## Precalculus (6th Edition) Blitzer

The value of $\left( f+g \right)=4x-6$, $\left( f-g \right)=2\left( x+2 \right)$, $\left( fg \right)=3{{x}^{2}}-16x+5$, and $\left( \frac{f}{g} \right)=\frac{3x-1}{x-5}$. The domain of $f+g,f-g,fg,\text{ and }\frac{f}{g}$ are $\left( -\infty,\infty \right)$, $\left( -\infty,\infty \right)$, $\left( -\infty,\infty \right)$, and $\left( -\infty,5 \right)\cup \left( 5,\infty \right)$ respectively.
Consider the function, \begin{align} & f+g=3x-1+x-5 \\ & =4x-6 \end{align} Since, the function contains neither division nor the square root, the domain of this function is the set of all real numbers. So, $\text{domain of }\left( f+g \right)=\left( -\infty,\infty \right)$. Now, consider the function, \begin{align} & f-g=3x-1-x+5 \\ & =2x+4 \\ & =2\left( x+2 \right) \end{align} Since, the function contains neither division nor the square root, the domain of this function is the set of all real numbers. So, $\text{domain of }\left( f-g \right)=\left( -\infty,\infty \right)$. Consider the function, \begin{align} & fg=\left( 3x-1 \right)\left( x-5 \right) \\ & =3{{x}^{2}}-15x-x+5 \\ & =3{{x}^{2}}-16x+5 \end{align} Since the function contains neither division nor the square root, the domain of this function is the set of all real numbers. So, $\text{domain of }\left( fg \right)=\left( -\infty,\infty \right)$. Consider the function, $\frac{f}{g}=\frac{3x-1}{x-5}$ As this function contains division and division by 0 is not defined, exclude those values of $x$ from the domain that cause the denominator to be zero. For this, set the denominator equal to 0, that is, $x-5=0$. This implies that $x=5$. Therefore, the domain of the function $\frac{f}{g}$ is the set of all real numbers excluding the number 5. So, $\text{domain of }\left( \frac{f}{g} \right)=\left( -\infty,5 \right)\cup \left( 5,\infty \right)$. Hence, the value of $\left( f+g \right)=4x-6$, $\left( f-g \right)=2\left( x+2 \right)$, $\left( fg \right)=3{{x}^{2}}-16x+5$, and $\left( \frac{f}{g} \right)=\frac{3x-1}{x-5}$. The domain of $f+g,f-g,fg,\text{ and }\frac{f}{g}$ are $\left( -\infty,\infty \right)$, $\left( -\infty,\infty \right)$, $\left( -\infty,\infty \right)$, and $\left( -\infty,5 \right)\cup \left( 5,\infty \right)$ respectively.