The set of all x for which the none of the functions is defined `f(x)=log_((x-1)//(x+3))2 and g(x)=(1)/(sqrt(x^(2)-9))`, is

Options

A. `[-3,1]`
B. [-3,2)
C. (-3,2]
D. (-3,-2)

Answer

To solve the problem of finding the set of all \( x \) for which none of the functions \( f(x) = \log_{\frac{x-1}{x+3}} 2 \) and \( g(x) = \frac{1}{\sqrt{x^2 - 9}} \) are defined, we need to analyze the conditions under which each function is defined.

### Step 1: Analyze the function \( f(x) \)

The function \( f(x) = \log_{\frac{x-1}{x+3}} 2 \) is defined under the following conditions:
1. The base \( \frac{x-1}{x+3} \) must be greater than 0.
2. The base \( \frac{x-1}{x+3} \) must not equal 1.

#### Condition 1: Base greater than 0
\[ \frac{x-1}{x+3} > 0 \]
This inequality holds when both the numerator and denominator are either both positive or both negative.

- **Case 1**: Both \( x-1 > 0 \) and \( x+3 > 0 \)
- \( x - 1 > 0 \Rightarrow x > 1 \)
- \( x + 3 > 0 \Rightarrow x > -3 \)
- Thus, \( x > 1 \) satisfies both.

- **Case 2**: Both \( x-1 < 0 \) and \( x+3 < 0 \)
- \( x - 1 < 0 \Rightarrow x < 1 \)
- \( x + 3 < 0 \Rightarrow x < -3 \)
- Thus, \( x < -3 \) satisfies both.

From these cases, we have:
\[ x < -3 \quad \text{or} \quad x > 1 \]

#### Condition 2: Base not equal to 1
\[ \frac{x-1}{x+3} \neq 1 \]
Cross-multiplying gives:
\[ x - 1 \neq x + 3 \Rightarrow -1 \neq 3 \quad \text{(always true)} \]
Thus, this condition does not impose any additional restrictions.

### Step 2: Analyze the function \( g(x) \)

The function \( g(x) = \frac{1}{\sqrt{x^2 - 9}} \) is defined when:
1. \( x^2 - 9 > 0 \)

This inequality can be factored:
\[ (x-3)(x+3) > 0 \]
The critical points are \( x = -3 \) and \( x = 3 \).

Using a number line analysis:
- The intervals to test are \( (-\infty, -3) \), \( (-3, 3) \), and \( (3, \infty) \).
- Testing these intervals:
- For \( x < -3 \): both factors are negative, so the product is positive.
- For \( -3 < x < 3 \): one factor is negative and one is positive, so the product is negative.
- For \( x > 3 \): both factors are positive, so the product is positive.

Thus, \( g(x) \) is defined for:
\[ x < -3 \quad \text{or} \quad x > 3 \]

### Step 3: Find the intersection of the conditions

We need to find the set of \( x \) where both functions are not defined.

From \( f(x) \):
- Not defined for \( x < -3 \) or \( x > 1 \).

From \( g(x) \):
- Not defined for \( -3 < x < 3 \).

### Step 4: Combine the conditions

The values for which both functions are not defined can be found by taking the intersection of the intervals:
- For \( f(x) \): \( (-\infty, -3) \cup (1, \infty) \)
- For \( g(x) \): \( (-\infty, -3) \cup (3, \infty) \)

The intersection yields:
- \( (-\infty, -3) \) (as both functions are not defined here).
- The interval \( (1, \infty) \) does not overlap with \( g(x) \) since \( g(x) \) is defined for \( x > 3 \).

### Final Answer
Thus, the set of all \( x \) for which none of the functions is defined is:
\[ (-\infty, -3] \]