Answer
The values of the functions are as follows:
$f+g=\frac{5x-3}{{{x}^{2}}-25}$ , $f-g=\frac{1}{x-5}$ , $fg=\frac{6{{x}^{2}}-10x-4}{\left( {{x}^{2}}-25 \right)\left( {{x}^{2}}-25 \right)}$ and $\frac{f}{g}=\frac{3x+1}{2x-4}$
The domain of $\frac{f}{g}$ is $\left( -\infty ,-5 \right)\cup \left( -5,2 \right)\cup \left( 2,5 \right)\cup \left( 5,\infty \right)$
The domain of $f-g,f+g$, and $fg$ is $\left( -\infty ,-5 \right)\cup \left( -5,5 \right)\cup \left( 5,\infty \right)$.
Work Step by Step
Calculate the value of $f+g$ as shown below to get,
$\begin{align}
& \left( f+g \right)\left( x \right)=f\left( x \right)+g\left( x \right) \\
& =\frac{3x+1}{{{x}^{2}}-25}+\frac{2x-4}{{{x}^{2}}-25} \\
& =\frac{3x+1+2x-4}{{{x}^{2}}-25} \\
& =\frac{5x-3}{{{x}^{2}}-25}
\end{align}$
Calculate the value of $f-g$ as shown below to get,
$\begin{align}
& \left( f-g \right)\left( x \right)=f\left( x \right)-g\left( x \right) \\
& =\frac{3x+1}{{{x}^{2}}-25}-\frac{2x-4}{{{x}^{2}}-25} \\
& =\frac{3x+1-2x+4}{{{x}^{2}}-25} \\
& =\frac{x+5}{{{x}^{2}}-25} \\
& =\frac{x+5}{\left( x-5 \right)\left( x+5 \right)} \\
& =\frac{1}{x-5}
\end{align}$
Calculate the value of $fg$ as shown below to get,
$\begin{align}
& \left( fg \right)\left( x \right)=f\left( x \right)g\left( x \right) \\
& =\left( \frac{3x+1}{{{x}^{2}}-25} \right)\left( \frac{2x-4}{{{x}^{2}}-25} \right) \\
& =\frac{\left( 3x+1 \right)\left( 2x-4 \right)}{\left( {{x}^{2}}-25 \right)\left( {{x}^{2}}-25 \right)} \\
& =\frac{6{{x}^{2}}-12x+2x-4}{\left( {{x}^{2}}-25 \right)\left( {{x}^{2}}-25 \right)} \\
& =\frac{6{{x}^{2}}-10x-4}{\left( {{x}^{2}}-25 \right)\left( {{x}^{2}}-25 \right)}
\end{align}$
Calculate the value of $\frac{f}{g}$ as shown below to get,
$\begin{align}
& \left( \frac{f}{g} \right)\left( x \right)=\frac{f\left( x \right)}{g\left( x \right)} \\
& =\frac{\left( \frac{3x+1}{{{x}^{2}}-25} \right)}{\left( \frac{2x-4}{{{x}^{2}}-25} \right)} \\
& =\frac{3x+1}{2x-4}
\end{align}$
If the functions $f+g$ , $f-g$ and $fg$ contain division by zero, they will be undefined. So, put the denominator ${{x}^{2}}-25$ equal to zero.
$\begin{align}
& {{x}^{2}}-25=0 \\
& x=\pm \sqrt{25} \\
& x=\pm 5
\end{align}$
Now, the domain of the functions $f+g$ $f-g$ and $fg$ is all the real numbers except 5 and -5.
Therefore, the domain of functions $f-g$ and $fg$ is $\left( -\infty ,-5 \right)\cup \left( -5,5 \right)\cup \left( 5,\infty \right)$.
If the function $\frac{f}{g}$ is divided by zero, it will be undefined. So, put the denominator $2x-4$ equal to zero.
$\begin{align}
& 2x-4=0 \\
& 2x=4 \\
& x=2
\end{align}$
Now, the domain of the function $\frac{f}{g}$ is all the real numbers except $2$.
Therefore, the domain of the function $\frac{f}{g}$ is $\left( -\infty ,-5 \right)\cup \left( -5,2 \right)\cup \left( 2,5 \right)\cup \left( 5,\infty \right)$.
