`f(x)=x|log_ex|,x gt 0` is monotonically decreasig in

Options

A. `(e,oo)`
B. `(0,1//e)`
C. `(1//e,1)`
D. (1,e)

Explanation

To determine the intervals where the function \( f(x) = x \log_e x \) (for \( x > 0 \)) is monotonically decreasing, we will follow these steps:

### Step 1: Identify the function and its domain
The function is given as:
\[ f(x) = x \log_e x \]
with the domain \( x > 0 \).

### Step 2: Find the critical points
To find where the function is increasing or decreasing, we need to differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = \frac{d}{dx}(x \log_e x) \]
Using the product rule:
\[ f'(x) = \log_e x + x \cdot \frac{1}{x} = \log_e x + 1 \]

### Step 3: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:
\[ \log_e x + 1 = 0 \]
This simplifies to:
\[ \log_e x = -1 \]
Exponentiating both sides gives:
\[ x = e^{-1} = \frac{1}{e} \]

### Step 4: Determine the sign of the derivative
We will analyze the sign of \( f'(x) \) in the intervals \( (0, \frac{1}{e}) \) and \( (\frac{1}{e}, \infty) \).

1. **For \( 0 < x < \frac{1}{e} \)**:
- Since \( \log_e x < -1 \), we have:
\[ f'(x) < 0 \quad \text{(function is decreasing)} \]

2. **For \( x > \frac{1}{e} \)**:
- Since \( \log_e x > -1 \), we have:
\[ f'(x) > 0 \quad \text{(function is increasing)} \]

### Step 5: Conclusion
The function \( f(x) = x \log_e x \) is monotonically decreasing on the interval:
\[ (0, \frac{1}{e}) \]

### Final Answer
The function \( f(x) = x \log_e x \) is monotonically decreasing in the interval \( (0, \frac{1}{e}) \).

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