## Precalculus (6th Edition) Blitzer

The values of the functions are: $f+g=\frac{9{{x}^{2}}+79x-28}{\left( x-4 \right)\left( x+8 \right)}$ , $f-g=\frac{9{{x}^{2}}+65x+28}{\left( x-4 \right)\left( x+8 \right)}$ , $fg=\frac{63x}{\left( x-4 \right)\left( x+8 \right)},$ and $\frac{f}{g}=\frac{9{{x}^{2}}+72x}{7x-28}$. The domain of $\frac{f}{g}$ is $\left( -\infty ,-8 \right)\cup \left( -8,4 \right)\cup \left( 4,\infty \right)$ and that of $f-g,f+g$, and $fg$ is $\left( -\infty ,-8 \right)\cup \left( -8,4 \right)\cup \left( 4,\infty \right)$.
Calculate the value of $f+g$ as shown below to get, \begin{align} & f+g=f\left( x \right)+g\left( x \right) \\ & =\frac{9x}{x-4}+\frac{7}{x+8} \\ & =\frac{9x\left( x+8 \right)+7\left( x-4 \right)}{\left( x-4 \right)\left( x+8 \right)} \\ & =\frac{9{{x}^{2}}+72x+7x-28}{\left( x-4 \right)\left( x+8 \right)} \end{align} $=\frac{9{{x}^{2}}+79x-28}{\left( x-4 \right)\left( x+8 \right)}$ Calculate the value of $f-g$ as shown below to get, \begin{align} & f-g=f\left( x \right)-g\left( x \right) \\ & =\frac{9x}{x-4}-\frac{7}{x+8} \\ & =\frac{9x\left( x+8 \right)-7\left( x-4 \right)}{\left( x-4 \right)\left( x+8 \right)} \\ & =\frac{9{{x}^{2}}+72x-7x+28}{\left( x-4 \right)\left( x+8 \right)} \end{align} $=\frac{9{{x}^{2}}+65x+28}{\left( x-4 \right)\left( x+8 \right)}$ Calculate the value of $fg$ as shown below to get, \begin{align} & fg=f\left( x \right)g\left( x \right) \\ & =\left( \frac{9x}{x-4} \right)\left( \frac{7}{x+8} \right) \\ & =\frac{63x}{\left( x-4 \right)\left( x+8 \right)} \end{align} 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{\left( \frac{9x}{x-4} \right)}{\left( \frac{7}{x+8} \right)} \\ & =\frac{9x}{x-4}\times \frac{x+8}{7} \\ & =\frac{9x\left( x+8 \right)}{7\left( x-4 \right)} \\ & =\frac{9{{x}^{2}}+72x}{7x-28} \end{align} If the functions $f+g$ , $f-g$ and $fg$ contain division by zero, they will be undefined. So, put the denominator $\left( x-4 \right)\left( x+8 \right)$ equal to zero. $\left( x-4 \right)\left( x+8 \right)=0$ \begin{align} & \left( x-4 \right)\left( x+8 \right)=0 \\ & x-4=0,x+8=0 \\ \end{align} This implies that $x=4$ and $x=-8$. Now, the domain of the functions $f+g$ $f-g$ and $fg$ is all the real numbers except $-8$ and 4. Therefore, the domain of functions $f-g$ and $fg$ is $\left( -\infty ,-8 \right)\cup \left( -8,4 \right)\cup \left( 4,\infty \right)$. In the function $\frac{f}{g}$, we check for division by zero. So, put the denominator $7x-28$ equal to zero. \begin{align} & 7x-28=0 \\ & 7x=28 \\ & x=4 \end{align} Now, the domain of the function $\frac{f}{g}$ is all the real numbers except $4$. Therefore, the domain of the functions $\frac{f}{g}$ is $\left( -\infty ,-8 \right)\cup \left( -8,4 \right)\cup \left( 4,\infty \right)$.