University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Section 4.2 - The Mean Value Theorem - Exercises - Page 224: 54

Answer

See proof below.

Work Step by Step

Let $s(t)$ be the function of his position, (s is in miles, t in hours) and let us assume that the function $s$ is continuous and differentiable over $[0,2.2].$ The function of velocity is the derivative of s, and we are given $s'(0)=s'(2.2)=0$ We also assume that the runner does not change direction (he runs always towards the finish line), so we treat velocity as speed. By the Mean Value Theorem, there exists a moment in time $t_{0}\in(0,2.2)$ where $s'(t_{0})=\displaystyle \frac{s(2.2)-s(0)}{2.2-0}=\frac{26.2}{2.2}=\frac{131}{11}\approx 11.09$ miles per hour We have $s'(0)=0,\ s'(t_{0})=11.09,$ and, since $s'$ is continuous, by the Intermediate Value Theorem, it assumes all values between $0 $ and $11.09$ on $[0,t_{0}]$. Thus, at least once in this interval, his speed was $11$ mph. We also have $s'(t_{0})=11.09,\ s'(2.2)=0$, So on the interval $[t_{0},2.2]$, his speed was $11$ mph at least once. Thus, the runner's speed was $11$ mph at least twice.
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