I had an argument with my maths teacher today... He says, along with another classmate of mine that $x^3$ is increasing for all intervals. I argue that it isn't. If we look at conditions for increasing, decreasing and stable states of a function, for a function $f(x)$:
function is increasing for $f'(x) > 0$
function is decreasing for $f'(x) < 0$
(as per my knowledge) function is neither increasing nor decreasing (i.e. it's stable) for $f'(x) = 0$
so for $x = 0$, function $f(x) = x^3$ should be stable, NOT INCREASING. Hence the interval for which the function must be increasing, according to me, would be $\mathbb{R}$ - {$0$}.
This means, the function $f(x)=x^3$ should be increasing for all real values EXCEPT $x=0$.
My teacher denied, saying that we consider values of $f'(x) >= 0$. My response was to look at the function $f(x) = x^2$ which is increasing for interval $(0, \infty]$ and decreasing for interval $(\infty, 0]$, without a doubt.
again, I reason that this time too, $x=0$ is neither decreasing nor increasing hence it is not included in either of the intervals. My teacher, had nothing to say on this, simply telling me that these two examples are completely incomparable.
[another reasoning that my classmate and my teacher gave was that $x=0$ is a single point. Thus the function has to be constant there. Since the function is increasing before and after the point $x=0$, the function must be increasing for value $x=0$ as well.
In my opinion, that is completely against the very idea of calculus. When we define the rate of change at a point, it is solely attributed to that point alone. The limit of the rate of change, as points around $x=0$ close in on 0 is 0, hence in my opinion, the function should be constant at $x=0$, while increasing elsewhere in the set of real numbers]
Am I correct?