Single Variable Calculus, Chapter 1, 1.3, Section 1.3, Problem 30

Find the functions $f+g$, $f-g$, $f \cdot g$, and $f/g$ and their domains

$f(x) = \sqrt{3-x} , \qquad g(x) \sqrt{x^2-1}$



$
\begin{equation}
\begin{aligned}

@ f+g\\
f+g =& f(x)+g(x) && \text{ Substitute the given values of the function $f(x)$ and $g(x)$}\\

\end{aligned}
\end{equation}
$


$\boxed{f+g = \sqrt{3-x}+\sqrt{x^2-1}}$

$\boxed{ \text{The domain of this function is} (-\infty,3]}$


$
\begin{equation}
\begin{aligned}

@f-g\\
f-g =& f(x)-g(x) && \text{ Substitute the given values of the function $f(x)$ and $g(x)$}

\end{aligned}
\end{equation}
$



$\boxed{f-g = \sqrt{3-x} - \sqrt{x^2-1}}$

$\boxed{ \text{ The domain of this function is} (-\infty,3]}$



$
\begin{equation}
\begin{aligned}

@f \cdot g\\
f \cdot g =& f(x) \cdot g(x) && \text{ Substitute the given values of the function $f(x)$ and $g(x)$}\\
f \cdot g =& (3-x)^{\frac{1}{2}} (x^2-1)^{\frac{1}{2}} && \text{ Using FOIL method}\\
f \cdot g =& (3x^2-3-x^3+x)^{\frac{1}{2}}

\end{aligned}
\end{equation}
$




$\boxed{ f \cdot g = (3x^2-x^3+x-3)^{\frac{1}{2}}   or \sqrt{-x^3+3x^2+x-3}}$

By factoring $\sqrt{-x^3+3x^2+x-3}$, we get $\sqrt{(3-x)(x+1)(x-1)}$

We must ensure that the radicand is not less than 0.

Therefore, The domain of the function is $(-\infty,-1) \bigcup (1,3)$


$
\begin{equation}
\begin{aligned}
@\frac{f}{g}\\
\frac{f}{g} =& \frac{f(x)}{g(x)}
&& \text{ Substitute the given values of the function $f(x)$ and $g(x)$}

\end{aligned}
\end{equation}
$



$\boxed{\displaystyle \frac{f}{g} = \frac{\sqrt{3-x}}{\sqrt{x^2-1}} \text{ or } \frac{\sqrt{3 - x}}{\sqrt{(x + 1)(x - 1)}}}$

We must ensure that the radicand is not less than 0.

Therefore, The domain of the function is $(-\infty,-1) \bigcup (1,3)$