The set of all x for which there are no functions ` f(x) = log_((x-2)/(x+3)) 2 and g(x) =(1)/(sqrt(x^2-9))`

Options

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

Answer

To solve the problem, we need to determine the values of \( x \) for which the functions \( f(x) = \log_2\left(\frac{x-2}{x+3}\right) \) and \( g(x) = \frac{1}{\sqrt{x^2 - 9}} \) are undefined.

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

The function \( f(x) \) is defined when the argument of the logarithm is positive:
\[ \frac{x-2}{x+3} > 0 \]

This inequality holds when both the numerator and denominator are either both positive or both negative.

#### Step 1.1: Find the critical points

Set the numerator and denominator to zero:
- Numerator: \( x - 2 = 0 \) → \( x = 2 \)
- Denominator: \( x + 3 = 0 \) → \( x = -3 \)

#### Step 1.2: Test intervals

The critical points divide the number line into intervals:
1. \( (-\infty, -3) \)
2. \( (-3, 2) \)
3. \( (2, \infty) \)

We will test each interval to see where the fraction is positive.

- **Interval 1: \( (-\infty, -3) \)**
Choose \( x = -4 \):
\[ \frac{-4-2}{-4+3} = \frac{-6}{-1} = 6 > 0 \]

- **Interval 2: \( (-3, 2) \)**
Choose \( x = 0 \):
\[ \frac{0-2}{0+3} = \frac{-2}{3} < 0 \]

- **Interval 3: \( (2, \infty) \)**
Choose \( x = 3 \):
\[ \frac{3-2}{3+3} = \frac{1}{6} > 0 \]

#### Step 1.3: Conclusion for \( f(x) \)

The function \( f(x) \) is undefined in the interval \( (-3, 2) \).

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

The function \( g(x) \) is defined when the expression inside the square root is positive:
\[ x^2 - 9 > 0 \]
This can be factored as:
\[ (x - 3)(x + 3) > 0 \]

#### Step 2.1: Find the critical points

Set the factors to zero:
- \( x - 3 = 0 \) → \( x = 3 \)
- \( x + 3 = 0 \) → \( x = -3 \)

#### Step 2.2: Test intervals

The critical points divide the number line into intervals:
1. \( (-\infty, -3) \)
2. \( (-3, 3) \)
3. \( (3, \infty) \)

We will test each interval to see where the product is positive.

- **Interval 1: \( (-\infty, -3) \)**
Choose \( x = -4 \):
\[ (-4 - 3)(-4 + 3) = (-7)(-1) = 7 > 0 \]

- **Interval 2: \( (-3, 3) \)**
Choose \( x = 0 \):
\[ (0 - 3)(0 + 3) = (-3)(3) = -9 < 0 \]

- **Interval 3: \( (3, \infty) \)**
Choose \( x = 4 \):
\[ (4 - 3)(4 + 3) = (1)(7) = 7 > 0 \]

#### Step 2.3: Conclusion for \( g(x) \)

The function \( g(x) \) is undefined in the interval \( (-3, 3) \).

### Step 3: Combine the intervals

We need to find the set of all \( x \) for which both functions are undefined. The intervals where both functions are undefined are:
- \( f(x) \) is undefined in \( (-3, 2) \)
- \( g(x) \) is undefined in \( (-3, 3) \)

The combined interval where both functions are undefined is:
\[ (-3, 2) \]

### Final Answer

The set of all \( x \) for which there are no functions \( f(x) \) and \( g(x) \) is:
\[ \boxed{(-3, 2)} \]