I am trying to prove or refute the following statement. $$O({{n}^{2}}+4n)\subseteq O({{n}^{2}}-30n)$$ How do I prove or refute it?
So far I've taken $f(n)$ from $O({{n}^{2}}+4n)$ and then tried to prove it from there by showing $f(n)\le c({{n}^{2}}+4n)\,,\,n\ge {{n}_{o}}$ and then I am stuck. The $-30n$ for me is the problem.
The strategy is to multiply $n^2-30n$ by $2$ so that it's eventually bigger than $n^2+4n$.
When is $2(n^2-30n)>n^2+4n$? Using algebra: $$2n^2-60n>n^2+4n \iff n^2>64n \iff n>64$$
Recalling the fact that $f(n) = O(n^2 + 4)$, we have that $$f(n) \leq c(n^2+4n) < 2c(n^2-30n)\text{ when $n>\max(n_0,64)$} $$
So the new $c$ is $2c$ and the new $n_0$ is $\max(n_0, 64)$