Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Appendices - Section A.1 - Real Numbers and the Real Line - Exercises A.1 - Page AP-6: 19



Work Step by Step

The square root function is defined for nonnegative real numbers, and is an increasing function. This means that if a and b are nonnegative, $a \lt b\Rightarrow\sqrt{a} \lt \sqrt{b}$ An alternate definition of the absolute value is $|x|=\sqrt{x^{2}}$. All parts of the compound inequality sign are nonnegative, so we may take the square root of all parts, with the direction of inequality being unchanged: $\sqrt{4} \lt \sqrt{x^{2}} \lt \sqrt{9}$ $2 \lt |x| \lt 3$ Now we have two inequalities: $\left[\begin{array}{lll} & 2 \lt |x| & \\ & \swarrow\searrow & \\ x \lt -2 & ..or.. & x \gt 2 \end{array}\right]\qquad$and$\qquad \left[\begin{array}{l} |x| \lt 3\\ -3 \lt x \lt 3\\ \end{array}\right]$ $x $ is in the intersection of $(-3,3)$ with $(-\infty,-2)\cup(2,\infty)$ From the image below, this solution set in interval form is $x\in(-3,-2)\cup(2,3)$
Small 1583193122
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.