## Precalculus (6th Edition) Blitzer

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 $\left( -\infty ,-5 \right)\cup \left( -5,3 \right)\cup \left( 3,\infty \right)$.
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, $\left( -\infty ,\infty \right)$.