Answer
$f+g=-1$
$D:\{x|x\in R\}$ or $D:(-\infty,\infty)$
$f-g=-2x^{2}+7$
$D:\{x|x\in R\}$ or $D:(-\infty,\infty)$
$fg=-x^{4}+7x^{2}-12$
$D:\{x|x\in R\}$ or $D:(-\infty,\infty)$
$f/g=\dfrac{3-x^{2}}{x^{2}-4}$
$D:\{x|x\ne2,-2\}$ or $D:(-\infty,-2)\cup(-2,2)\cup(2,\infty)$
Work Step by Step
$f(x)=3-x^{2},$ $g(x)=x^{2}-4$
$f+g$
$f(x)+g(x)=3-x^{2}+x^{2}-4=-1$
The domain of this function includes all real numbers. It can be expressed as follows:
$D=\{x|x\in R\}$ or $D=(-\infty,\infty)$
$f-g$
$f(x)-g(x)=3-x^{2}-(x^{2}-4)=3-x^{2}-x^{2}+4=-2x^{2}+7$
The domain of this function includes all real numbers. It can be expressed as follows:
$D=\{x|x\in R\}$ or $D=(-\infty,\infty)$
$fg$
$f(x)\cdot g(x)=(3-x^{2})(x^{2}-4)=3x^{2}-12-x^{4}+4x^{2}=...$
$...=-x^{4}+7x^{2}-12$
The domain of this function includes all real numbers. It can be expressed as follows:
$D=\{x|x\in R\}$ or $D=(-\infty,\infty)$
$f/g$
$\dfrac{f(x)}{g(x)}=\dfrac{3-x^{2}}{x^{2}-4}$
This function is undefined for the values of $x$ that make the denominator equal to $0$. Let's find these values solving the following equation:
$x^{2}-4=0$
$x^{2}=4$
$\sqrt{x^{2}}=\sqrt{4}$
$x=\pm2$
Now we know the values of $x$ for which the function is undefined. The domain can be expressed as follows:
$D:\{x|x\ne2,-2\}$ or $D:(-\infty,-2)\cup(-2,2)\cup(2,\infty)$