Precalculus (6th Edition) Blitzer

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

Chapter 2 - Section 2.3 - Polynomial Functions and Their Graphs - Exercise Set - Page 349: 33

Answer

\[f\left( x \right)={{x}^{3}}-x-1\] has a zero between 1 and 2.

Work Step by Step

Find $f\left( 1 \right)$ and $f\left( 2 \right)$. If the sign change in values of $f\left( 1 \right)$ and $f\left( 2 \right)$ takes place, then there must be at least one zero between 1 and 2. That is, $\begin{align} & f\left( x \right)={{x}^{3}}-x-1 \\ & f\left( 1 \right)=1-1-1 \\ & f\left( 1 \right)=-1, \\ \end{align}$ Which is negative. And $\begin{align} & f\left( x \right)={{x}^{3}}-x-1 \\ & f\left( 2 \right)=8-2-1 \\ & f\left( 2 \right)=5, \\ \end{align}$ Which is positive. Therefore, there is a sign change in the value of $f\left( 1 \right)$ and $f\left( 2 \right)$ and so there is at least one zero between 1 and 2.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.