Prove that $n^3 - 19n^2 - 10n - 2$ = $ \Omega(n^3)$.

Question

The definition of f(n) and g(n) where f(n) = $\Omega$g(n) is that $\exists_{n_0}$ > 0 and c > 0, such that f(n) $\ge$ cg(n), and $\forall_n \ge n_0$.

However, when I look at both the graphs, it seems that $n^3$ grows much faster than $n^3 - 19n^2 - 10n - 2$. Is there a piece of the definition that I must missing such that the above statement is true or am I just kidding myself?

If there statement is correct, what would be an example of the constants c and $ n_0$ in this case to make it work?

Any help would be appreciated. Thank you.

Answer 1

for every c<1 we can find no such that for every n>no, f(n)>=cg(n).. for example let c=1/2 and no=1000.

From graphs perspective it is true that n^3 grows much faster than f(n) but for c<1, c*n^3 does not grow faster than f(n)