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: 58

Answer

$c=\displaystyle \frac{a+b}{2}$

Work Step by Step

$f9x)=x^{2}$ is continuous and differentiable over the interval $[a,b]$. $f(a)= a^{2},\qquad f(b)=b^{2},\qquad $ $f'(x)=2x,\qquad $ The Mean Value Theorem tells us that there exists a $c\in(a,b)$ such that $f'(c)=\displaystyle \frac{f(b)-f(a)}{b-a}$ $2c=\displaystyle \frac{b^{2}-a^{2}}{b-a}$ $2c=\displaystyle \frac{(b-a)(b+a)}{(b-a)}$ $c=\displaystyle \frac{a+b}{2}$
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