To find the domain of the function \( f(x) = \frac{9}{9 - x^2} + \log_{10}(x^3 - x) \), we need to ensure that both parts of the function are defined.
### Step 1: Identify the restrictions from the first term \( \frac{9}{9 - x^2} \)
The denominator \( 9 - x^2 \) must not be equal to zero.
Setting the denominator to zero gives:
\[
9 - x^2 = 0 \implies x^2 = 9 \implies x = 3 \text{ or } x = -3
\]
Thus, we have the restriction:
\[
x \neq 3 \text{ and } x \neq -3
\]
### Step 2: Identify the restrictions from the second term \( \log_{10}(x^3 - x) \)
The argument of the logarithm must be greater than zero:
\[
x^3 - x > 0
\]
Factoring gives:
\[
x(x^2 - 1) > 0 \implies x(x - 1)(x + 1) > 0
\]
### Step 3: Analyze the inequality \( x(x - 1)(x + 1) > 0 \)
To solve this inequality, we find the critical points:
- \( x = 0 \)
- \( x = 1 \)
- \( x = -1 \)
We can test intervals determined by these points:
- \( (-\infty, -1) \)
- \( (-1, 0) \)
- \( (0, 1) \)
- \( (1, \infty) \)
Testing these intervals:
1. For \( x < -1 \) (e.g., \( x = -2 \)): \( (-)(-)(-) = - \) (not valid)
2. For \( -1 < x < 0 \) (e.g., \( x = -0.5 \)): \( (-)(-)(+) = + \) (valid)
3. For \( 0 < x < 1 \) (e.g., \( x = 0.5 \)): \( (+)(-)(+) = - \) (not valid)
4. For \( x > 1 \) (e.g., \( x = 2 \)): \( (+)(+)(+) = + \) (valid)
Thus, the solution to the inequality is:
\[
x \in (-1, 0) \cup (1, \infty)
\]
### Step 4: Combine the restrictions
Now, we combine the restrictions from both parts:
1. From \( \frac{9}{9 - x^2} \): \( x \neq 3, -3 \)
2. From \( \log_{10}(x^3 - x) \): \( x \in (-1, 0) \cup (1, \infty) \)
The intersection of these sets gives us:
\[
x \in (-1, 0) \cup (1, 3) \cup (3, \infty)
\]
### Final Domain
Thus, the domain of the function \( f(x) \) is:
\[
\boxed{(-1, 0) \cup (1, 3) \cup (3, \infty)}
\]
