Let `f(x)={(x^(2)-4x+3",",x lt 3),(x-4",",x ge 3):}`and
`g(x)={(x-3",",x lt 4),(x^(2)+2x+2",",x ge 4):}`.
Describe the function `f//g` and find its domain.

Explanation

To solve the problem, we need to analyze the functions \( f(x) \) and \( g(x) \) given in the piecewise format, and then find the function \( \frac{f}{g} \) and its domain.

### Step 1: Define the functions \( f(x) \) and \( g(x) \)

The function \( f(x) \) is defined as:
- \( f(x) = x^2 - 4x + 3 \) for \( x < 3 \)
- \( f(x) = x - 4 \) for \( x \geq 3 \)

The function \( g(x) \) is defined as:
- \( g(x) = x - 3 \) for \( x < 4 \)
- \( g(x) = x^2 + 2x + 2 \) for \( x \geq 4 \)

### Step 2: Write the expression for \( \frac{f}{g} \)

Now, we will write \( \frac{f(x)}{g(x)} \) for the different intervals of \( x \):

1. For \( x < 3 \):
\[ \frac{f(x)}{g(x)} = \frac{x^2 - 4x + 3}{x - 3} \]

2. For \( 3 \leq x < 4 \):
\[ \frac{f(x)}{g(x)} = \frac{x - 4}{x - 3} \]

3. For \( x \geq 4 \):
\[ \frac{f(x)}{g(x)} = \frac{x - 4}{x^2 + 2x + 2} \]

### Step 3: Identify the domain of \( \frac{f}{g} \)

The domain of \( \frac{f}{g} \) is determined by the values of \( x \) for which \( g(x) \neq 0 \), since division by zero is undefined.

- For \( x < 3 \):
- \( g(x) = x - 3 \) which is zero at \( x = 3 \). However, since \( x < 3 \), this part does not affect the domain.

- For \( 3 \leq x < 4 \):
- \( g(x) = x - 3 \) which is zero at \( x = 3 \). Therefore, \( x = 3 \) is not included in the domain.

- For \( x \geq 4 \):
- \( g(x) = x^2 + 2x + 2 \) is a quadratic function that is always positive (its discriminant is negative), so it does not affect the domain.

### Conclusion: Domain of \( \frac{f}{g} \)

The only restriction on the domain comes from \( g(x) \) being zero at \( x = 3 \). Thus, the domain of \( \frac{f}{g} \) is all real numbers except \( 3 \).

\[ \text{Domain of } \frac{f}{g} = \mathbb{R} \setminus \{3\} \]