College Algebra, Chapter 3, 3.6, Section 3.6, Problem 10

Evaluate $f + g$, $f - g$, $fg$ and $\displaystyle \frac{f}{g}$ of the function $\displaystyle f(x) = \frac{2}{x+1}$ and $\displaystyle g(x) = \frac{x}{x+1}$ and find their domain

For $f+g$,

$
\begin{equation}
\begin{aligned}
f+g &= f(x) + g(x)\\
\\
f+g &= \frac{2}{x+1} + \frac{x}{x+1} && \text{Substitute } f(x) = \frac{2}{x+1} \text{ and } g(x) = \frac{x}{x+1}\\
\\
f+g &= \frac{2+x}{x+1}
\end{aligned}
\end{equation}
$

The function can't have a denominator equal to zero. So the domain of $f(x) + g(x)$ is $(-\infty,-1)\bigcup(-1,\infty)$

For $f-g$

$
\begin{equation}
\begin{aligned}
f-g &= f(x) - g(x) \\
\\
f-g &= \frac{2}{x+1} - \frac{x}{x+1} && \text{Substitute } f(x) = \frac{2}{x+1} \text{ and } g(x) = \frac{x}{x+1}\\
\\
f-g &= \frac{2-x}{x+1}
\end{aligned}
\end{equation}
$

The function can't have a denominator equal to zero. So the domain of $f(x) - g(x)$ is $(-\infty,-1)\bigcup(-1,\infty)$


For $fg$

$
\begin{equation}
\begin{aligned}
fg &= f(x) \cdot g(x) \\
\\
fg &= \left( \frac{2}{x+1} \right) \left( \frac{x}{x+1} \right) && \text{Substitute } f(x) = \frac{2}{x+1} \text{ and } g(x) = \frac{x}{x+1}\\
\\
fg &= \frac{2x}{(x+1)^2}
\end{aligned}
\end{equation}
$

The function can't have a denominator equal to zero. So the domain of $f(x) \cdot g(x)$ is $(-\infty,-1)\bigcup(-1,\infty)$

For $\displaystyle \frac{f}{g}$

$
\begin{equation}
\begin{aligned}
\frac{f}{g} &= \frac{f(x)}{g(x)}\\
\\
\frac{f}{g} &= \frac{\frac{2}{x+1}}{\frac{x}{x+1}} && \text{Substitute } f(x) = \frac{2}{x+1} \text{ and } g(x) = \frac{x}{x+1}\\
\\
\frac{f}{g} &= \frac{2}{\cancel{x+1}} \cdot \frac{\cancel{x+1}}{x} && \text{Revert the divisor and multiply}\\
\\
\frac{f}{g} &= \frac{2}{x}
\end{aligned}
\end{equation}
$

The function can't have a denominator equal to zero. So the domain of $\displaystyle \frac{f(x)}{g(x)}$ is $(-\infty, 0) \bigcup (0, \infty)$