I just started a course on algorithm complexity and I tried to understand what was the big-o notation about. I think I understand it means a function is dominated by another, more simple. But what I don't understand is the following:
A function $f$ is bounded on an interval $\Leftrightarrow f(x)=\underset{x\to a}O(1)$
$1$ will never dominate $x^2$, yet $x^2$ is has a lower bound...
First of all, let me clarify the big-o notation. If $f(x)=O(g(x))$, then it means that $f(x)\leq c\cdot g(x)\,\forall x$.
Now, if the function is bounded, it means that it has both a finite minimum and a finite maximum. Say this maximum is $m$, at point (I assume) $x=a$. Since this is the maximum value the function can take, we have that
$$f(x)\leq m\,\forall x.$$
Furthermore, you can write $m$ as $m\cdot 1=O(1)$, so you end up having $$f(x)=O(1).$$
Essentially, what this means is that the function can be upper-bounded in the interval of interest by a sufficiently high constant.