Pick a "c" and "n" for the problem below.

f(n) = 2nlgn, g(n) = n^2-n

My answer is below, but I am not sure if I am correct. I've been trying to figure out this same problem for a while. **The real solution is at the bottom, but I don't know how 2nlgn turned into 2n^2**.Any help or suggestions is appreciated.

f(n) = 2nlgn, g(n) = n^2-n

2nlgn <= cn^2-n

2nlgn/n^2 + n/n^2 <= cn^2/n^2

2nlgn/n^2 + 1/n <= c

1 <= 1

c = 1, n = 1

Real solution:

2n lg n <= 2nn <= 2n^2 <= 3(n^2-n) n >= 3