Answer
$(f+g)(x)=-1 \rightarrow$ $D=(-\infty,+\infty)$
$(f-g)(x)=7 -2x^{2} \rightarrow$ $D=(-\infty,+\infty)$
$(f.g)(x)=3x^{2}-12 -x^{4}+4x^{2} = 4x^{2}-x^{4}-12\rightarrow$ $D=(-\infty,+\infty)$
$(\frac{f}{g})(x)=\frac{3-x^{2}}{x^{2}-4}\rightarrow$ $D=(-\infty,-2)\cup(2,+\infty)$
Work Step by Step
We are given $f(x)=3-x^{2}$ and $g(x)=x^{2}-4$
$(f+g)(x)=-1 \rightarrow$ the domain is $(-\infty,+\infty)$
$(f-g)(x)=7 -2x^{2} \rightarrow$ the domain is $(-\infty,+\infty)$
$(f.g)(x)=3x^{2}-12 -x^{4}+4x^{2} = 4x^{2}-x^{4}-12\rightarrow$ the domain is $(-\infty,+\infty)$
$(\frac{f}{g})(x)=\frac{3-x^{2}}{x^{2}-4}\rightarrow$ the domain is $(-\infty,-2)\cup(2,+\infty)$
