So I'm reading Introductions to Algorithms and sometimes I wish it would be a little bit more friendly with how it explains topics. One of these topics is proving big-theta. I understand that the definition of $\Theta$ is as follows: $\Theta(g(n)) = f(n)$ if there exists positive constants $c_1$, $c_2$, and $n_0$ such that $0 \le c_1 g(n) \le f(n) \le c_2 g(n)$ for all $n \ge n_0$.In other words, for $f(n)$ and $g(n)$, $f(n)$ can be bounded by $c_2 g(n)$ from above and bounded below by $c_1 g(n)$. So the example in the book goes: Show that $\frac{1}{2} n^2 - 3 n = \Theta(n^2)$.From the definition of big theta: $$ c_1 n^2 \le \frac{1}{2} n^2 - 3 n \le c_2 n^2$$CLRS begins by dividing by the largest order term of $n$ which is $n^2$ to get $$ c_1 \le \frac{1}{2} - \frac{3}{n} \le c_2 $$From here we split the problem into two parts, the right-hand inequality and the left-hand inequality. On the right hand side: CLRS chooses $c_2 \ge \frac{1}{2}$ because for $n \gt 1$, $\frac{1}{2} - \frac{3}{n}$ can never be less than $\frac{1}{2}$ since $\frac{3}{n}$ goes to $0$ as $n$ goes to infinity. (I'm assuming they choose $n \gt 1$ here because if $n=0$ then we would be dividing by $0$.)Now this is where I start to get lost. On the left hand side: CLRS chooses  $c_1 = \frac{1}{14}$ (not sure why) and $n \ge 7$. I'm not sure what the significance of these choices is. At $n \le 6$, $\frac{1}{2}-\frac{3}{n}$ becomes negative. But why $\frac{1}{14}$ for $c_2$? I'm just not sure how they arrived at solving the left hand side and the book doesn't really explain it well for me.   
Trouble understanding how to pick constants to prove big theta
complexity theory;asymptotics
For $n \leq 6$, $$1/2 - 3/n \le0,$$so if you are taking integer value of $n$, put $n = 7$ as it gives $1/14$. This is how he arrived in the conclusion i.e., for $n \geq 7$, $$ 1/2 - 3/n \ge1/14,$$because $n$ is in the denominator it will only decrease the value of -ve term, so overall value will increase with $n$, which is the constant $c_1$.