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 222: 7


$c=\dfrac{1+\sqrt 7}{3}, \dfrac{1-\sqrt 7}{3}$

Work Step by Step

Here, the function $f(x)=x^3-x^2$ is continuous on $[-1,2]$ and differentiable on $(-1,2)$ when $f'(x)=3x^2-2x$ . The Mean value Theorem states that there is a point $c \in (-1,2)$ such that $f'(c)=\dfrac{f(2)-f(-1)}{2-(-1)}=\dfrac{6}{3}=2$ Then $f'(c)=3c^2-2c=2 \implies 3c^2-2c-2=0$ This implies $c=\dfrac{1+\sqrt 7}{3}, \dfrac{1-\sqrt 7}{3}$
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