Calculus Concepts: An Informal Approach to the Mathematics of Change 5th Edition

Published by Brooks Cole
ISBN 10: 1-43904-957-2
ISBN 13: 978-1-43904-957-0

Chapter 4 - Analyzing Change: Applications of Derivatives - 4.1 Activities - Page 255: 6

Answer

$${\text{a}}){g_L}\left( 7 \right) = - 12.9x + 94.3{\text{ b}})\,0.775$$

Work Step by Step

$$\eqalign{ & {\text{We have}}: \cr & g\left( 7 \right) = 4,{\text{ }}g'\left( 7 \right) = - 12.9;{\text{ }}x = 7.25 \cr & \cr & \left( {\bf{a}} \right){\text{Using the formula }}\left( {{\text{see page 253}}} \right){\text{ }} \cr & {g_L}\left( x \right) = g\left( c \right) + g'\left( c \right)\left( {x - c} \right) \cr & \cr & {\text{Let }}c = 7 \cr & {g_L}\left( x \right) = g\left( 7 \right) + g'\left( 7 \right)\left( {x - 7} \right) \cr & {\text{Substitute }}g\left( 7 \right){\text{ and }}g'\left( 7 \right) \cr & {g_L}\left( 7 \right) = 4 - 12.9\left( {x - 7} \right) \cr & {\text{multiply}} \cr & {g_L}\left( 7 \right) = 4 - 12.9x + 90.3 \cr & {g_L}\left( 7 \right) = - 12.9x + 94.3 \cr & \cr & \left( {\bf{b}} \right){\text{use the linearization to estomate }}g\left( x \right){\text{ at }}x = 7.25 \cr & {g_L}\left( {7.25} \right) = - 12.9\left( {7.25} \right) + 94.3 \cr & {\text{simplifying}} \cr & {g_L}\left( {7.25} \right) = 0.775 \cr} $$
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