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: 22

Answer

$0$, $3$ ,$-3$

Work Step by Step

$x^{3}-9x=0\Rightarrow x\left( x^{2}-9\right) =0$ $1)x=0;$ $2)x^{2}-9=0\Rightarrow x=\pm \sqrt {9}=\pm 3$
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