Answer
The value of $f+g=6{{x}^{2}}-2,f-g=6{{x}^{2}}-2x,fg=6{{x}^{3}}-7{{x}^{2}}+1$ and $\frac{f}{g}=\frac{6{{x}^{2}}-x-1}{x-1}$.The domain of the functions $f+g,f-g$ , and $fg$ is $\left( -\infty ,\infty \right)$ and $\frac{f}{g}$ is $\left( -\infty ,1 \right)\cup \left( 1,\infty \right)$.
Work Step by Step
Calculate the value of $f+g$ as shown below to get:
$\begin{align}
& f+g=f\left( x \right)+g\left( x \right) \\
& =\ 6{{x}^{2}}-x-1+x-1 \\
& =6{{x}^{2}}-2
\end{align}$
Calculate the value of $f-g$ as shown below to get:
$\begin{align}
& f-g=f\left( x \right)-g\left( x \right) \\
& =\ \left( 6{{x}^{2}}-x-1 \right)-\left( x-1 \right) \\
& =6{{x}^{2}}-x-1-x+1 \\
& =6{{x}^{2}}-2x
\end{align}$
Calculate the value of $fg$ as shown below to get:
$\begin{align}
& fg=f\left( x \right)g\left( x \right) \\
& =\ \left( 6{{x}^{2}}-x-1 \right)\left( x-1 \right) \\
& =x\left( 6{{x}^{2}}-x-1 \right)-1\left( 6{{x}^{2}}-x-1 \right) \\
& =6{{x}^{3}}-{{x}^{2}}-x-6{{x}^{2}}+x+1
\end{align}$
$=6{{x}^{3}}-7{{x}^{2}}+1$
Calculate the value of $\frac{f}{g}$ as shown below to get:
$\begin{align}
& \frac{f}{g}=\frac{f\left( x \right)}{g\left( x \right)} \\
& =\frac{6{{x}^{2}}-x-1}{x-1}
\end{align}$
If the function $\frac{f}{g}$ is divided by zero, it will be undefined. So, put the denominator $x-1$ equal to zero.
$\begin{align}
& x-1=0 \\
& \ \ x=1 \\
\end{align}$
Now, the domain of the function $\frac{f}{g}$ is all the real numbers except $1$.
Therefore, the domain of $\frac{f}{g}$ is $\left( -\infty ,1 \right)\cup \left( 1,\infty \right)$.
The functions $f+g,f-g,fg$ do not have division or even roots, so the domain of the functions $f+g,f-g,fg$ is the set of real numbers -- that is, $\left( -\infty ,\infty \right)$.
