Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.1 Real Numbers, Functions, and Graphs - Exercises - Page 10: 30

Answer

a. 9 is the maximum value b. $x^2 − 16 ≤ 9$ is true. See proof

Work Step by Step

Solve for x: $x-4\leq1$ and $x-4\geq-1$ $x\leq5$ and $x\geq3$ a. The maximum value can be found by using the greatest value of x, which is 5 $|5+4|=9$ b. $|x^2 − 16| ≤ 9$ solve for $x$: $x^2 − 16 ≤ 9$ and $x^2 − 16 \geq -9$ $x\leq5$ and $x\geq\sqrt{7}$, this does not contradict the previous conditions that we found for x, which is $x\leq5$ and $x\geq3$. Therefore, $x^2 − 16 ≤ 9$ is true.
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