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: 30

Answer

$$\left( {1,\infty } \right)$$

Work Step by Step

$$\eqalign{ & x\sqrt {x - 1} > 0 \cr & \sqrt {x - 1} {\text{ is always positive and }}x > 1 \cr & {\text{Then}} \cr & {\text{The solution of the inequality is }}\left( {1,\infty } \right) \cr} $$
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