Prove that $$f(x) = \frac{x^3-7x^2+6x+4}{x^2-5x+4}$$ is bounded on the the domain $D = \{x: 2 \le x \le 3\}.$
The answer key has the line
When $2 \le x \le 3$, we have $|x| < 3, 1 \le x-1 \le 2$ and $−4 \le x − 6 \le −3$, so that $|x−1| \le 2$ and $|x − 6| \le 4.$
Where do the $|x| < 3$ and the other absolute values come from?
Also, doesn't $|x|<3$ introduce additional values of $x$ that don't satisfy the original inequality?
On
The answer key is claiming not that $$2 \le x \le 3\iff |x| < 3,$$ but that $$2 \le x \le 3\implies|x| < 3;$$ similarly, it is claiming also that $$1 \le x-1 \le 2\implies |x−1| \le 2$$ and that $$−4 \le x − 6 \le −3\implies |x − 6| \le 4.$$
Basically, the goal is to investigate the consequence of $x$ lying in the domain $[2,3],$ in particular, that $f$ is bounded there. This is why the proof starts with the hypothesis that $2 \le x \le 3,$ then runs with it.
EDIT to add nickalh's complementary explanation
The extra values introduced by $|x|<3$ vs. the original $2≤x≤3$ have no effect on whether or not bounds exist for the rational function $f.$ The introduction of $|x|<3$ is merely a simplification that makes it easier to bound the denominator away from zero.
As a basic intuition, $|x|$ measures the distance of $x$ from the origin. As a result, if $2 \le x \le 3$ (or, similarly, if $-3 < x < -2$), the distance of $x$ from the origin is certainly less than $3$.
You can think about it in a different way, too. Construct a ball of length $3$ around the origin, which we will call $B_3(0)$ and note that $B_3(0)$ are the points $-3 \le x \le 3$. If $x$ obeys your inequality $2 \le x \le 3$, it will certainly obey the wider inequality.