The domain of the function `f(x) = sqrt({(-log _(0.3) (x-1))/(-x^2+ 3x +18)})` is

Options

A. `[2,6]`
B. `]2,6[`
C. `[2,6[`
D. None of these

Answer

To find the domain of the function \( f(x) = \sqrt{\frac{-\log_{0.3}(x-1)}{-x^2 + 3x + 18}} \), we need to ensure that the expression inside the square root is non-negative. This means we need to solve the inequality:

\[ \frac{-\log_{0.3}(x-1)}{-x^2 + 3x + 18} \geq 0 \]

### Step 1: Analyze the denominator
First, we need to factor the denominator \( -x^2 + 3x + 18 \):

\[ -x^2 + 3x + 18 = -(x^2 - 3x - 18) \]

Next, we factor \( x^2 - 3x - 18 \):

\[ x^2 - 3x - 18 = (x - 6)(x + 3) \]

Thus, the denominator can be rewritten as:

\[ -(x - 6)(x + 3) \]

### Step 2: Determine when the denominator is positive
The expression \( -(x - 6)(x + 3) \) is positive when \( (x - 6)(x + 3) < 0 \). To find the intervals where this is true, we find the roots:

- \( x - 6 = 0 \) gives \( x = 6 \)
- \( x + 3 = 0 \) gives \( x = -3 \)

Now, we test the intervals:

1. For \( x < -3 \): Both factors are negative, so the product is positive.
2. For \( -3 < x < 6 \): One factor is negative and the other is positive, so the product is negative.
3. For \( x > 6 \): Both factors are positive, so the product is positive.

Thus, the denominator is positive for \( x < -3 \) and \( x > 6 \).

### Step 3: Analyze the numerator
Next, we analyze the numerator \( -\log_{0.3}(x-1) \). The logarithm is defined for \( x - 1 > 0 \), which gives:

\[ x > 1 \]

Since the base of the logarithm \( 0.3 \) is less than 1, \( -\log_{0.3}(x-1) \) is positive when \( x - 1 < 1 \) (i.e., \( x < 2 \)).

Thus, the numerator is positive for:

\[ 1 < x < 2 \]

### Step 4: Combine the conditions
Now, we combine the conditions from the numerator and denominator:

1. For the numerator: \( 1 < x < 2 \)
2. For the denominator: \( x < -3 \) or \( x > 6 \)

The only overlapping interval is \( 1 < x < 2 \).

### Step 5: Conclusion
Since the numerator is positive in the interval \( (1, 2) \) and the denominator is negative in this interval, the overall expression is non-negative. Therefore, the domain of \( f(x) \) is:

\[ \text{Domain of } f(x) = [2, 6) \]