Domain of definition of the function
`f(x)=(9)/(9-x^(2)) +log_(10) (x^(3)-x)`, is

Options

A. `(-1, 0) cup (1, oo)`
B. `(1, 3) cup (3, oo)`
C. `(-1, 0) cup (1, 3) cup (3, oo)`
D. `(-1, 0) cup (3, oo)`

Explanation

To find the domain of the function \( f(x) = \frac{9}{9 - x^2} + \log_{10}(x^3 - x) \), we need to determine the values of \( x \) for which the function is defined. This involves two parts: the rational function and the logarithmic function.

### Step 1: Analyze the Rational Function
The first part of the function is \( \frac{9}{9 - x^2} \). This expression is undefined when the denominator is zero.

Set the denominator equal to zero:
\[ 9 - x^2 = 0 \]
Solving for \( x \):
\[ x^2 = 9 \implies x = \pm 3 \]
Thus, the rational function is undefined at \( x = 3 \) and \( x = -3 \).

### Step 2: Analyze the Logarithmic Function
The second part of the function is \( \log_{10}(x^3 - x) \). The logarithm is defined only for positive arguments, so we need:
\[ x^3 - x > 0 \]
Factoring the expression:
\[ x(x^2 - 1) > 0 \implies x(x - 1)(x + 1) > 0 \]
To find the intervals where this product is positive, we identify the critical points: \( x = -1, 0, 1 \). We can test the intervals formed by these points: \( (-\infty, -1) \), \( (-1, 0) \), \( (0, 1) \), and \( (1, \infty) \).

- For \( x < -1 \) (e.g., \( x = -2 \)):
\[ (-2)(-2 - 1)(-2 + 1) = (-2)(-3)(-1) < 0 \]
- For \( -1 < x < 0 \) (e.g., \( x = -0.5 \)):
\[ (-0.5)(-0.5 - 1)(-0.5 + 1) = (-0.5)(-1.5)(0.5) > 0 \]
- For \( 0 < x < 1 \) (e.g., \( x = 0.5 \)):
\[ (0.5)(0.5 - 1)(0.5 + 1) = (0.5)(-0.5)(1.5) < 0 \]
- For \( x > 1 \) (e.g., \( x = 2 \)):
\[ (2)(2 - 1)(2 + 1) = (2)(1)(3) > 0 \]

### Step 3: Combine the Conditions
From the analysis:
1. The rational function is defined for \( x \in (-\infty, -3) \cup (-3, 3) \cup (3, \infty) \).
2. The logarithmic function is defined for \( x \in (-1, 0) \cup (1, \infty) \).

Now, we find the intersection of these intervals:
- For \( (-\infty, -3) \): No overlap with logarithmic function.
- For \( (-3, 3) \): Overlaps with \( (-1, 0) \).
- For \( (3, \infty) \): Overlaps with \( (1, \infty) \).

Thus, the domain of \( f(x) \) is:
\[ (-3, -1) \cup (1, 3) \cup (3, \infty) \]

### Final Domain
The domain of the function \( f(x) \) is:
\[ (-3, -1) \cup (1, 3) \cup (3, \infty) \]