Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Section 2.5 - Continuity - 2.5 Exercises: 54

Answer

There is at least one root in the specified interval.

Work Step by Step

$f(x) = \ln x - x + \sqrt x$ Now use the intervals $(2, 3)$: $f(2) = \ln 2 - 2 + \sqrt 2$ $f(2) = -1.307 + 1.414$ $f(2) \approx 0.107$ $f(3) = \ln 3 - 3 + \sqrt 3$ $f(2) = -1.901 + 1.732$ $f(2) \approx -0.169$ So by definition of the Intermediate Value Theorem, $f(x)$ will take all values between $0.107$ and $-0.169$. So there is at least one root/value of $x$ in $(2,3)$ for which $f(x)$ is zero.
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