Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises - Page 362: 37

Answer

$f\left( x \right) = 2{x^4} + \ln x - 5$

Work Step by Step

$$\eqalign{ & f'\left( x \right) = 8{x^3} + \frac{1}{x},{\text{ }}x > 0,{\text{ }}f\left( 1 \right) = - 3 \cr & {\text{Find the general antiderivative }}f'\left( x \right) \cr & {\text{Using the formulas in Table 2 }}\left( {{\text{see page 358}}} \right) \cr & {\text{Function: }}cf\left( x \right) \to {\text{Particular antiderivative: }}cF\left( x \right) \cr & {\text{Function: }}{x^n}\left( {n \ne - 1} \right) \to {\text{Particular antiderivative: }}\frac{{{x^{n + 1}}}}{{n + 1}} \cr & {\text{Function: }}\frac{1}{x} \to {\text{Particular antiderivative: }}\ln \left| x \right| \cr & {\text{we obtain}} \cr & f\left( x \right) = 8\left( {\frac{{{x^4}}}{4}} \right) + \ln \left| x \right| + {C_1} \cr & f\left( x \right) = 2{x^4} + \ln \left| x \right| + {C_1}{\text{ }}\left( {\bf{1}} \right) \cr & {\text{Find }}{C_1}{\text{ using the given information }}f\left( 1 \right) = - 3 \cr & - 3 = 2{\left( 1 \right)^4} + \ln \left| 1 \right| + {C_1} \cr & - 5 = {C_1} \cr & {\text{Substitute }}{C_1}{\text{ into }}\left( {\bf{1}} \right) \cr & f\left( x \right) = 2{x^4} + \ln \left| x \right| - 5 \cr & {\text{We know that }}x > 0,{\text{ then}} \cr & f\left( x \right) = 2{x^4} + \ln x - 5 \cr} $$
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