## Precalculus (6th Edition) Blitzer

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