<s>[INST] Understanding the solution to this polynomial estimation problem Show that there exist $K, N > 0$  such that for $x ∈\mathbb{R}$,
$$x ≥ N \implies
\frac{3x^
2 − 4x + 8}{
5x + 6}
≥ Kx.$$
Solution:
For $x ≥ 4,$ we have $4x ≤ x^
2$
so that
$$3x^
2 − 4x + 8 ≥ 3x^
2 − x^
2 = 2x^
2.$$
Similarly, for $x ≥ 1,$ we have
$$5x + 6 ≤ 11x.$$
Therefore, when $x ≥ 4,$ we have
$$\frac{3x^
2 − 4x + 8}{
5x + 6}
≥\frac{
2x^
2}{
11x}
=
\frac{2}
{11} x,$$
and so we can take $N = 4$ and 
$K = \frac{2}{
11} .$
I'm struggling to follow the logical steps in this solution. I'm not sure why the inequality of $x ≥ 4$ and $x ≥ 1$ has been chosen. Can someone please break down this solution and show where the steps come from?
 [/INST] Essentially the method of attack that was used in this proof is to find a quadratic function that is less than the numerator and a linear function that is greater than the denominator. Then when you divide these you get a linear function which is less than the original $\frac{3x^2-4x+8}{5x+6}$. 
In layman's terms the question is asking if it is possible to find some real number $N$ such that $\frac{3x^2-4x+8}{5x+6}$ is greater than some linear function $Kx$ for all $x$ greater than or equal to $N$. 
So again we are looking for a quadratic equation which is less than $3x^2-4x+8$ for some values of $x > 0$. Notice that $3x^2 + 8 \geq 3x^2 $ for all $x > 0$. Now we need to account for the $-4x$ term still in the numerator. There's actually an infinite amount of choices we could do for the next step. In  the proof you presented the author realized that $x^2$ would be greater than or equal to $4x$ eventually (namely at $x = 4$). So from the inequality above we note that $3x^2 - 4x + 8 \geq 3x^2 - x^2 = 2x^2$ when $x \geq 4$ (because we subtracted something greater from the RHS than the LHS). 
Now for the next step again we have an infinite amount of choices, however the author realized that $5x +6 \leq 11x$ when $x \geq 1$. Of course they could just have easily said $5x+6 \leq 12x$ or an infinite number of other choices for that matter (thus finding a different value of $K$) but they chose this. Now since $5x + 6 \leq 11x$ when $x \geq 1, $ it is certainly also true when $x \geq4.$ 
Now for the final step, we can obtain the equation $\frac{3x^2 - 4x + 8}{5x+6} \geq \frac{2x^2}{11x} = \frac{2}{11}x$ for $x \geq 4$ because the numerator of the LHS is greater than the numerator of the RHS and the denominator of the LHS is less than the numerator of the RHS for $x\geq4$ as we showed above. 
Thus we can pick $N = 4$ and $K = \frac{2}{11}$. To make sure you understand it try and find another value of $N$ and $K$, for instance by realizing that $3x^2 - 4x + 8 \geq 3x^2 -2x^2 = x^2$ for $x \geq 4$ in the first step.  
</s>