Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 4 - Polynomial and Rational Functions - 4.4 Polynomial and Rational Inequalities - 4.4 Assess Your Understanding - Page 220: 67

Answer

$(-\infty,-2] \cup [2,\infty)$

Work Step by Step

The equation according to the text: $x^4-16\geq0\\(x-2)(x+2)(x^2+4)\geq0$ The zeros: $x+2=0\\x=-2\\$ or $x-2=0\\x=2$. Then, according to the table ($x^2+4$ is non-negative because $x^2$ is non-negative) (where I used the zeros ($x=-2,2$) to divide the number line into different parts), the solution set: $(-\infty,-2] \cup [2,\infty)$ because the function is non-negative in this interval.
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