1. ## Big Omega question

Show that f(x) = 21x^2 - 17x - 3 is Big omega (x^2) .

Thanks

2. Hi buddy!

Originally Posted by micky_577
Show that f(x) = 21x^2 - 17x - 3 is Big omega (x^2) .

You have to show:
$\exists\ c > 0\ : |f(x)| \ge c\cdot |g(x)|$

g(x) is in this case equal to x^2

So,

$21x^2-17x - 3 \ge c*x^2$

This is what we want to show

$\frac{21x^2-17x - 3}{x^2} \ge c$

or easier

$21 - \frac{17}{x} - \frac{3}{x^2} \ge c$

Now we determine x to infty and get

$lim_{x \to \infty} [ 21 - \frac{17}{x} - \frac{3}{x^2} ] \ge c$

$lim_{x \to \infty} [ 21 - 0 - 0 ] \ge c$

So the claim is correct, it is in big omega(x²), because it is

$0 < c \le 21$

Any questions?

Yours
Rapha