Answer is: option1

Solution:

Determine the Domain:

The function \( f(x) = x \ln x \) is defined for \( x > 0 \).

Find the First Derivative:

Using the product rule:

\[ f'(x) = \frac{d}{dx} [x \ln x] = \ln x + x \cdot \frac{1}{x} = \ln x + 1 \]

Find Critical Points:

Set \( f'(x) = 0 \):

\[ \ln x + 1 = 0 \Rightarrow \ln x = -1 \Rightarrow x = e^{-1} = \frac{1}{e} \]

Second Derivative Test:

\[ f''(x) = \frac{d}{dx} [\ln x + 1] = \frac{1}{x} \]

Evaluate at \( x = \frac{1}{e} \):

\[ f''\left(\frac{1}{e}\right) = e > 0 \]

Since \( f''(x) > 0 \), the function has a local minimum at \( x = \frac{1}{e} \).

Compute the Minimum Value:

Evaluate \( f(x) \) at \( x = \frac{1}{e} \):

\[ f\left(\frac{1}{e}\right) = \frac{1}{e} \ln \left(\frac{1}{e}\right) = \frac{1}{e}(-1) = -\frac{1}{e} \]

Conclusion:

The minimum value attained by \( f(x) = x \ln x \) is \( -\frac{1}{e} \), which corresponds to option (A).