Calculus (3rd Edition)

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

Chapter 2 - Limits - 2.8 Intermediate Value Theorem - Exercises - Page 86: 6



Work Step by Step

One can see that the function $ f(x)=x^3+2x+1 $ is continuous on $[-\frac{1}{2},0]$ and $ f(-\frac{1}{2})=-\frac{1}{8}\neq f(0)=1$ and $0$ is between $ f(\frac{1}{2})$ and $ f(0)$. Then, by the Intermediate Value Theorem, the function $f(x)$ takes on the value $0$ for some $x\in (-\frac{1}{2},0)$. Hence, $ f(x)$ has a root in the interval $[-\frac{1}{2},0]$ which has length $\frac{1}{2}$.
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