Can a function be increasing *at a point*?

From what I understand we say that a function is increasing on an interval $I$ if $$ x_1 < x_2 \quad\Rightarrow\quad f(x_1) < f(x_2). $$ for all $x_1,x_2\in I$. I understand that some might call this strictly increasing and allow for equality when saying increasing.

We know that when a function is differentiable on the interval $I$, then if $f'(x) > 0$ on $I$ then $f$ is increasing on $I$. But here we have to be careful because if $f(x) = x^3$ then $f$ indeed is increasing on $\mathbb{R}$ even though $f'(0) = 0$. So we can't conclude that if $f$ is increasing, then $f'(x)$ must be positive on the interval.

Also, even though $f(x) = x^2$, then $f$ is increasing on $[0,\infty)$ even though $f'(0) = 0$.

My question is

Does it make sense to talk about a function being increasing at a point?

The only definition I can think of making is to say that $f$ is increasing at a point $a$ if $f'(a) > 0$. But the problems with saying this is that then the function given by $f(x) = x^3$ is not increasing at $0$ even thought it is increasing on all of $\mathbb{R}$.

I am aware (from here) of the example with $$ f(x) = \begin{cases}x^2\sin\left(\frac{1}{x}\right) + \frac{x}{2} & \text{if }x\neq 0 \\ 0 & \text{if } x = 0\end{cases}. $$ This function is differentiable at $0$ with $f'(0) = \frac{1}{2} > 0$. For $x\neq 0$ we have $f'(x) = 2x\sin(1/x) - \cos(1/x) + 1/2$. And so for all $x_k = \frac{1}{2k\pi}$ we have $f'(\frac{1}{2k\pi}) = -\frac{1}{2} < 0$. That is, there is not interval around $0$ where $f$ is increasing. And so it shouldn't really make sense to say that $f$ is increasing at $0$.

One example I can think of where it maybe should make sense to take about a function being increasing at a point is when when we take the derivative of function $s(t)$ that gives the position of a particle. The derivative is then the velocity $v(t) = s'(t)$. And maybe this is just a language issue, but here it sounds reasonable saying that if $v(a)>0$ then the velocity is positive at time $t=a$ and so the rate of change is positive at $a$. This is almost saying that the position is increasing at that point.

Again, my question is just if it makes sense to talk about a function being increasing (or decreasing) at a point. If so, how is this defined?

Solution 1:

A function $f$ is increasing at $x$ if $f(t)\gt f(x)$ for every $t\gt x$ close enough to $x$ and $f(t)\lt f(x)$ for every $t\lt x$ close enough to $x$. More rigorously, one asks that there exists $\varepsilon\gt0$ such that, for every $(t,s)$ such that $x-\varepsilon\lt t\lt x\lt s\lt x+\varepsilon$, $f(t)\lt f(x)\lt f(s)$.

No notion of differentiability is needed. Consider for example the function $f$ defined by $f(t)=2t$ for every rational $t$ and $f(t)=t$ for every irrational $t$. Then $f$ is increasing at $x=0$ and only at $x=0$ while $f$ is nowhere differentiable.