{:\begin{align} $\newline$ &f(x)=\frac{(x+ $9$ )}{(x^{ $2$ }+ $16$ x+ $63$ )},(\newline\)&g(x)=\frac{(x^{ $2$ })}{(x $-2$ )}. $\newline$ \end{align}:} $\newline$ For each function, find the domain. $\newline$ Write each answer as an interval or union of intervals. $\newline$ Domain of $\newline$ $f$ : $\newline$ Domain of $\newline$ $g$ :

Full solution

Q. {:\begin{align*}

\newline

&f(x)=\frac{(x+

9

)}{(x^{

2

16

63

)},(\newline\)&g(x)=\frac{(x^{

2

})}{(x

-2

)}.

\newline

\end{align*}:}

\newline

For each function, find the domain.

\newline

Write each answer as an interval or union of intervals.

\newline

Domain of

\newline

f

\newline

Domain of

\newline

g

Find Domain of $f(x)$ : To find the domain of the function $f(x) = \frac{x+9}{x^2+16x+63}$ , we need to determine the values of $x$ for which the function is defined. The function is undefined where the denominator is zero, so we need to find the roots of the denominator. $\newline$ $x^2 + 16x + 63 = 0$ $\newline$ We can factor this quadratic equation to find the roots. $\newline$ $(x + 7)(x + 9) = 0$ $\newline$ So, the roots are $x = -7$ and $x = -9$ . $\newline$ The domain of $f(x)$ is all real numbers except for the roots of the denominator.
Write Domain of (x): The domain of (x) can be written as an interval or union of intervals. Since the roots are $-7$ and $-9$ , the domain is all real numbers except these two points. $\newline$ The domain of (x) is (-, $-9$ ) ( $-9$ , $-7$ ) ( $-7$ , ).
Find Domain of $\newline$ $g(x)$ : Now, let's find the domain of the function $\newline$ $g(x) = \frac{x^2}{(x - 2)}$ . Similar to $\newline$ $f(x)$ , $\newline$ $g(x)$ is undefined where the denominator is zero. $\newline$ $\newline$ $x - 2 = 0$ $\newline$ Solving for $\newline$ $x$ gives us $\newline$ $x = 2$ . $\newline$ The domain of $\newline$ $g(x)$ is all real numbers except for $\newline$ $x = 2$ .
Write Domain of $g(x)$ : The domain of $g(x)$ can also be written as an interval or union of intervals. Since the function is undefined at $x = 2$ , the domain is all real numbers except this point. $\newline$ The domain of $g(x)$ is $(-\infty, 2) \cup (2, \infty)$ .