Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Review Exercises - Page 430: 32

Answer

See explanations.

Work Step by Step

Step 1. Evaluate the given function at $x=1$; we have $f(1)=1^3-2(1)-1=-2$ Step 2. Evaluate the given function at $x=2$; we have $f(2)=2^3-2(2)-1=3$ Step 3. As $f(1)$ and $f(2)$ have opposite signs, based on the Intermediate Value Theorem, there exists a value $1\lt x\lt2$ such that $f(x)=0$
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