The domain of the definition of the function `f(x)=sqrt(1+log_(e)(1-x))` is

Options

A. `-ooltxle0`
B. `-ooltxle(e-1)/(e)`
C. `-ooltxle1`
D. `xge1-e`

Answer

To find the domain of the function \( f(x) = \sqrt{1 + \log_e(1 - x)} \), we need to ensure that the expression under the square root is non-negative and that the logarithm is defined.

### Step 1: Ensure the logarithm is defined
The logarithmic function \( \log_e(1 - x) \) is defined only when its argument is positive. Therefore, we require:
\[ 1 - x > 0 \]
This simplifies to:
\[ x < 1 \]

### Step 2: Ensure the expression under the square root is non-negative
Next, we need the expression inside the square root to be non-negative:
\[ 1 + \log_e(1 - x) \geq 0 \]
This can be rearranged to:
\[ \log_e(1 - x) \geq -1 \]
Exponentiating both sides gives:
\[ 1 - x \geq e^{-1} \]
This simplifies to:
\[ 1 - \frac{1}{e} \geq x \]
or equivalently:
\[ x \leq 1 - \frac{1}{e} \]

### Step 3: Combine the inequalities
We now have two inequalities:
1. \( x < 1 \)
2. \( x \leq 1 - \frac{1}{e} \)

Since \( 1 - \frac{1}{e} < 1 \), the more restrictive condition is \( x \leq 1 - \frac{1}{e} \).

### Conclusion
Thus, the domain of the function \( f(x) \) is:
\[ (-\infty, 1 - \frac{1}{e}] \]