Calculus, 10th Edition (Anton)

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

Chapter 3 - The Derivative In Graphing And Applications - 3.8 Rolle's Theorem; Mean-Value Theorem - Exercises Set 3.8 - Page 257: 5

Answer

$$c = 1$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = {x^2} - x;{\text{ }}\left[ { - 3,5} \right],{\text{ }}a = - 3,{\text{ }}b = 5 \cr & {\text{Calculate }}f'\left( c \right) \cr & f'\left( x \right) = 2x - 1 \cr & f'\left( c \right) = 2c - 1 \cr & f\left( x \right){\text{ is continuous and differentiable on the interval }}\left[ { - 3,5} \right], \cr & {\text{Using the Mean - Value Theorem}} \cr & f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}} \cr & 2c - 1 = \frac{{f\left( 5 \right) - f\left( { - 3} \right)}}{{5 - \left( { - 3} \right)}} \cr & 2c - 1 = \frac{{\left[ {{{\left( 5 \right)}^2} - \left( 5 \right)} \right] - \left[ {{{\left( { - 3} \right)}^2} - \left( { - 3} \right)} \right]}}{8} \cr & 2c - 1 = \frac{{20 - 12}}{8} \cr & 2c - 1 = 1 \cr & c = 1 \cr} $$
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