Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 4: Applications of Derivatives - Practice Exercises - Page 244: 19

Answer

Global minimum $y=\frac{1}{2}$ at $x=2$

Work Step by Step

From the graph of the given function, we can identify a global (absolute) minimum of $y=\frac{1}{2}$ at $x=2$. Point $(1,1)$ is not a maximum because the point is not in the domain of the function.
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