Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.5 Inequalities - R.5 Exercises - Page R-21: 37

Answer

$9-x^2\leq0$ $(3-x)(3+x)\leq0$ $-(x-3)(x+3)\leq0$ $(x-3)(x+3)\geq0$ $x \leq-3$ and $3\leq x$
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Work Step by Step

Step 1: Factorize inequality (difference of squares) Step 2: Factorize $-1$ out of first Bracket (common factor) Step 3: Multiply with $-1$ on both sides of the inequality Step 4: Since the inequality was multiplied with a negative the direction of the inequality flips. Step 5: The inequality function is greater than $0$ for $x \leq-3$ and $3\leq x$
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