Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises - Page 356: 48


$f(x)=-\sin x+\frac{3}{2}x^2 +3x +1$

Work Step by Step

$$f'''(x)=\cos x$$. $$\text{then} \ f''(x)= \sin x +C$$. $$3= \sin0 +C \Rightarrow C=3.$$ $$f'(x)=-\cos x+3x+C$$ $$2=-\cos0+3\cdot0 +C \Rightarrow C=3$$ $$f(x)=-\sin x +\frac{3}{2}x^2+3x+C$$ $$1=-\sin0+\frac{3}{2}\cdot0^2+3\cdot0+C \Rightarrow 1=C$$ Thus $f(x)=-\sin x+\frac{3}{2}x^2 +3x +1$.
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