Find the domain and range of each of the following real functions: `f(x)=(1)/(sqrt(x^(2)-1))`

Explanation

To find the domain and range of the function \( f(x) = \frac{1}{\sqrt{x^2 - 1}} \), we will follow these steps:

### Step 1: Determine the Domain
To find the domain, we need to ensure that the expression under the square root is non-negative and that the denominator is not zero.

1. **Set the expression under the square root greater than zero:**
\[ x^2 - 1 > 0 \]
This can be factored as:
\[ (x - 1)(x + 1) > 0 \]

2. **Find the critical points:**
The critical points are \( x = -1 \) and \( x = 1 \).

3. **Test intervals around the critical points:**
- For \( x < -1 \) (e.g., \( x = -2 \)):
\[ (-2 - 1)(-2 + 1) = (-3)(-1) > 0 \quad \text{(True)} \]
- For \( -1 < x < 1 \) (e.g., \( x = 0 \)):
\[ (0 - 1)(0 + 1) = (-1)(1) < 0 \quad \text{(False)} \]
- For \( x > 1 \) (e.g., \( x = 2 \)):
\[ (2 - 1)(2 + 1) = (1)(3) > 0 \quad \text{(True)} \]

4. **Combine the intervals where the inequality holds:**
The solution to \( (x - 1)(x + 1) > 0 \) is:
\[ x < -1 \quad \text{or} \quad x > 1 \]

Thus, the domain of \( f(x) \) is:
\[ \text{Domain} = (-\infty, -1) \cup (1, \infty) \]

### Step 2: Determine the Range
To find the range, we analyze the behavior of the function \( f(x) \).

1. **Understand the function behavior:**
Since \( f(x) = \frac{1}{\sqrt{x^2 - 1}} \), we know that \( \sqrt{x^2 - 1} \) is always positive for the values in the domain.

2. **Find the minimum value of \( f(x) \):**
- As \( x \) approaches \( -1 \) or \( 1 \), \( \sqrt{x^2 - 1} \) approaches \( 0 \), which means \( f(x) \) approaches \( \infty \).
- As \( |x| \) increases (i.e., \( x \to \pm \infty \)), \( \sqrt{x^2 - 1} \) approaches \( |x| \), and thus \( f(x) \) approaches \( 0 \).

3. **Determine the range:**
Since \( f(x) \) is always positive and approaches \( 0 \) but never reaches it, the range is:
\[ \text{Range} = (0, \infty) \]

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

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