The functions $f$ and $g$ are defined as follows. \[ \begin{array}{l} f(x)=\frac{x+9}{x^{2}-18 x+81} \\ g(x)=\frac{x+7}{x^{2}-49} \end{array} \] For each function, find the domain. Write each answer as an interval or union of intervals. Domain of $f$ : П Domain of $g$ :

Step 1 ：The functions $f$ and $g$ are defined as follows: \[f(x)=\frac{x+9}{x^{2}-18 x+81}\] \[g(x)=\frac{x+7}{x^{2}-49}\]

Step 2 ：The domain of a function is the set of all possible input values (often the 'x' variable), which produce a valid output from a particular function. The domain of a function is the complete set of possible values of the independent variable. In plain English, this definition means: The domain is the set of all possible x-values which will make the function 'work', and will output real y-values.

Step 3 ：For the function $f(x)=\frac{x+9}{x^{2}-18 x+81}$, the denominator cannot be equal to zero because division by zero is undefined. Therefore, we need to find the values of x that make the denominator zero and exclude them from the domain.

Step 4 ：For the function $g(x)=\frac{x+7}{x^{2}-49}$, the denominator cannot be equal to zero because division by zero is undefined. Therefore, we need to find the values of x that make the denominator zero and exclude them from the domain.

Step 5 ：The values that make the denominator of $f$ zero are [9] and the values that make the denominator of $g$ zero are [-7, 7]. Therefore, these values should be excluded from the domain of the functions. The domain of a function is usually all real numbers except the values that make the denominator zero. Therefore, the domain of $f$ is all real numbers except 9 and the domain of $g$ is all real numbers except -7 and 7.

Step 6 ：Final Answer: The domain of $f$ is \(\boxed{(-\infty, 9) \cup (9, \infty)}\) and the domain of $g$ is \(\boxed{(-\infty, -7) \cup (-7, 7) \cup (7, \infty)}\).