Thomas' Calculus 13th Edition

$\displaystyle \frac{2}{7}\lt x\lt \frac{2}{5},\qquad$ or, $x\displaystyle \in(\frac{2}{7},\frac{2}{5})$
Use properties from the table 'Absolute Values and Intervals". Applying Property 6: $|x|\lt a\quad\Leftrightarrow\quad -a\lt x\lt a, \quad$ for a = positive number We have: $|3-\displaystyle \frac{1}{x}|\lt \frac{1}{2}\quad\Leftrightarrow\quad -\frac{1}{2}\lt 3-\frac{1}{x}\lt \frac{1}{2}$ Applying rule 1 for inequalities, add $-3$ to each part of the compound inequality $-3-\displaystyle \frac{1}{2}\lt -\frac{1}{x}\lt -3+\frac{1}{2}$ $-\displaystyle \frac{7}{2}\lt -\frac{1}{x}\lt -\frac{5}{2}$ Apply rule 4: multiply with a negative number, $(-1)$ $\displaystyle \frac{7}{2}\gt \frac{1}{x}\gt \frac{5}{2}$ Apply rule 6 (reciprocals cause a change of inequality direction) $\displaystyle \frac{2}{7}\lt x\lt \frac{2}{5}$ This is an open interval, $(\displaystyle \frac{2}{7},\frac{2}{5})$ (see table: "Types of intervals".)