The domain of the function `f(x)=sqrt(-log_.3 (x-1))/(sqrt(x^2+2x+8))` is

Options

A. `(1,4)`
B. `(-2,4)`
C. `(2,4)`
D. `[2,oo)`

Explanation

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 the function is defined, which means both the numerator and denominator must be valid.

### Step 1: Conditions for the Numerator
The numerator is \( \sqrt{-\log_{0.3}(x-1)} \). For this square root to be defined and real, the expression inside must be non-negative:
\[ -\log_{0.3}(x-1) \geq 0 \]
This simplifies to:
\[ \log_{0.3}(x-1) \leq 0 \]

### Step 2: Understanding the Logarithmic Inequality
Since the base of the logarithm (0.3) is less than 1, the logarithmic function is decreasing. Therefore, the inequality \( \log_{0.3}(x-1) \leq 0 \) implies:
\[ x - 1 \geq 1 \]
This leads to:
\[ x \geq 2 \]

### Step 3: Conditions for the Denominator
The denominator is \( \sqrt{x^2 + 2x + 8} \). For this square root to be defined, the expression inside must be positive:
\[ x^2 + 2x + 8 > 0 \]

### Step 4: Analyzing the Quadratic Expression
To analyze \( x^2 + 2x + 8 \), we can complete the square:
\[ x^2 + 2x + 1 + 7 = (x + 1)^2 + 7 \]
Since \( (x + 1)^2 \) is always non-negative, the minimum value of \( (x + 1)^2 + 7 \) is 7 (when \( x = -1 \)). Thus, \( x^2 + 2x + 8 > 0 \) for all real \( x \).

### Step 5: Combining Conditions
From Step 2, we found that \( x \geq 2 \). From Step 4, we established that the denominator is positive for all real \( x \). Therefore, the only restriction on the domain comes from the numerator.

### Conclusion
The domain of the function \( f(x) \) is:
\[ \text{Domain} = [2, \infty) \]