The maximum value of |x log x| for `0 lt x le 1` is

Options

A. 0
B. `1//e`
C. `2e^(-1)`
D. none of these

Answer

To find the maximum value of \( |x \log x| \) for \( 0 < x \leq 1 \), we can follow these steps:

### Step 1: Define the function
We define the function \( f(x) = x \log x \). Since \( \log x \) is negative for \( 0 < x < 1 \), we have:
\[ |x \log x| = -x \log x \quad \text{for } 0 < x \leq 1 \]

### Step 2: Differentiate the function
Next, we need to find the derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}(x \log x) \]
Using the product rule, where \( u = x \) and \( v = \log x \):
\[ f'(x) = u'v + uv' = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1 \]

### Step 3: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:
\[ \log x + 1 = 0 \]
Solving for \( x \):
\[ \log x = -1 \implies x = e^{-1} = \frac{1}{e} \]

### Step 4: Determine the nature of the critical point
To determine whether this critical point is a maximum or minimum, we can analyze the sign of \( f'(x) \):
- For \( 0 < x < \frac{1}{e} \), \( \log x < -1 \) so \( f'(x) < 0 \) (decreasing).
- For \( \frac{1}{e} < x < 1 \), \( \log x > -1 \) so \( f'(x) > 0 \) (increasing).

Thus, \( x = \frac{1}{e} \) is a minimum point. Therefore, the function increases from \( 0 \) to \( \frac{1}{e} \) and then decreases from \( \frac{1}{e} \) to \( 1 \).

### Step 5: Evaluate the function at the endpoints and critical point
Now we evaluate \( f(x) \) at the endpoints and the critical point:
1. At \( x = 1 \):
\[ f(1) = 1 \log 1 = 0 \]
2. At \( x = \frac{1}{e} \):
\[ f\left(\frac{1}{e}\right) = \frac{1}{e} \log\left(\frac{1}{e}\right) = \frac{1}{e} \cdot (-1) = -\frac{1}{e} \]
Therefore, \( |f\left(\frac{1}{e}\right)| = \frac{1}{e} \).
3. As \( x \) approaches \( 0 \):
\[ \lim_{x \to 0^+} x \log x = 0 \]

### Step 6: Conclusion
The maximum value of \( |x \log x| \) on the interval \( (0, 1] \) occurs at \( x = \frac{1}{e} \):
\[ \text{Maximum value} = \frac{1}{e} \]

### Final Answer
The maximum value of \( |x \log x| \) for \( 0 < x \leq 1 \) is \( \frac{1}{e} \).

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