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.
