Calculus with Applications (10th Edition)

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

Chapter 6 - Applications of the Derivative - 6.6 Differentials: Linear Approximation - 6.6 Exercises - Page 348: 3

Answer

$$\,\,\,\,\,dy = 0.1$$

Work Step by Step

$$\eqalign{ & y = {x^3} - 2{x^2} + 3;\,\,\,\,\,\,x = 1{\text{ and }}\Delta x = - 0.1 \cr & {\text{The differential }}dy{\text{ is the product of }}f'\left( x \right){\text{ and }}dx,{\text{ or}} \cr & \,\,\,\,\,dy = f'\left( x \right)dx \cr & {\text{Find }}f'\left( x \right) \cr & f'\left( x \right) = \left( {{x^3} - 2{x^2} + 3} \right)' \cr & f'\left( x \right) = 3{x^2} - 4x \cr & {\text{Then}}{\text{, the differential }}dy{\text{ is}} \cr & \,\,\,\,\,dy = \left( {3{x^2} - 4x} \right)dx \cr & {\text{As }}dx{\text{ approaches 0}}{\text{, the value of }}dy{\text{ gets closer and closer that of }}\Delta y,{\text{ }} \cr & {\text{so that for small nonzero values of }}dx, \cr & \,\,\,\,\,dy \approx \Delta y, \cr & {\text{or}} \cr & \,\,\,\,\,dy = f'\left( x \right)\Delta x \cr & {\text{Then }} \cr & \,\,\,\,\,dy = \left( {3{x^2} - 4x} \right)\Delta x \cr & {\text{Substituting }}x = 1{\text{ and }}\Delta x = dx = - 0.1 \cr & \,\,\,\,\,dy = \left( {3{{\left( 1 \right)}^2} - 4\left( 1 \right)} \right)\left( { - 0.1} \right) \cr & {\text{simplifying}} \cr & \,\,\,\,\,dy = 0.1 \cr} $$
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