## Precalculus (6th Edition) Blitzer

The value of the functions is as follows: $f+g=2{{x}^{2}}-2,f-g=2{{x}^{2}}-2x-4,fg=\ 2{{x}^{3}}+{{x}^{2}}-4x-3$ and $\frac{f}{g}=\frac{2{{x}^{2}}-x-3}{x+1}$. The domain of the functions $f+g,f-g$ , and $fg$ is $\left( -\infty ,\infty \right)$ and $\frac{f}{g}$ is $\left( -\infty ,-1 \right)\cup \left( -1,\infty \right)$.
Calculate the value of $f+g$ as shown below to get: \begin{align} & f+g=f\left( x \right)+g\left( x \right) \\ & =\ 2{{x}^{2}}-x-3+x+1 \\ & =2{{x}^{2}}-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( 2{{x}^{2}}-x-3 \right)-\left( x+1 \right) \\ & =2{{x}^{2}}-x-3-x-1 \\ & =2{{x}^{2}}-2x-4 \end{align} Calculate the value of $fg$ as shown below to get: \begin{align} & fg=f\left( x \right)g\left( x \right) \\ & =\ \left( 2{{x}^{2}}-x-3 \right)\left( x+1 \right) \\ & =x\left( 2{{x}^{2}}-x-3 \right)+1\left( 2{{x}^{2}}-x-3 \right) \\ & =2{{x}^{3}}-{{x}^{2}}-3x+2{{x}^{2}}-x-3 \end{align} $=2{{x}^{3}}+{{x}^{2}}-4x-3$ 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{2{{x}^{2}}-x-3}{x+1} \end{align} If the function $\frac{f}{g}$ is divided by zero, it will be undefined. So, put the denominator $x+1$ equal to zero. \begin{align} & x+1=0 \\ & \ \ x=-1 \end{align} Now, the domain of the function $\frac{f}{g}$ is all the real numbers except $-1$. Therefore, the domain of $\frac{f}{g}$ is $\left( -\infty ,-1 \right)\cup \left( -1,\infty \right)$. The functions $f+g,f-g,fg$ do not have any division or even roots, so the domain of the functions $f+g,f-g,fg$ is the set of real numbers -- that is, $\left( -\infty ,\infty \right)$.