Answer
$f+g,f-g,fg$: all real numbers; $\frac{f}{g}$: all real numbers except $-6$ and $2$
Work Step by Step
For all but $\frac{f}{g}$, the domain includes all real numbers. To find the domain of $\frac{f}{g}$, set the denominator equal to $0$.
$f+g=(5-x^2)+(x^2+4x-12)$
$f+g=4x-7$
$(-∞,∞)$
$f-g=(5-x^2)-(x^2+4x-12)$
$f-g=-x^2-4x+17$
$(-∞,∞)$
$fg=(5-x^2)(x^2+4x-12)$
$fg=-x^4-4x^3+17x^2+20x-60$
$(-∞,∞)$
$\frac{f}{g}=\frac{5-x^2}{(x+6)(x-2)}$
$x\ne-6,2$
