The functions $$ f $$ and $$ g $$ are defined as follows. $$ f(x)=\frac{x-1}{x^{2}+6 x-7} $$ $$ g(x)=\frac{x}{x^{2}+64} $$ For each function, find the domain. Write each answer as an interval or union of intervals. Domain of $$ f: $$ D Domain of $$ g: $$ :

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We start by finding the values of $$ x $$ that make the denominator zero because those values must be excluded from the domain.

### For $$ f(x)=\frac{x-1}{x^{2}+6x-7} $$

1. **Factor the denominator:**  
   The quadratic $$ x^2+6x-7 $$ factors as  
   $$
   x^2+6x-7=(x+7)(x-1)
   $$

2. **Find the zeros of the denominator:**  
   Set each factor equal to zero:  
   $$
   x+7=0 \quad \Longrightarrow \quad x=-7
   $$
   $$
   x-1=0 \quad \Longrightarrow \quad x=1
   $$

3. **State the domain:**  
   The values $$ x=-7 $$ and $$ x=1 $$ must be excluded. Therefore, the domain of $$ f $$ in interval notation is  
   $$
   (-\infty,-7) \cup (-7,1) \cup (1,\infty)
   $$

### For $$ g(x)=\frac{x}{x^{2}+64} $$

1. **Analyze the denominator:**  
   The denominator is  
   $$
   x^2+64
   $$
   Since $$ x^2 $$ is always nonnegative, the smallest value of $$ x^2+64 $$ is $$ 64 $$ (when $$ x=0 $$), which is never zero.

2. **State the domain:**  
   Since the denominator never equals zero for any real $$ x $$, the domain of $$ g $$ is all real numbers, expressed in interval notation as  
   $$
   (-\infty,\infty)
   $$

---

**Domain of $$ f: $$**
$$
(-\infty,-7) \cup (-7,1) \cup (1,\infty)
$$

**Domain of $$ g: $$**
$$
(-\infty,\infty)
$$
**Domain of $$ f: $$** $$ (-\infty,-7) \cup (-7,1) \cup (1,\infty) $$ **Domain of $$ g: $$** $$ (-\infty,\infty) $$

The functions and are defined as follows.

For each function, find the domain.
Write each answer as an interval or union of intervals.
Domain of D
Domain of :

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The functions and are defined as follows. For each function, find the domain. Write each answer as an interval or union of intervals. Domain of D Domain of :