## Precalculus (6th Edition) Blitzer

The value of $f+g=6{{x}^{2}}-2,f-g=6{{x}^{2}}-2x,fg=6{{x}^{3}}-7{{x}^{2}}+1$ and $\frac{f}{g}=\frac{6{{x}^{2}}-x-1}{x-1}$.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) \\ & =\ 6{{x}^{2}}-x-1+x-1 \\ & =6{{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( 6{{x}^{2}}-x-1 \right)-\left( x-1 \right) \\ & =6{{x}^{2}}-x-1-x+1 \\ & =6{{x}^{2}}-2x \end{align} Calculate the value of $fg$ as shown below to get: \begin{align} & fg=f\left( x \right)g\left( x \right) \\ & =\ \left( 6{{x}^{2}}-x-1 \right)\left( x-1 \right) \\ & =x\left( 6{{x}^{2}}-x-1 \right)-1\left( 6{{x}^{2}}-x-1 \right) \\ & =6{{x}^{3}}-{{x}^{2}}-x-6{{x}^{2}}+x+1 \end{align} $=6{{x}^{3}}-7{{x}^{2}}+1$ 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{6{{x}^{2}}-x-1}{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 division or even roots, so the domain of the functions $f+g,f-g,fg$ is the set of real numbers -- that is, $\left( -\infty ,\infty \right)$.