Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.1 Maxima and Minima - 4.1 Exercises - Page 243: 25

Answer

$$\eqalign{ & \left( a \right){\text{Critical points }}x = - 7,\,\,x = - 3,\,\,x = 3,\,\,x = 7 \cr & \left( b \right)f\left( 7 \right) = \frac{{154}}{3}{\text{ Is the absolute maximum}} \cr & \,\,\,\,\,\,f\left( { - 7} \right) = \frac{{154}}{3}{\text{is the absolute}}\,{\text{minimum}} \cr & \,\,\,\,\,\,\,f\left( { - 3} \right) = 18{\text{ is the local maximum}} \cr & \,\,\,\,\,\,\,f\left( 3 \right) = - 18{\text{ is the local minimum}} \cr} $$

Work Step by Step

$$\eqalign{ & f\left( x \right) = \frac{{{x^3}}}{3} - 9x,{\text{ on the interval }}\left[ { - 7,7} \right] \cr & \cr & {\text{Because }}f{\text{ is a polynomial}}{\text{, its derivative exists everywhere}} \cr & f'\left( x \right) = \frac{{3{x^2}}}{3} - 9 \cr & f'\left( x \right) = {x^2} - 9 \cr & {\text{Setting the derivative equal to zero}}{\text{, we have}} \cr & {x^2} - 9 = 0 \cr & {\text{Solving this equation gives the critical points}} \cr & x = 3{\text{ and }}x = - 3 \cr & {\text{Both of which lie in the given interval }}\left[ { - 7,7} \right]. \cr & {\text{These points and the endpoints are candidates for the location}} \cr & {\text{of absolute extrema:}} \cr & {\text{Points }}x = \left\{ { - 7, - 3,7,7} \right\} \cr & \cr & {\text{Evaluating these points}} \cr & f\left( { - 7} \right) = \frac{{{{\left( { - 7} \right)}^3}}}{3} - 9\left( { - 7} \right) = - \frac{{154}}{3} \cr & f\left( { - 3} \right) = \frac{{{{\left( { - 3} \right)}^3}}}{3} - 9\left( { - 3} \right) = 18 \cr & f\left( 3 \right) = \frac{{{{\left( 3 \right)}^3}}}{3} - 9\left( 3 \right) = - 18 \cr & f\left( 7 \right) = \frac{{{{\left( 7 \right)}^3}}}{3} - 9\left( 7 \right) = \frac{{154}}{3} \cr & \cr & {\text{The largest of these function values is }}f\left( 7 \right) = \frac{{154}}{3},{\text{ which is the}} \cr & {\text{absolute maximum on the interval }}\left[ { - 7,7} \right]. \cr & \cr & {\text{The smallest of these values is }}f\left( { - 7} \right) = \frac{{154}}{3}{\text{which is the absolute}} \cr & {\text{minimum on the interval }}\left[ { - 7,7} \right]. \cr & \cr & {\text{The next graph shows that the critical points corresponds to neither}} \cr & {\text{a local maximum nor a local minimum}}{\text{.}} \cr} $$
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