Can a function be increasing *at a point*? From what I understand we say that a function is increasing on an interval $I$ if
$$
x_1 &lt; x_2 \quad\Rightarrow\quad f(x_1) &lt; f(x_2).
$$
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 $f'(x) &gt; 0$ 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 $f'(a) &gt; 0$. 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 
$$
f(x) = \begin{cases}x^2\sin\left(\frac{1}{x}\right) + \frac{x}{2} &amp; \text{if }x\neq 0 \\ 0 &amp; \text{if } x = 0\end{cases}.
$$
This function is differentiable at $0$ with $f'(0) = \frac{1}{2} &gt; 0$. 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 $f'(\frac{1}{2k\pi}) = -\frac{1}{2} &lt; 0$. 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 $v(a)&gt;0$ 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?
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.
The truth is that it doesn't really make sense to say that a function is increasing or decreasing at point. However, due to culture and "bad lexicon,"
people usually say that a function is increasing at point when they really mean to say that the point is an interval in which the function is increasing. Most people who know math are aware of this, so they usually aren't bothered by this language. For students learning calculus for the first time this is unfortunate. But English speakers say many things that don't really make sense like driving on parkways and parking in driveways, etc. Also, you could try arguing with someone that a so-called combination lock should really be called a permutation lock. Please don't take everything you hear too seriously.