College Algebra, Chapter 3, 3.6, Section 3.6, Problem 8

Evaluate $f + g$, $f - g$, $fg$ and $\displaystyle \frac{f}{g}$ of the function $f(x) = \sqrt{9-x^2}$ and $g(x) = \sqrt{x^2-4}$ and find their domain

For $f+g$,

$
\begin{equation}
\begin{aligned}
f+g &= f(x) + g(x)\\
\\
f+g &= \sqrt{9-x^2} + \sqrt{x^2 - 4}
\end{aligned}
\end{equation}
$

The radicand can't be a negative value. So we factor $9- x^2 = (3-x)(3+x)$ and $x^2 - 4 = (x- 2)(x+2)$. Thus the domain of $f(x) + g(x)$ is $[-3,-2]\bigcup[2,3]$

For $f-g$

$
\begin{equation}
\begin{aligned}
f-g &= f(x) - g(x) \\
\\
f-g &= \sqrt{9-x^2} - \sqrt{x^2 - 4}
\end{aligned}
\end{equation}
$

The radicand can't be a negative value. So we factor $9- x^2 = (3-x)(3+x)$ and $x^2 - 4 = (x- 2)(x+2)$. Thus the domain of $f(x) - g(x)$ is $[-3,-2]\bigcup[2,3]$


For $fg$

$
\begin{equation}
\begin{aligned}
fg &= f(x) \cdot g(x) \\
\\
fg &= \left( \sqrt{9-x^2} \right) \left( \sqrt{x^2 - 4} \right) && \text{Substitute } f(x) = \sqrt{9-x^2} \text{ and } g(x) = \sqrt{x^2 - 4}\\
\\
fg &= \sqrt{(9-x^2)(x^2-4)} && \text{Apply FOIL method}\\
\\
fg &= \sqrt{9x^2 - 36 - x^4 + 4x^2} && \text{Combine like terms}\\
\\
fg &= \sqrt{13x^2 - x^4 - 36}
\end{aligned}
\end{equation}
$

The radicand can't be a negative value. So we factor $9- x^2 = (3-x)(3+x)$ and $x^2 - 4 = (x- 2)(x+2)$. Thus the domain of $f(x) \cdot g(x)$ is $[-3,-2]\bigcup[2,3]$


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

$
\begin{equation}
\begin{aligned}
\frac{f}{g} &= \frac{f(x)}{g(x)}\\
\\
\frac{f}{g} &= \frac{\sqrt{9-x^2}}{\sqrt{x^2-4}} && \text{Substitute } f(x) = \sqrt{9-x} \text{ and } g(x) = \sqrt{x^2 - 4}\\
\\
\frac{f}{g} &= \sqrt{\frac{9-x^2}{x^2-4}}
\end{aligned}
\end{equation}
$

The function $\displaystyle \frac{f}{g}$ can't have a denominator equal to zero and the radicand can't be a negative value. So we factor $9-x^2 = (3-x)(3+x)$ and $x^2 - 4 = (x-2)(x+2)$. Thus, the domain of $\displaystyle \frac{f}{g}$ is $[-3,-2) \bigcup (2,3]$