To find the domain of the function \( f(x) = \frac{1}{4 - x^2} + \log_{10}(x^3 - x) \), we need to determine the conditions under which each part of the function is defined.
### Step 1: Find the domain of \( \frac{1}{4 - x^2} \)
The function \( \frac{1}{4 - x^2} \) is defined as long as the denominator is not zero. Therefore, we set up the equation:
\[
4 - x^2 \neq 0
\]
Solving this gives:
\[
x^2 \neq 4 \implies x \neq \pm 2
\]
Thus, the domain for this part is:
\[
D_1 = \mathbb{R} \setminus \{-2, 2\}
\]
### Step 2: Find the domain of \( \log_{10}(x^3 - x) \)
The logarithmic function is defined only for positive arguments. Therefore, we need:
\[
x^3 - x > 0
\]
Factoring the expression:
\[
x(x^2 - 1) > 0 \implies x(x - 1)(x + 1) > 0
\]
Next, we find the critical points by setting the factors to zero:
\[
x = 0, \quad x = 1, \quad x = -1
\]
Now we will test the intervals determined by these points: \( (-\infty, -1) \), \( (-1, 0) \), \( (0, 1) \), and \( (1, \infty) \).
- **Interval \( (-\infty, -1) \)**: Choose \( x = -2 \):
\[
(-2)(-2 - 1)(-2 + 1) = (-2)(-3)(-1) = -6 < 0 \quad \text{(negative)}
\]
- **Interval \( (-1, 0) \)**: Choose \( x = -0.5 \):
\[
(-0.5)(-0.5 - 1)(-0.5 + 1) = (-0.5)(-1.5)(0.5) = 0.375 > 0 \quad \text{(positive)}
\]
- **Interval \( (0, 1) \)**: Choose \( x = 0.5 \):
\[
(0.5)(0.5 - 1)(0.5 + 1) = (0.5)(-0.5)(1.5) = -0.375 < 0 \quad \text{(negative)}
\]
- **Interval \( (1, \infty) \)**: Choose \( x = 2 \):
\[
(2)(2 - 1)(2 + 1) = (2)(1)(3) = 6 > 0 \quad \text{(positive)}
\]
From this analysis, \( x^3 - x > 0 \) in the intervals:
\[
D_2 = (-1, 0) \cup (1, \infty)
\]
### Step 3: Combine the domains
The overall domain \( D \) of the function \( f(x) \) is the intersection of \( D_1 \) and \( D_2 \):
\[
D = D_1 \cap D_2
\]
1. From \( D_1 = \mathbb{R} \setminus \{-2, 2\} \), we exclude \( -2 \) and \( 2 \).
2. From \( D_2 = (-1, 0) \cup (1, \infty) \).
Now we find the intersection:
- For the interval \( (-1, 0) \): This is valid as it does not include \( -2 \) or \( 2 \).
- For the interval \( (1, \infty) \): This is also valid as it does not include \( -2 \) or \( 2 \).
Thus, the final domain is:
\[
D = (-1, 0) \cup (1, 2) \cup (2, \infty)
\]
### Final Answer:
The domain of the function \( f(x) \) is:
\[
(-1, 0) \cup (1, 2) \cup (2, \infty)
\]
