Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 5 - Graphs and the Derivative - 5.3 Higher Derivatives, Concavity, and the Second Derivative Test - 5.3 Exercises - Page 283: 12

Answer

$$f''\left( x \right) = \frac{4}{{3{x^{5/3}}}},f''\left( 0 \right) = {\text{Undefined}},f''\left( 2 \right) = \frac{4}{{3{{\left( 2 \right)}^{5/3}}}}$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = - 6{x^{1/3}} \cr & {\text{find the derivative of }}f\left( x \right) \cr & {\text{ }}f'\left( x \right) = \frac{d}{{dx}}\left[ { - 6{x^{1/3}}} \right] \cr & {\text{ }}f'\left( x \right) = - 6\frac{d}{{dx}}\left[ {{x^{1/3}}} \right] \cr & {\text{use power rule }}\frac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{n - 1}} \cr & {\text{ }}f'\left( x \right) = - 6\left( {\frac{1}{3}{x^{ - 2/3}}} \right) \cr & {\text{ }}f'\left( x \right) = - 2{x^{ - 2/3}} \cr & \cr & {\text{find the derivative of }}f'\left( x \right) \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ { - 2{x^{ - 2/3}}} \right] \cr & f''\left( x \right) = - 2\frac{d}{{dx}}\left[ {{x^{ - 2/3}}} \right] \cr & {\text{use power rule}} \cr & f''\left( x \right) = - 2\left( { - \frac{2}{3}{x^{ - 5/3}}} \right) \cr & f''\left( x \right) = \frac{4}{3}{x^{ - 5/3}} \cr & f''\left( x \right) = \frac{4}{{3{x^{5/3}}}} \cr & \cr & {\text{find }}f''\left( 0 \right){\text{ and }}f''\left( 2 \right) \cr & f''\left( 0 \right) = \frac{4}{{3{{\left( 0 \right)}^{5/3}}}} = \frac{4}{0}.{\text{ then the second derivative is not defined for }}x = 0 \cr & f''\left( 2 \right) = \frac{4}{{3{{\left( 2 \right)}^{5/3}}}} \cr} $$
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