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 - Chapter 3 Review Exercises - Page 260: 36

Answer

$${\text{ }}f\left( x \right){\text{ has a relative maximum at }}x = - \frac{{27}}{{64}}$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = 2x + {x^{2/3}} \cr & {\text{Calculate the derivative of }}f\left( x \right) \cr & f'\left( x \right) = 2 + \frac{3}{2}{x^{ - 1/3}} \cr & {\text{Find the critical points, set }}f'\left( x \right) = 0 \cr & 2 + \frac{3}{2}{x^{ - 1/3}} = 0 \cr & 2 + \frac{3}{{2\root 3 \of x }} = 0 \cr & \frac{3}{{2\root 3 \of x }} = - 2 \cr & \frac{3}{2} = - 2\root 3 \of x \cr & \root 3 \of x = - \frac{3}{4} \cr & x = {\left( { - \frac{3}{4}} \right)^3} \cr & x = - \frac{{27}}{{64}} \cr & {\text{Calculate the second derivative}} \cr & f''\left( x \right) = \frac{3}{2}\left( { - \frac{1}{3}} \right){x^{ - 4/3}} \cr & f''\left( x \right) = - \frac{1}{2}{x^{ - 4/3}} \cr & {\text{Use the second test derivative into }}x = - \frac{{27}}{{64}} \cr & f''\left( { - \frac{{27}}{{64}}} \right) = - \frac{1}{2}{\left( { - \frac{{27}}{{64}}} \right)^{ - 4/3}} \cr & f''\left( { - \frac{{27}}{{64}}} \right) = - \frac{{128}}{{81}} \cr & f''\left( { - \frac{{27}}{{64}}} \right) < 0,{\text{ then }}f\left( x \right){\text{ has a relative maximum at }}x = - \frac{{27}}{{64}} \cr} $$
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