The domain of `f(x)=3/(4-x^2)+log_(10) (x^3-x)` (1) `(-1,0)uu(1,2)uu(3,oo)` (2) `(-2,-1)uu(-1,0)uu(2,oo)` (3) `(-1,0)uu(1,2)uu(2,oo)` (4) `(1,2)uu(2,oo)`

Options

A. `(-1,0)uu(1,2)uu(3,oo)`
B. `(-2,-1)uu(-1,0)uu(2,oo)`
C. `(-1,0)uu(1,2)uu(2,oo)`
D. `(1,2)uu(2,oo)`

Answer

To find the domain of the function \( f(x) = \frac{3}{4 - x^2} + \log_{10}(x^3 - x) \), we need to ensure that both components of the function are defined.

### Step 1: Determine where the first term is defined
The first term is \( \frac{3}{4 - x^2} \). This term is undefined when the denominator is zero:
\[ 4 - x^2 = 0 \]
Solving for \( x \):
\[ x^2 = 4 \implies x = 2 \quad \text{or} \quad x = -2 \]
Thus, \( x \) cannot be \( 2 \) or \( -2 \).

### Step 2: Determine where the second term is defined
The second term is \( \log_{10}(x^3 - x) \). The logarithm is defined only for positive arguments, so we need:
\[ x^3 - x > 0 \]
Factoring gives:
\[ x(x^2 - 1) > 0 \implies x(x - 1)(x + 1) > 0 \]
The critical points are \( -1, 0, \) and \( 1 \). We will analyze the sign of the expression in the intervals determined by these points.

### Step 3: Test intervals
1. **Interval \( (-\infty, -1) \)**: Choose \( x = -2 \):
\[ (-2)(-2 - 1)(-2 + 1) = (-2)(-3)(-1) = -6 \quad (\text{negative}) \]

2. **Interval \( (-1, 0) \)**: Choose \( x = -0.5 \):
\[ (-0.5)(-0.5 - 1)(-0.5 + 1) = (-0.5)(-1.5)(0.5) = 0.375 \quad (\text{positive}) \]

3. **Interval \( (0, 1) \)**: Choose \( x = 0.5 \):
\[ (0.5)(0.5 - 1)(0.5 + 1) = (0.5)(-0.5)(1.5) = -0.375 \quad (\text{negative}) \]

4. **Interval \( (1, \infty) \)**: Choose \( x = 2 \):
\[ (2)(2 - 1)(2 + 1) = (2)(1)(3) = 6 \quad (\text{positive}) \]

### Step 4: Combine results
From the sign analysis, we find that \( x^3 - x > 0 \) in the intervals:
- \( (-1, 0) \)
- \( (1, \infty) \)

### Step 5: Exclude points from the domain
Now we combine the intervals from both terms:
- From \( \frac{3}{4 - x^2} \): \( x \neq 2, -2 \)
- From \( \log_{10}(x^3 - x) \): \( x \in (-1, 0) \cup (1, \infty) \)

Since \( -2 \) is not in the interval \( (-1, 0) \) and \( 2 \) is in \( (1, \infty) \), we exclude \( 2 \) from the domain.

### Final Domain
The domain of \( f(x) \) is:
\[ (-1, 0) \cup (1, 2) \cup (2, \infty) \]

### Conclusion
Thus, the correct option is:
(3) \( (-1, 0) \cup (1, 2) \cup (2, \infty) \)