Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 4 - Integration - 4.8 Average Value Of A Function And It's Applications - Exercises Set 4.8: 15

Answer

False.

Work Step by Step

Just because the average value of $g(x)$ is less than the average value of $f(x)$ doesn't mean that $g(x)$ can't be greater than $f(x)$ at a single point. For example, the average value of $f(x)=2$ on the interval $[0,3]$ is greater than the average value of $g(x)=x$ on $[0,3]$, but $g(3)$ $(y=3)$ is greater than $f(3)$ $(y=2).$
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