The domain of the function `f(x)=sqrt(-log_(0.3) (x-1))/(sqrt(-x^(2)+2x+8))" is"`

Options

A. (1,4)
B. (-2,4)
C. [2,4)
D. None of these

Answer

To find the domain of the function \( f(x) = \frac{\sqrt{-\log_{0.3}(x-1)}}{\sqrt{-x^2 + 2x + 8}} \), we need to ensure that both the numerator and denominator are defined and valid.

### Step 1: Analyze the Numerator
The numerator is \( \sqrt{-\log_{0.3}(x-1)} \). For this square root to be defined, the expression inside must be non-negative:
\[ -\log_{0.3}(x-1) \geq 0 \]
This implies:
\[ \log_{0.3}(x-1) \leq 0 \]
Since the base \( 0.3 < 1 \), the logarithm function is decreasing. Therefore, \( \log_{0.3}(x-1) \leq 0 \) means:
\[ x - 1 \geq 1 \quad \Rightarrow \quad x \geq 2 \]

### Step 2: Analyze the Denominator
The denominator is \( \sqrt{-x^2 + 2x + 8} \). For this square root to be defined and non-zero, we need:
\[ -x^2 + 2x + 8 > 0 \]
Rearranging gives:
\[ x^2 - 2x - 8 < 0 \]
Next, we can factor the quadratic:
\[ (x - 4)(x + 2) < 0 \]
To find the intervals where this inequality holds, we find the roots, which are \( x = 4 \) and \( x = -2 \).

### Step 3: Test Intervals
We will test the intervals determined by the roots:
1. \( (-\infty, -2) \)
2. \( (-2, 4) \)
3. \( (4, \infty) \)

- For \( x < -2 \) (e.g., \( x = -3 \)):
\[ (-3 - 4)(-3 + 2) = (-7)(-1) = 7 > 0 \quad \text{(not valid)} \]

- For \( -2 < x < 4 \) (e.g., \( x = 0 \)):
\[ (0 - 4)(0 + 2) = (-4)(2) = -8 < 0 \quad \text{(valid)} \]

- For \( x > 4 \) (e.g., \( x = 5 \)):
\[ (5 - 4)(5 + 2) = (1)(7) = 7 > 0 \quad \text{(not valid)} \]

Thus, the valid interval for the denominator is \( (-2, 4) \).

### Step 4: Combine Conditions
Now we combine the conditions from the numerator and denominator:
1. From the numerator: \( x \geq 2 \)
2. From the denominator: \( -2 < x < 4 \)

The intersection of these conditions is:
\[ [2, 4) \]

### Final Answer
The domain of the function \( f(x) \) is:
\[ \boxed{[2, 4)} \]