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 $[-3,-2) \bigcup (2,3]$