The domain of `f(x)=(x)/(16-x^(2))+log_(2)(x^(3)-2x)` is

Options

A. `(-sqrt2,0)uu(sqrt2, oo)`
B. `(-sqrt2, 0)uu(sqrt2, 4)`
C. `(-sqrt2, 0)uu(4, 00)`
D. `(-sqrt2,0)uu(sqrt2, 4)uu(4,oo)`

Answer

To find the domain of the function \( f(x) = \frac{x}{16 - x^2} + \log_2(x^3 - 2x) \), we need to analyze the restrictions imposed by each part of the function.

### Step 1: Analyze the first part \( \frac{x}{16 - x^2} \)

1. The function is defined for all \( x \) except where the denominator is zero.
2. Set the denominator equal to zero:
\[ 16 - x^2 = 0 \]
This simplifies to:
\[ x^2 = 16 \implies x = \pm 4 \]
3. Therefore, the function is undefined at \( x = 4 \) and \( x = -4 \).
4. The intervals where the function is defined are:
\[ (-\infty, -4) \cup (-4, 4) \cup (4, \infty) \]

### Step 2: Analyze the second part \( \log_2(x^3 - 2x) \)

1. The logarithm is defined when its argument is positive:
\[ x^3 - 2x > 0 \]
2. Factor the expression:
\[ x(x^2 - 2) > 0 \implies x(x - \sqrt{2})(x + \sqrt{2}) > 0 \]
3. Identify the critical points:
- \( x = 0 \)
- \( x = \sqrt{2} \)
- \( x = -\sqrt{2} \)

4. We will test the sign of \( x(x - \sqrt{2})(x + \sqrt{2}) \) in the intervals determined by these critical points:
- **Interval 1:** \( (-\infty, -\sqrt{2}) \)
- Choose \( x = -2 \): \( (-2)(-2 - \sqrt{2})(-2 + \sqrt{2}) \) is positive.
- **Interval 2:** \( (-\sqrt{2}, 0) \)
- Choose \( x = -1 \): \( (-1)(-1 - \sqrt{2})(-1 + \sqrt{2}) \) is negative.
- **Interval 3:** \( (0, \sqrt{2}) \)
- Choose \( x = 1 \): \( (1)(1 - \sqrt{2})(1 + \sqrt{2}) \) is negative.
- **Interval 4:** \( (\sqrt{2}, \infty) \)
- Choose \( x = 2 \): \( (2)(2 - \sqrt{2})(2 + \sqrt{2}) \) is positive.

5. The intervals where \( x^3 - 2x > 0 \) are:
\[ (-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty) \]

### Step 3: Combine the intervals from both parts

1. From the first part, we have:
\[ (-\infty, -4) \cup (-4, 4) \cup (4, \infty) \]
2. From the second part, we have:
\[ (-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty) \]
3. Now we need to find the intersection of these intervals:
- For \( (-\infty, -4) \) and \( (-\infty, -\sqrt{2}) \): The intersection is \( (-\infty, -4) \).
- For \( (-4, 4) \) and \( (-\sqrt{2}, 0) \): The intersection is \( (-\sqrt{2}, 0) \).
- For \( (4, \infty) \) and \( (\sqrt{2}, \infty) \): The intersection is \( (4, \infty) \).

### Final Domain

Combining all the intervals, we get the final domain of \( f(x) \):
\[ \text{Domain of } f(x) = (-\infty, -4) \cup (-\sqrt{2}, 0) \cup (4, \infty) \]