Answer
$f+g=2x+3$; domain: $(-\infty,\infty)$
$f-g=-4x+1$; domain: $(-\infty,\infty)$
$f\cdot g=-3x^2+5x+2$; domain: $(-\infty,\infty)$
$\frac{f}{g}=\dfrac{2-x}{3x+1}$; domain: $(-\infty,-\frac{1}{3})\cup(-\frac{1}{3},\infty)$
Work Step by Step
Given:
$f(x)=2-x$
$g(x)=3x+1$
Step $1$. Since $f+g=f(x)+g(x)$, then:
$\begin{align*}
f+g&=f(x)+g(x)\\
&=(2-x)+(3x+1)\\
&=2x+3
\end{align*}$
Since both $f$ and $g$ have the set of all real numbers as their domain, then the domain of $f+g$ is also the set of all real numbers. Hence, the domain is $(-\infty,\infty)$.
Step $2$. Since $f-g=f(x)-g(x)$, then:
$\begin{align*}
f-g&=f(x)-g(x)\\
&(2-x)-(3x+1)\\
&=2-x-3x-1\\
&=-4x+1
\end{align*}$
Since both $f$ and $g$ have the set of all real numbers as their domain, then the domain of $f-g$ is also the set of all real numbers. Hence, the domain is $(-\infty,\infty)$.
Step $3$. Since $f\cdot g=f(x)\cdot g(x)$, then:
$\begin{align*}
f\cdot g=f(x)\cdot g(x)\\
&=(2-x)(3x+1)\\
&=2(3x+1)-x(3x+1)\\
&=6x+2-3x^2-x\\
&=-3x^2+5x+2
\end{align*}$
Since both $f$ and $g$ have the set of all real numbers as their domain, then the domain of $f\cdot g$ is also the set of all real numbers. Hence, the domain is $(-\infty,\infty)$.
Step $4$. Since $\frac{f}{g}=\frac{f(x)}{g(x)}$, then:
$\begin{align*}
\frac{f}{g}&=\frac{2-x}{3x+1}
\end{align*}$
The expression $\dfrac{2-x}{3x+1}$ is undefined when $x=-\frac{1}{3}$. Thus, the domain of $\frac{f}{g}$ is $(-\infty, -\frac{1}{3})\cup (-\frac{1}{3}, +\infty)$.
