The domain of the function `f(x)=(1)/(4-x^(2))+log_(10)(x^(2)-x)` is

Options

A. `(-oo,0)uu(1,oo)`
B. `(-oo,0)uu(1,oo)-{-2,2}`
C. `R-(-2,2)`
D. `(-oo,0]uu[1,oo)-{-2,2}`

Answer

To find the domain of the function \( f(x) = \frac{1}{4 - x^2} + \log_{10}(x^2 - x) \), we need to ensure that both parts of the function are defined.

### Step 1: Analyze the first term \( \frac{1}{4 - x^2} \)

For the first term to be defined, the denominator must not be zero:
\[ 4 - x^2 \neq 0 \]
This implies:
\[ x^2 \neq 4 \]
Thus, we have:
\[ x \neq 2 \quad \text{and} \quad x \neq -2 \]

### Step 2: Analyze the second term \( \log_{10}(x^2 - x) \)

For the logarithmic function to be defined, the argument must be positive:
\[ x^2 - x > 0 \]
Factoring gives:
\[ x(x - 1) > 0 \]

### Step 3: Solve the inequality \( x(x - 1) > 0 \)

To solve this inequality, we find the critical points by setting the expression to zero:
\[ x(x - 1) = 0 \implies x = 0 \quad \text{or} \quad x = 1 \]
Now we test the intervals determined by these critical points: \( (-\infty, 0) \), \( (0, 1) \), and \( (1, \infty) \).

1. **Interval \( (-\infty, 0) \)**:
- Choose \( x = -1 \): \( (-1)(-1 - 1) = (-1)(-2) = 2 > 0 \) (satisfies the inequality)

2. **Interval \( (0, 1) \)**:
- Choose \( x = 0.5 \): \( (0.5)(0.5 - 1) = (0.5)(-0.5) = -0.25 < 0 \) (does not satisfy the inequality)

3. **Interval \( (1, \infty) \)**:
- Choose \( x = 2 \): \( (2)(2 - 1) = (2)(1) = 2 > 0 \) (satisfies the inequality)

### Step 4: Combine the results

From the analysis:
- The first term \( \frac{1}{4 - x^2} \) is defined for \( x \in (-\infty, -2) \cup (-2, 2) \cup (2, \infty) \).
- The second term \( \log_{10}(x^2 - x) \) is defined for \( x \in (-\infty, 0) \cup (1, \infty) \).

### Step 5: Find the intersection of the two domains

The domain of \( f(x) \) is the intersection of the two sets:
1. From \( \frac{1}{4 - x^2} \): \( (-\infty, -2) \cup (-2, 2) \cup (2, \infty) \)
2. From \( \log_{10}(x^2 - x) \): \( (-\infty, 0) \cup (1, \infty) \)

The intersection is:
- From \( (-\infty, -2) \) and \( (-\infty, 0) \): \( (-\infty, -2) \)
- From \( (-2, 2) \) and \( (1, \infty) \): \( (1, 2) \) (but we must exclude 2)
- From \( (2, \infty) \) and \( (1, \infty) \): \( (2, \infty) \) (but we must exclude 2)

Thus, the final domain of \( f(x) \) is:
\[ (-\infty, -2) \cup (1, \infty) \]

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