Find the domain and the range of the function `y=f(x)`, where `f(x)` is given by
`sqrt(x^(2)-2x-3)`

Answer

To find the domain and range of the function \( y = f(x) = \sqrt{x^2 - 2x - 3} \), we will follow these steps:

### Step 1: Determine the Domain
The expression under the square root must be non-negative for the function to be defined. Therefore, we need to solve the inequality:

\[ x^2 - 2x - 3 \geq 0 \]

### Step 2: Factor the Quadratic
To solve the inequality, we can factor the quadratic expression:

\[ x^2 - 2x - 3 = (x - 3)(x + 1) \]

### Step 3: Find Critical Points
The critical points occur when the expression is equal to zero:

\[ (x - 3)(x + 1) = 0 \]

This gives us the points \( x = 3 \) and \( x = -1 \).

### Step 4: Test Intervals
We will test the intervals determined by the critical points: \( (-\infty, -1) \), \( (-1, 3) \), and \( (3, \infty) \).

1. **For \( x < -1 \)** (e.g., \( x = -2 \)):
\[ (-2 - 3)(-2 + 1) = (-5)(-1) = 5 \quad (\text{positive}) \]

2. **For \( -1 < x < 3 \)** (e.g., \( x = 0 \)):
\[ (0 - 3)(0 + 1) = (-3)(1) = -3 \quad (\text{negative}) \]

3. **For \( x > 3 \)** (e.g., \( x = 4 \)):
\[ (4 - 3)(4 + 1) = (1)(5) = 5 \quad (\text{positive}) \]

### Step 5: Write the Domain
From our tests, the expression is non-negative in the intervals \( (-\infty, -1] \) and \( [3, \infty) \). Thus, the domain of \( f(x) \) is:

\[ \text{Domain} = (-\infty, -1] \cup [3, \infty) \]

### Step 6: Determine the Range
Next, we need to find the range of the function \( f(x) \). The minimum value of the expression under the square root occurs at the vertex of the quadratic \( x^2 - 2x - 3 \).

### Step 7: Find the Vertex
The vertex \( x \) can be found using the formula \( x = -\frac{b}{2a} \):

\[ x = -\frac{-2}{2 \cdot 1} = 1 \]

### Step 8: Evaluate at the Vertex
Now we evaluate \( f(1) \):

\[ f(1) = \sqrt{1^2 - 2 \cdot 1 - 3} = \sqrt{1 - 2 - 3} = \sqrt{-4} \quad (\text{not valid, since it's negative}) \]

Since \( x = 1 \) is not in the domain, we check the values at the endpoints of the domain.

1. **At \( x = -1 \)**:
\[ f(-1) = \sqrt{(-1)^2 - 2(-1) - 3} = \sqrt{1 + 2 - 3} = \sqrt{0} = 0 \]

2. **At \( x = 3 \)**:
\[ f(3) = \sqrt{3^2 - 2 \cdot 3 - 3} = \sqrt{9 - 6 - 3} = \sqrt{0} = 0 \]

### Step 9: Behavior as \( x \to \pm \infty \)
As \( x \) approaches \( \pm \infty \), \( f(x) \) approaches \( \infty \).

### Step 10: Write the Range
Thus, the range of \( f(x) \) is:

\[ \text{Range} = [0, \infty) \]

### Final Answer
- **Domain**: \( (-\infty, -1] \cup [3, \infty) \)
- **Range**: \( [0, \infty) \)