Answer
$(f+g)(x)=\sqrt{16-x^{2}}+\sqrt{x^{2}-1}$ , domain: $[-4,-1]\cup[1,4].$
$(f-g)(x)=\sqrt{16-x^{2}}-\sqrt{x^{2}-1}$ , domain: $[-4,-1]\cup[1,4].$
$(fg)(x)==\sqrt{16-x^{2}}\cdot\sqrt{x^{2}-1}$ , domain: $[-4,-1]\cup[1,4].$
$(\displaystyle \frac{f}{g})(x)=\frac{\sqrt{16-x^{2}}}{\sqrt{x^{2}-1}}$ , domain: $[-4,-1)\cup(1,4]$
Work Step by Step
The domain of f is $\{x|-4\leq x\leq 4\}=[-4,4]$, because
f is defined only for x that yield a nonnegative number under the square root.
The domain of g is $\{x|x^{2}\geq 1\}=(-\infty,-1]\cup[1,+\infty)$, for the same reason.
Their intersection is $[-4,-1]\cup[1,4].$
$(f+g)(x)=\sqrt{16-x^{2}}+\sqrt{x^{2}-1}$ , domain: $[-4,-1]\cup[1,4].$
$(f-g)(x)=\sqrt{16-x^{2}}-\sqrt{x^{2}-1}$ , domain: $[-4,-1]\cup[1,4].$
$(fg)(x)==\sqrt{16-x^{2}}\cdot\sqrt{x^{2}-1}$ , domain: $[-4,-1]\cup[1,4].$
$(\displaystyle \frac{f}{g})(x)=\frac{\sqrt{16-x^{2}}}{\sqrt{x^{2}-1}}$ , domain: $[-4,-1)\cup(1,4]$ , because $g(\pm 1)=0$, so $\pm 1$ are excluded from the domain.
