To find the domain of the function \( f(x) = \frac{1}{\sqrt{4x^2 - 1}} + \log_e(x(x^2 - 1)) \), we need to analyze the two components of the function separately.
### Step 1: Analyze the square root in the denominator
The first part of the function is \( \frac{1}{\sqrt{4x^2 - 1}} \). For this expression to be defined, the quantity inside the square root must be positive:
\[
4x^2 - 1 > 0
\]
This can be rearranged to:
\[
4x^2 > 1
\]
Dividing both sides by 4 gives:
\[
x^2 > \frac{1}{4}
\]
Taking the square root of both sides results in:
\[
|x| > \frac{1}{2}
\]
This means:
\[
x < -\frac{1}{2} \quad \text{or} \quad x > \frac{1}{2}
\]
### Step 2: Analyze the logarithmic function
The second part of the function is \( \log_e(x(x^2 - 1)) \). For the logarithm to be defined, the argument must be positive:
\[
x(x^2 - 1) > 0
\]
Factoring gives:
\[
x(x - 1)(x + 1) > 0
\]
The critical points are \( x = -1, 0, 1 \). We will test the intervals determined by these points: \( (-\infty, -1) \), \( (-1, 0) \), \( (0, 1) \), and \( (1, \infty) \).
- **Interval \( (-\infty, -1) \)**: Choose \( x = -2 \):
\[
-2(-2 - 1)(-2 + 1) = -2(-3)(-1) = -6 \quad (\text{negative})
\]
- **Interval \( (-1, 0) \)**: Choose \( x = -0.5 \):
\[
-0.5(-0.5 - 1)(-0.5 + 1) = -0.5(-1.5)(0.5) = 0.375 \quad (\text{positive})
\]
- **Interval \( (0, 1) \)**: Choose \( x = 0.5 \):
\[
0.5(0.5 - 1)(0.5 + 1) = 0.5(-0.5)(1.5) = -0.375 \quad (\text{negative})
\]
- **Interval \( (1, \infty) \)**: Choose \( x = 2 \):
\[
2(2 - 1)(2 + 1) = 2(1)(3) = 6 \quad (\text{positive})
\]
From this analysis, we find that \( x(x^2 - 1) > 0 \) in the intervals \( (-1, 0) \) and \( (1, \infty) \).
### Step 3: Find the intersection of the two conditions
Now, we need to find the intersection of the two conditions:
1. From the square root: \( (-\infty, -\frac{1}{2}) \cup (\frac{1}{2}, \infty) \)
2. From the logarithm: \( (-1, 0) \cup (1, \infty) \)
The intersection is:
- For \( (-\infty, -\frac{1}{2}) \) and \( (-1, 0) \): The intersection is \( (-1, -\frac{1}{2}) \).
- For \( (\frac{1}{2}, \infty) \) and \( (1, \infty) \): The intersection is \( (1, \infty) \).
### Final Domain
Thus, the domain of \( f(x) \) is:
\[
\text{Domain: } x \in (-1, -\frac{1}{2}) \cup (1, \infty)
\]
