Answer
The values of the functions are as follows:
$f+g=2x-12,f-g=-2{{x}^{2}}-2x+18,f\cdot g=-{{x}^{4}}-2{{x}^{3}}+18{{x}^{2}}+6x-45$ and $\frac{f}{g}=\frac{3-{{x}^{2}}}{{{x}^{2}}+2x-15}$.
The domain of the functions $f+g,f-g$ , and $fg$ is $\left( -\infty ,\infty \right)$ and that of $\frac{f}{g}$ is $\left( -\infty ,-5 \right)\cup \left( -5,3 \right)\cup \left( 3,\infty \right)$.
Work Step by Step
Calculate the value of $f+g$ as shown below to get:
$\begin{align}
& f+g=f\left( x \right)+g\left( x \right) \\
& =\ 3-{{x}^{2}}+{{x}^{2}}+2x-15 \\
& =2x-12
\end{align}$
Calculate the value of $f-g$ as shown below to get:
$\begin{align}
& f-g=f\left( x \right)-g\left( x \right) \\
& =\ \left( 3-{{x}^{2}} \right)-\left( {{x}^{2}}+2x-15 \right) \\
& =3-{{x}^{2}}-{{x}^{2}}-2x+15 \\
& =-2{{x}^{2}}-2x+18
\end{align}$
Calculate the value of $fg$ as shown below to get:
$\begin{align}
& fg=f\left( x \right)g\left( x \right) \\
& =\ \left( 3-{{x}^{2}} \right)\left( {{x}^{2}}+2x-15 \right) \\
& =3\left( {{x}^{2}}+2x-15 \right)-{{x}^{2}}\left( {{x}^{2}}+2x-15 \right) \\
& =3{{x}^{2}}+6x-45-{{x}^{4}}-2{{x}^{3}}+15{{x}^{2}}
\end{align}$
$=-{{x}^{4}}-2{{x}^{3}}+18{{x}^{2}}+6x-45$
Calculate the value of $\frac{f}{g}$ as shown below to get:
$\begin{align}
& \frac{f}{g}=\frac{f\left( x \right)}{g\left( x \right)} \\
& =\frac{3-{{x}^{2}}}{{{x}^{2}}+2x-15}
\end{align}$
If the function $\frac{f}{g}$ is divided by zero, it will be undefined. So, put the denominator ${{x}^{2}}+2x-15$ equal to zero.
$\begin{align}
& {{x}^{2}}+2x-15=0 \\
& {{x}^{2}}+5x-3x-15=0 \\
& x\left( x+5 \right)-3\left( x+5 \right)=0 \\
& \left( x-3 \right)\left( x+5 \right)=0
\end{align}$
Case 1:
$\begin{align}
& x-3=0 \\
& x=-3
\end{align}$
Case 2:
$\begin{align}
& x+5=0 \\
& x=-5
\end{align}$
This implies that $x=3$ and $\ x=-5$.
Now, the domain of the function $\frac{f}{g}$ is all the real numbers except $3$ and $-5$.
Therefore, the domain of $\frac{f}{g}$ is $\left( -\infty ,-5 \right)\cup \left( -5,3 \right)\cup \left( 3,\infty \right)$.
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, $\left( -\infty ,\infty \right)$.
