Domain of `f(x) = sqrt(log_(0.3)(x !))` is

Options

A. `[0,1]`
B. `{0,1}`
C. `[0, infty)`
D. {0,1,2,3…..}

Answer

To find the domain of the function \( f(x) = \sqrt{\log_{0.3}(x!)} \), we need to determine the values of \( x \) for which the function is defined. The function is defined when the expression inside the square root is non-negative, which means:

1. **Condition for the square root**: The argument of the square root must be greater than or equal to zero:
\[ \log_{0.3}(x!) \geq 0 \]

2. **Condition for the logarithm**: The logarithm is defined only for positive arguments, so we also need:
\[ x! > 0 \]

### Step 1: Analyze the logarithmic condition
The logarithm \( \log_{0.3}(x!) \) is non-negative when \( x! \leq 1 \). This is because the base \( 0.3 \) is less than 1, and the logarithm of a number less than 1 is negative, while the logarithm of 1 is zero.

### Step 2: Find when \( x! \leq 1 \)
The factorial function \( x! \) is defined for non-negative integers \( x \). We can evaluate \( x! \) for small values of \( x \):
- For \( x = 0 \): \( 0! = 1 \)
- For \( x = 1 \): \( 1! = 1 \)
- For \( x = 2 \): \( 2! = 2 \) (which is greater than 1)

From this, we see that \( x! \) is less than or equal to 1 only for \( x = 0 \) and \( x = 1 \).

### Step 3: Compile the domain
Thus, the only values of \( x \) that satisfy both conditions (where \( x! > 0 \) and \( \log_{0.3}(x!) \geq 0 \)) are \( x = 0 \) and \( x = 1 \).

### Conclusion
The domain of the function \( f(x) = \sqrt{\log_{0.3}(x!)} \) is:
\[ \{0, 1\} \]