Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.1 - Maximum and Minimum Values - 4.1 Exercises - Page 283: 25

Answer

There is no absolute minimum. The absolute maximum is $f(0) = 1$ There is no local minimum. There is no local maximum.

Work Step by Step

$f(x) = 1-\sqrt{x}$ Note that this function is only defined for values of $x$ such that $0 \leq x$ There is no absolute minimum. The absolute maximum is $f(0) = 1$ There is no local minimum. There is no local maximum.
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