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 the denominator are defined and valid.

### Step 1: Analyze the Denominator
The denominator is \( \sqrt{x^2 + 2x + 8} \). For this square root to be defined and non-zero, we need:
\[ x^2 + 2x + 8 > 0 \]
This is a quadratic expression. We can analyze it by finding its discriminant:
\[ D = b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot 8 = 4 - 32 = -28 \]
Since the discriminant is negative, the quadratic does not have real roots and is always positive. Therefore, \( x^2 + 2x + 8 > 0 \) for all \( x \in \mathbb{R} \).

### Step 2: Analyze the Numerator
The numerator is \( \sqrt{-\log_{0.3}(x-1)} \). For this square root to be defined, we need:
\[ -\log_{0.3}(x-1) \geq 0 \]
This simplifies to:
\[ \log_{0.3}(x-1) \leq 0 \]
The logarithm \( \log_{0.3}(x-1) \) is less than or equal to zero when \( x-1 \geq 1 \) (since \( 0.3 < 1 \)). Thus:
\[ x - 1 \leq 1 \implies x \leq 2 \]
Also, since we are taking the logarithm, we need \( x - 1 > 0 \) which gives:
\[ x > 1 \]
Combining these conditions, we find:
\[ 1 < x \leq 2 \]

### Step 3: Combine Conditions
From Step 1, we found that the denominator is valid for all \( x \in \mathbb{R} \). From Step 2, the numerator is valid for \( 1 < x \leq 2 \). Therefore, the domain of the function \( f(x) \) is:
\[ (1, 2] \]

### Final Answer
The domain of the function \( f(x) = \frac{\sqrt{-\log_{0.3}(x-1)}}{\sqrt{x^2 + 2x + 8}} \) is \( (1, 2] \).

---