Show that f(x) = 21x^2 - 17x - 3 is Big omega (x^2) .
I really dont know where to start. Please help.
Thanks
Hi buddy!
You have to show:
$\displaystyle \exists\ c > 0\ : |f(x)| \ge c\cdot |g(x)|$
g(x) is in this case equal to x^2
So,
$\displaystyle 21x^2-17x - 3 \ge c*x^2$
This is what we want to show
Deviding by x² leads to
$\displaystyle \frac{21x^2-17x - 3}{x^2} \ge c$
or easier
$\displaystyle 21 - \frac{17}{x} - \frac{3}{x^2} \ge c$
Now we determine x to infty and get
$\displaystyle lim_{x \to \infty} [ 21 - \frac{17}{x} - \frac{3}{x^2} ] \ge c$
$\displaystyle lim_{x \to \infty} [ 21 - 0 - 0 ] \ge c$
So the claim is correct, it is in big omega(x²), because it is
$\displaystyle 0 < c \le 21$
Any questions?
Yours
Rapha