Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 3 - Applications of Differentiation - Review Exercises - Page 238: 41

Answer

$${\text{Relative minimum at }}\left( { - 9,0} \right)$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = {\left( {x + 9} \right)^2} \cr & {\text{*Calculate the first derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{{\left( {x + 9} \right)}^2}} \right] \cr & f'\left( x \right) = 2\left( {x + 9} \right) \cr & {\text{Set }}f'\left( x \right) = 0 \cr & 2\left( {x + 9} \right) = 0 \cr & x = - 9 \cr & \cr & *{\text{Calculate the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {f'\left( x \right)} \right] \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {2\left( {x + 9} \right)} \right] \cr & f'\left( x \right) = 2 \cr & \cr & {\text{Evaluate the second derivative at }}x = - 9 \cr & f''\left( { - 9} \right) = 2 \cr & f''\left( { - 9} \right) > 0,{\text{ Then by the second derivative test }}\left( {{\text{Theorem 3}}{\text{.9}}} \right) \cr & f\left( x \right){\text{ has a relative minimum at }}\left( { - 9,f\left( { - 9} \right)} \right) \cr & f\left( { - 9} \right) = {\left( { - 9 + 9} \right)^2} \cr & {\text{Relative minimum at }}\left( { - 9,0} \right) \cr} $$
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