Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Appendix A - Appendix A Exercises - Page 1158: 33

Answer

$$\left( { - 2, - 1} \right) \cup \left( {2,3} \right)$$

Work Step by Step

$$\eqalign{ & 3 < \left| {2x - 1} \right| < 5 \cr & {\text{The solution of the inequality is any number that is a solution }} \cr & {\text{of both of these inequalities}}: \cr & 3 < \left| {2x - 1} \right|{\text{ and }}\left| {2x - 1} \right| < 5 \cr & \cr & {\text{Solving }}3 < \left| {2x - 1} \right|,{\text{use the property }}\left| x \right| \geqslant a \Leftrightarrow x \leqslant - a{\text{ or }}x \geqslant a \cr & 2x - 1 < - 3{\text{ or }}2x - 1 > 3 \cr & 2x < - 2{\text{ or }}2x > 4 \cr & x < - 1{\text{ or }}x > 2 \cr & {\text{Express in form of intervals}} \cr & \left( { - \infty , - 1} \right) \cup \left( {2,\infty } \right) \cr & \cr & {\text{Solving }}\left| {2x - 1} \right| < 5,{\text{use the property }}\left| x \right| \leqslant a \Leftrightarrow - a \leqslant x \leqslant a \cr & - 5 < 2x - 1 < 5 \cr & - 4 < 2x < 6 \cr & - 2 < x < 3 \cr & {\text{Express in form of intervals}} \cr & \left( { - 2,3} \right) \cr & \cr & {\text{Intersecting both solutions}} \cr & \left[ {\left( { - \infty , - 1} \right) \cup \left( {2,\infty } \right)} \right] \cap \left( { - 2,3} \right) \cr & {\text{We obtain}} \cr & \left( { - 2, - 1} \right) \cup \left( {2,3} \right) \cr & {\text{The solution set is }} \cr & - 2 < x < - 1{\text{ or }}2 < x < 3 \cr} $$
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