Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.3* The Natural Exponential Function - 6.3* Exercises: 96

Answer

$f(x)=3e^{x}-5sinx+4x-2'$

Work Step by Step

Integrating $f''(x)=3e^{x}+5sinx$ with respect to x. $f'(x)=3e^{x}-5cosx+C$; where C is an arbitary constant. Put $f'(0)=2$ Then $C=4$ After putting the value of C, we have $f'(x)=3e^{x}-5cosx+4$ Now, again integrate this expression, we get $f(x)=3e^{x}-5sinx+4x+C'$ Put $f(0)=1$ $C'=-2$ After putting the value of C', we have Hence, $f(x)=3e^{x}-5sinx+4x-2'$
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