Answer
$f+g=x^{2}-4x+5$
$D:\{x|x\in R\}$ or $D=(-\infty,\infty)$
$f-g=-x^{2}+2x+5$
$D:\{x|x\in R\}$ or $D=(-\infty,\infty)$
$fg=-x^{3}+8x^{2}-15x$
$D:\{x|x\in R\}$ or $D=(-\infty,\infty)$
$f/g=\dfrac{5-x}{x^{2}-3x}$
$D:\{x|x\ne0,3\}$ or $D:(-\infty,0)\cup(0,3)\cup(3,\infty)$
Work Step by Step
$f(x)=5-x,$ $g(x)=x^{2}-3x$
$f+g$
$f(x)+g(x)=5-x+x^{2}-3x=x^{2}-4x+5$
The domain of this function includes all real numbers. It can be expressed as follows:
$D:\{x|x\in R\}$ or $D=(-\infty,\infty)$
$f-g$
$f(x)-g(x)=5-x-(x^{2}-3x)=5-x-x^{2}+3x=...$
$...=-x^{2}+2x+5$
The domain of this function includes all real numbers. It can be expressed as follows:
$D:\{x|x\in R\}$ or $D=(-\infty,\infty)$
$fg$
$f(x)\cdot g(x)=(5-x)(x^{2}-3x)=5x^{2}-15x-x^{3}+3x^{2}=...$
$...=-x^{3}+8x^{2}-15x$
The domain of this function includes all real numbers. It can be expressed as follows:
$D:\{x|x\in R\}$ or $D=(-\infty,\infty)$
$f/g$
$\dfrac{f(x)}{g(x)}=\dfrac{5-x}{x^{2}-3x}$
This function is undefined for the values of $x$ that make the denominator equal to $0$. To find said values, solve the following equation:
$x^{2}-3x=0$
$x(x-3)=0$
$x=0$ and $x=3$
Knowing the values of $x$ for which the function is not defined, the domain can be expressed as follows:
$D:\{x|x\ne0,3\}$ or $D:(-\infty,0)\cup(0,3)\cup(3,\infty)$
