Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 4 - Section 4.1 - Maximum and Minimum Values - 4.1 Exercises - Page 287: 31

Answer

$x = 0,{\text{ }}x = - 4,{\text{ }}x = 2$

Work Step by Step

$$\eqalign{ & f\left( x \right) = 3{x^4} + 8{x^3} - 48{x^2} \cr & {\text{Differentiate the given function}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {3{x^4} + 8{x^3} - 48{x^2}} \right] \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {3{x^4}} \right] + \frac{d}{{dx}}\left[ {8{x^3}} \right] - \frac{d}{{dx}}\left[ {48{x^2}} \right] \cr & f'\left( x \right) = 3\frac{d}{{dx}}\left[ {{x^4}} \right] + 8\frac{d}{{dx}}\left[ {{x^3}} \right] - 48\frac{d}{{dx}}\left[ {{x^2}} \right] \cr & {\text{Recall that }}\frac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{n - 1}} \cr & f'\left( x \right) = 3\left( {4{x^3}} \right) + 8\left( {3{x^2}} \right) - 48\left( {2x} \right) \cr & f'\left( x \right) = 12{x^3} + 24{x^2} - 96x \cr & {\text{The domain of }}f'\left( x \right){\text{ is }}\left( { - \infty ,\infty } \right),{\text{ }}f'\left( x \right){\text{ exists for all }}x,{\text{ then}} \cr & {\text{the critical numbers occur when }}f'\left( x \right) = 0 \cr & 12{x^3} + 24{x^2} - 96x = 0 \cr & {\text{Factoring}} \cr & 12x\left( {{x^2} + 2x - 8} \right) = 0 \cr & 12x\left( {x + 4} \right)\left( {x - 2} \right) = 0 \cr & {\text{Zero - factor property}} \cr & 12x = 0,{\text{ }}x + 4 = 0,{\text{ }}x - 2 = 0 \cr & x = 0,{\text{ }}x = - 4,{\text{ }}x = 2 \cr} $$
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