Thomas' Calculus 13th Edition

Published by Pearson

Appendices - Section A.1 - Real Numbers and the Real Line - Exercises A.1 - Page AP-6: 10

Answer

$-2\lt x\lt 2,\qquad$ or, $x\in(-2,2)$

Work Step by Step

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 $|x|\lt 2\quad\Leftrightarrow\quad -2\lt x\lt 2$ This is an open interval, $(-2,2)$ (see table: "Types of intervals".)

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