How can I prove $20n = O(n^6)$

Question

I am trying to understand how to prove or disprove this:

$$20n = O(n^6) $$

I think it is false, but I am lost on how to show it.

Answer 1

Remember that

$$f(x) = O(g(x)) \mbox{ as } x \rightarrow \infty$$

is the same as saying

$$\exists M > 0, x_0 \in \mathbb{R}: |f(x)| \leq Mg(x)\ \forall x \geq x_0$$

i.e. there is a cut-off for $x$, beyond which $g$ times a constant is always larger than $f$.

In this case, there's an easy pick: let $M = 20$ and $x_0 = 1$. Then for $x \geq x_0$ it's pretty simple to see that $20x \leq 20x^6$, and hence $20x = O(x^6)$.