## Precalculus (6th Edition) Blitzer

The value of $f+g=4x-2,f-g=2x-6,fg=\ 3{{x}^{2}}+2x-8\$ and $\frac{f}{g}=\frac{3x-4}{x+2}$. The domain of the functions $f+g,f-g$ , and $fg$ is $\left( -\infty ,\infty \right)$ and that of $\frac{f}{g}$ is $\left( -\infty ,-2 \right)\cup \left( -2,\infty \right)$.
Calculate the value of $f+g$ as shown below to get: \begin{align} & f+g=f\left( x \right)+g\left( x \right) \\ & =\ 3x-4+x+2 \\ & =4x-2 \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( 3x-4 \right)-\left( x+2 \right) \\ & =3x-4-x-2 \\ & =2x-6 \end{align} Calculate the value of $fg$ as shown below to get: \begin{align} & fg=f\left( x \right)g\left( x \right) \\ & =\ \left( 3x-4 \right)\left( x+2 \right) \\ & =3x\left( x+2 \right)-4\left( x+2 \right) \\ & =3{{x}^{2}}+6x-4x-8 \end{align} $=3{{x}^{2}}+2x-8$ 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{3x-4}{x+2} \end{align} If the function $\frac{f}{g}$ is divided by zero it will be undefined. So, put the denominator $x+2$ equal to zero. \begin{align} & x+2=0 \\ & x\ =-2 \end{align} Now, we see that the domain of function $\frac{f}{g}$ is all real numbers except -2. Therefore, the domain of $\frac{f}{g}$ is $\left( -\infty ,-2 \right)\cup \left( -2,\infty \right)$. The functions $f+g,f-g,fg$ do not involve any division and do not contain even roots, so the domain of the functions $f+g,f-g,fg$ is the set of real numbers -- that is, $\left( -\infty ,\infty \right)$.