?

I had a problem given to me in a math tutorial recently that asked:

Construct a function f satisfying all the following properties:

•Domain f = $\mathbb{R}$

•f is continuous

•$f^{\prime}(0)=0$

•f does not have a local extremum at $0$.

•There isn’t an interval centered at $0$ on which $f$ is increasing.

•There isn’t an interval centered at $0$ on which $f$ is decreasing.

The function we were shown that worked was $f(x) = \displaystyle \lim_{c \to x}c^{2}\sin(\frac{1}{c})$.

I was wondering if instead, the derivative of some number "$a$" was strictly larger than zero ($f^{\prime}(a)>0$) and the extremum was at "$a$" and the intervals centred at "$a$" if a function could be created to meet the criteria. Can you think of criteria that could be removed to make this work if it doesn't? Would love to learn more about these types of derivative questions, since this one was quite confusing and interesting.

Answer 1

The function $f$ defined by $$ f(x) = \begin{cases} x^2 \sin\left(\frac1x\right) & x\neq 0, \\ 0 & x = 0 \end{cases} $$ has some interesting properties beyond the listed ones. One of these is, no matter how small a neighborhood around $0$ you take, there is always an interval in that neighborhood on which $f'(x) > \frac12$ and an interval on which on which $f'(x) < -\frac12.$

So let $g(x) = 3f(x) + x.$ Then we have $g'(0) = 1 > 0.$ Moreover, if $N$ is a neighborhood of $0$ (that is, $N$ is an interval centered at $0$), then

• there is an interval within $N$ in which $g'(x) = 3f'(x) + 1 < 3 \left(-\frac12\right) + 1 = -\frac12,$ and therefore $g$ is decreasing in that interval and therefore not increasing in $N$;
• and similarly there is an interval within $N$ in which $g'(x) > \frac12,$ so $g$ is increasing in that interval and not decreasing in $N$.

So $g(x) = 3f(x) + x$ is a function with the following properties:

• $\operatorname{dom}(g) = \mathbb{R}$.
• $g$ is continuous
• $g'(0) > 0$.
• $g$ does not have a local extremum at $0$.
• There isn't an interval centered at $0$ on which $g$ is increasing.
• There isn't an interval centered at $0$ on which $g$ is decreasing.

If you want all those properties to apply at $x=a$ for some fixed non-zero number $a$ instead of at $x=0,$ just replace $g$ by the function $h$ defined by $h(x) = g(x - a).$

Answer 2

Sure, and we can get such a function by some usual transformation tricks on the example you already saw; let's call it $f_1$:

$$ f_1(x) = \begin{cases} x^2 \sin \left(\frac{1}{x}\right) & x \neq 0 \\ 0 & x = 0 \end{cases} $$

A real function with $f_2'(a) = b>0$ but not increasing (or decreasing) on any interval containing $a$ is:

$$ f_2(x) = \begin{cases} b(x-a) + (b+1)f_1(x-a) & x \neq a \\ 0 & x = a \end{cases} $$

Knowing that $f_1$ is continuous and differentiable at zero with $f_1'(0)=0$ makes it simple to show that $f_2$ is continuous and differentiable at $a$ with $f_2'(a)=b$.

At $x \neq a$, the derivative of $f_2$ is

$$ f_2'(x) = b + (b+1)\left[2(x-a) \sin\left(\frac{1}{x-a}\right) - \cos\left(\frac{1}{x-a}\right)\right] $$

When $\frac{1}{x-a}$ is an even (non-zero) multiple of $\pi$, we have:

$$ f_2'\left(a+\frac{1}{2n\pi}\right) = -1 $$

Any open interval which contains $a$ also contains $x_1 = a+\frac{1}{2n\pi}$ for some integer $n$. Since the derivative is negative at that point, the interval also contains some point $x_2$ with $x_1<x_2$ but $f_2(x_1)>f_2(x_2)$, so $f_2$ is not increasing in any interval.