Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 4 - Integration - 4.2 The Indefinite Integral - Exercises Set 4.2 - Page 280: 62

Answer

See explanation.

Work Step by Step

(a) $\left.G^{\prime}(x)=1 \text { and } F^{\prime}(x)=1 \text { (because } f(x)=3+x\right)$ (b) If $x>0$, then $G(x)=3+x=F(x)$ so $C=0$ If $x<0,$ then $G(x)=x=-3+F(x)$ so $C=-3$ So we note that we obtain two different $C$-values; there does not exist a $C$ such that $G(x)=C+F(x)$ for all $x \neq 0$ (c) No, because the anti-derivatives are not limited to one open interval since they are defined everywhere except for at zero.
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