Calculus: Early Transcendentals (2nd Edition)

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

Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises - Page 238: 78

Answer

$$f\left( x \right) = {x^3} - x + 2$$

Work Step by Step

$$\eqalign{ & f'\left( x \right) = 3{x^2} - 1;{\text{ }}f\left( 1 \right) = 2 \cr & {\text{Calculating the general solution}} \cr & f\left( x \right) = \int {f'\left( x \right)} dx \cr & f\left( x \right) = \int {\left( {3{x^2} - 1} \right)} dx \cr & f\left( x \right) = {x^3} - x + C \cr & {\text{Calculating the particular solution for }}f\left( 1 \right) = 2 \cr & 2 = {\left( 1 \right)^3} - \left( 1 \right) + C \cr & 2 = C \cr & {\text{The particular solution is}} \cr & f\left( x \right) = {x^3} - x + 2 \cr & \cr & {\text{Graphing general solutions for }}C = 0,{\text{ 1, 4 and the particular}} \cr & {\text{solution }}f\left( x \right) = {x^3} - x + 2 \cr} $$
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