## Precalculus (6th Edition) Blitzer

The values of the functions are: $f+g=\frac{8{{x}^{2}}+30x-12}{\left( x-2 \right)\left( x+3 \right)}$ , $f-g=\frac{8{{x}^{2}}+18x+12}{\left( x-2 \right)\left( x+3 \right)}$ , $fg=\frac{48x}{\left( x-2 \right)\left( x+3 \right)},$ and $\frac{f}{g}=\frac{8{{x}^{2}}+24x}{6x-12}$. The domain of $\frac{f}{g}$ is $\left( -\infty ,-3 \right)\cup \left( -3,2 \right)\cup \left( 2,\infty \right)$ and that of $f-g,f+g$ and $fg$ is $\left( -\infty ,-3 \right)\cup \left( -3,2 \right)\cup \left( 2,\infty \right)$.
Calculate the value of $f+g$ as shown below to get, \begin{align} & f+g=f\left( x \right)+g\left( x \right) \\ & =\frac{8x}{x-2}+\frac{6}{x+3} \\ & =\frac{8x\left( x+3 \right)+6\left( x-2 \right)}{\left( x-2 \right)\left( x+3 \right)} \\ & =\frac{8{{x}^{2}}+24x+6x-12}{\left( x-2 \right)\left( x+3 \right)} \end{align} $=\frac{8{{x}^{2}}+30x-12}{\left( x-2 \right)\left( x+3 \right)}$ Calculate the value of $f-g$ as shown below to get, \begin{align} & f-g=f\left( x \right)-g\left( x \right) \\ & =\frac{8x}{x-2}-\frac{6}{x+3} \\ & =\frac{8x\left( x+3 \right)-6\left( x-2 \right)}{\left( x-2 \right)\left( x+3 \right)} \\ & =\frac{8{{x}^{2}}+24x-6x+12}{\left( x-2 \right)\left( x+3 \right)} \end{align} $=\frac{8{{x}^{2}}+18x+12}{\left( x-2 \right)\left( x+3 \right)}$ Calculate the value of $fg$ as shown below to get, \begin{align} & fg=f\left( x \right)g\left( x \right) \\ & =\left( \frac{8x}{x-2} \right)\left( \frac{6}{x+3} \right) \\ & =\frac{48x}{\left( x-2 \right)\left( x+3 \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{8x}{x-2} \right)}{\left( \frac{6}{x+3} \right)} \\ & =\frac{8x}{x-2}\times \frac{x+3}{6} \\ & =\frac{8x\left( x+3 \right)}{6\left( x-2 \right)} \\ & =\frac{8{{x}^{2}}+24x}{6x-12} \end{align} If the functions $f+g$ , $f-g$ and $fg$ contain division by zero, they will be undefined. So, put the denominator $\left( x-2 \right)\left( x+3 \right)$ equal to zero. $\left( x-2 \right)\left( x+3 \right)=0$ This implies that $x=2$ and $x=-3$. Now, the domain of the functions $f+g$ $f-g$ and $fg$ is all the real numbers except $-3$ and $2$. Therefore, the domain of the functions $f+g$ $f-g$ and $fg$ is $\left( -\infty ,-3 \right)\cup \left( -3,2 \right)\cup \left( 2,\infty \right)$. If the function $\frac{f}{g}$ is divided by zero, it will be undefined. So, put the denominator $6x-12$ equal to zero. \begin{align} & 6x-12=0 \\ & 6x=12 \\ & 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 $\left( -\infty ,-3 \right)\cup \left( -3,2 \right)\cup \left( 2,\infty \right)$.