1.2.7 An Example:

Let f (n) = 2n² + 4n + 10. f (n) is O(n²). For,

f (n) $\leq$ 3n^2$\forall$  n$\geq$ 6

Thus, C = 3 and n₀ = 6

Also,

f (n) $\leq$ 4n^2$\forall$  n$\geq$ 4

Thus, C = 4 and n₀ = 4

f (n) is O(n³)

In fact, if f (n) is O(n^k) for some k, it is O(n^h) for h > k

f (n) is not O(n).

Suppose $\exists$ a constant C such that 
                            2n² + 4n + 10 $$\leq$$Cn$$\forall$$n$$\geq$$n₀
This can be easily seen to lead to a contradiction. Thus, we have that:

f (n) is $\Omega$(n²) and f (n) is $\Theta$(n²)