Find the domain of the real function `f(x)=sqrt(x^2 -4)`

Answer

To find the domain of the function \( f(x) = \sqrt{x^2 - 4} \), we need to determine the values of \( x \) for which the expression inside the square root is non-negative, since the square root of a negative number is not defined in the real number system.

### Step-by-Step Solution:

1. **Set up the inequality**:
We need to find when the expression under the square root is greater than or equal to zero:
\[ x^2 - 4 \geq 0 \]

2. **Rearrange the inequality**:
This can be rewritten as:
\[ x^2 \geq 4 \]

3. **Factor the inequality**:
We can factor the left side:
\[ (x - 2)(x + 2) \geq 0 \]

4. **Find the critical points**:
The critical points occur when the expression equals zero:
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \]

5. **Test intervals**:
We will test the intervals determined by the critical points \( -2 \) and \( 2 \):
- **Interval 1**: \( (-\infty, -2) \)
- **Interval 2**: \( (-2, 2) \)
- **Interval 3**: \( (2, \infty) \)

- For \( x < -2 \) (e.g., \( x = -3 \)):
\[ (-3 - 2)(-3 + 2) = (-5)(-1) = 5 \quad (\text{positive}) \]
- For \( -2 < x < 2 \) (e.g., \( x = 0 \)):
\[ (0 - 2)(0 + 2) = (-2)(2) = -4 \quad (\text{negative}) \]
- For \( x > 2 \) (e.g., \( x = 3 \)):
\[ (3 - 2)(3 + 2) = (1)(5) = 5 \quad (\text{positive}) \]

6. **Determine the intervals where the inequality holds**:
From our testing:
- The expression is non-negative in the intervals \( (-\infty, -2] \) and \( [2, \infty) \).

7. **Write the domain**:
Therefore, the domain of the function \( f(x) \) is:
\[ \text{Domain} = (-\infty, -2] \cup [2, \infty) \]

### Summary:
The domain of the function \( f(x) = \sqrt{x^2 - 4} \) is \( (-\infty, -2] \cup [2, \infty) \).