Just wanting to be sure I'm understanding this notation correctly. So if $f(x) \in O(x^2)$ then $f(x)$ is bounded by $x^2$. But that means it's also bounded by any function greater than $x^2$, for example any function $x^k$ where $k>2$.
Therefore $f(x) \in O(x^2) \Rightarrow f(x) \in O(x^k)$ for all $k>2$. Is this correct?
$f(x)=O(g(x))$ (or $f(x)\in O(g(x))$ as $x\to \infty$ does not mean $f$ is bounded by $g.$ It means there exists $m>0$ and there exists $x_0\in \Bbb R$ such that $x>x_0 \implies |f(x)|\le m|g(x)|.$ For example $10x^3+100 x^2=O(x^3)$ as $x\to \infty.$
So if $f(x)=O(x^2)$ take $m>0$ and take $x_1$ such that $x>x_1\implies |f(x)|\le m|x^2|=mx^2.$ Now let $x_0=\max (1,x_1).$ If $k>2$ we have $$x>x_0\implies |f(x)|\le mx^2\le mx^k=m|x^k| .$$ So $f(x)=O(x^k)$ for $k> 2.$
In general if $f(x)=O(g(x))$ and $g(x)=O(h(x))$ then $f(x)=O(h(x))$.