Calculus with Applications (10th Edition)

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

Chapter 5 - Graphs and the Derivative - Chapter Review - Review Exercises - Page 297: 33

Answer

$$f''\left( x \right) = 36{x^2} - 10,\,\,\,\,\,f''\left( 1 \right) = {\text{26}},\,\,\,\,f''\left( { - 3} \right) = 314$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = 3{x^4} - 5{x^2} - 11x \cr & {\text{find the derivative of }}f\left( x \right) \cr & {\text{ }}f'\left( x \right) = \frac{d}{{dx}}\left[ {3{x^4} - 5{x^2} - 11x} \right] \cr & {\text{ }}f'\left( x \right) = \frac{d}{{dx}}\left[ {3{x^4}} \right] - \frac{d}{{dx}}\left[ {5{x^2}} \right] - \frac{d}{{dx}}\left[ {11x} \right] \cr & {\text{use power rule }}\frac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{n - 1}} \cr & {\text{ }}f'\left( x \right) = 3\left( {4{x^3}} \right) - 5\left( {2x} \right) - 11\left( 1 \right) \cr & {\text{ }}f'\left( x \right) = 12{x^3} - 10x - 11 \cr & \cr & {\text{find the derivative of }}f'\left( x \right) \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {12{x^3} - 10x - 11} \right] \cr & {\text{ }}f''\left( x \right) = \frac{d}{{dx}}\left[ {12{x^3}} \right] - \frac{d}{{dx}}\left[ {10x} \right] - \frac{d}{{dx}}\left[ {11} \right] \cr & {\text{use power rule}} \cr & {\text{ }}f''\left( x \right) = 12\left( {3{x^2}} \right) - 10\left( 1 \right) - 0 \cr & f''\left( x \right) = 36{x^2} - 10 \cr & \cr & {\text{find }}f''\left( 1 \right){\text{ and }}f''\left( { - 3} \right) \cr & f''\left( 1 \right) = 36{\left( 1 \right)^2} - 10 \cr & f''\left( 1 \right) = 26 \cr & {\text{and}} \cr & f''\left( { - 3} \right) = 36{\left( { - 3} \right)^2} - 10 \cr & f''\left( { - 3} \right) = 324 - 10 \cr & f''\left( { - 3} \right) = 314 \cr} $$
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