Answer
The values of the functions are as follows:
$f+g=4x-7,f-g=-2{{x}^{2}}-4x+17,fg=-{{x}^{4}}-4{{x}^{3}}+17{{x}^{2}}+20x-60$ and $\frac{f}{g}=\frac{5-{{x}^{2}}}{{{x}^{2}}+4x-12}$.
The domain of the functions $f+g,f-g$ , and $fg$ is $\left( -\infty ,\infty \right)$ and that of $\frac{f}{g}$ is $\left( -\infty ,-6 \right)\cup \left( -6,2 \right)\cup \left( 2,\infty \right)$.
Work Step by Step
Calculate the value of $f+g$ as shown below to get:
$\begin{align}
& f+g=f\left( x \right)+g\left( x \right) \\
& =\ 5-{{x}^{2}}+{{x}^{2}}+4x-12 \\
& =4x-7
\end{align}$
Calculate the value of $f-g$ as shown below to get:
$\begin{align}
& f-g=f\left( x \right)-g\left( x \right) \\
& =\ \left( 5-{{x}^{2}} \right)-\left( {{x}^{2}}+4x-12 \right) \\
& =5-{{x}^{2}}-{{x}^{2}}-4x+12 \\
& =-2{{x}^{2}}-4x+17
\end{align}$
Calculate the value of $fg$ as shown below to get:
$\begin{align}
& fg=f\left( x \right)g\left( x \right) \\
& =\left( 5-{{x}^{2}} \right)\left( {{x}^{2}}+4x-12 \right) \\
& =5\left( {{x}^{2}}+4x-12 \right)-{{x}^{2}}\left( {{x}^{2}}+4x-12 \right) \\
& =5{{x}^{2}}+20x-60-{{x}^{4}}-4{{x}^{3}}+12{{x}^{2}}
\end{align}$
$=-{{x}^{4}}-4{{x}^{3}}+17{{x}^{2}}+20x-60$
Calculate the value of $\frac{f}{g}$ as shown below to get:
$\begin{align}
& \frac{f}{g}=\frac{f\left( x \right)}{g\left( x \right)} \\
& =\frac{5-{{x}^{2}}}{{{x}^{2}}+4x-12}
\end{align}$
If the function $\frac{f}{g}$ is divided by zero, it will be undefined. So, put the denominator ${{x}^{2}}+4x-12$ equal to zero.
$\begin{align}
& {{x}^{2}}+4x-12=0 \\
& {{x}^{2}}+6x-2x-12=0 \\
& x\left( x+6 \right)-2\left( x+6 \right)=0 \\
& \left( x-2 \right)\left( x+6 \right)=0
\end{align}$
Case 1:
$\begin{align}
& x-2=0 \\
& x=2
\end{align}$
Case 2:
$\begin{align}
& x+6=0 \\
& x=-6
\end{align}$
This implies that $x=2$ and $x=-6$.
Now, the domain of the function $\frac{f}{g}$ is all the real numbers except 2 and $-6$.
Therefore, the domain of $\frac{f}{g}$ is $\left( -\infty ,-6 \right)\cup \left( -6,2 \right)\cup \left( 2,\infty \right)$.
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, $\left( -\infty ,\infty \right)$.
Hence, the domain of the functions $f+g,f-g$ , and $fg$ is $\left( -\infty ,\infty \right)$ and that of $\frac{f}{g}$ is $\left( -\infty ,-6 \right)\cup \left( -6,2 \right)\cup \left( 2,\infty \right)$
