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To solve this problem, let’s go through each part carefully.

### Given function:
\[
f(x) = x \ln(x)
\]

### Part (a): Find the intervals on which the function is increasing and decreasing.

1. **Find the derivative \( f'(x) \):**

   Using the product rule for differentiation where \( f(x) = x \ln(x) \):
   \[
   f'(x) = \frac{d}{dx}(x) \cdot \ln(x) + x \cdot \frac{d}{dx}(\ln(x))
   \]
   Simplifying each term:
   \[
   f'(x) = 1 \cdot \ln(x) + x \cdot \frac{1}{x}
   \]
   \[
   f'(x) = \ln(x) + 1
   \]

2. **Set \( f'(x) = 0 \) to find critical points:**

   \[
   \ln(x) + 1 = 0
   \]
   \[
   \ln(x) = -1
   \]
   Solving for \( x \):
   \[
   x = e^{-1} = \frac{1}{e}
   \]

   So, we have a critical point at \( x = \frac{1}{e} \).

3. **Determine the sign of \( f'(x) \) on the intervals around the critical point \( x = \frac{1}{e} \):**

   - For \( x < \frac{1}{e} \): Choose a point like \( x = 0.1 \).
     - \( f'(0.1) = \ln(0.1) + 1 \), and since \( \ln(0.1) < -1 \), \( f'(0.1) < 0 \), so \( f(x) \) is **decreasing** on this interval.
   
   - For \( x > \frac{1}{e} \): Choose a point like \( x = 1 \).
     - \( f'(1) = \ln(1) + 1 = 0 + 1 = 1 \), so \( f'(1) > 0 \), meaning \( f(x) \) is **increasing** on this interval.

   Therefore, the function \( f(x) \) is increasing on the interval \( \left(\frac{1}{e}, \infty\right) \).

### Answer for Part (a):
   The correct answer is **A**: The function \( f \) is increasing on the open interval \( \left(\frac{1}{e}, \infty\right) \).

### Part (b): Identify local extreme values, if any.

Since \( f(x) \) changes from decreasing to increasing at \( x = \frac{1}{e} \), there is a local minimum at this point.

- **Local Minimum Value:**
  Substitute \( x = \frac{1}{e} \) into \( f(x) \):
  \[
  f\left(\frac{1}{e}\right) = \frac{1}{e} \ln\left(\frac{1}{e}\right)
  \]
  \[
  = \frac{1}{e} \cdot (-1) = -\frac{1}{e}
  \]

Therefore, there is a local minimum of \( -\frac{1}{e} \) at \( x = \frac{1}{e} \).

---

### Summary:
- **Intervals of Increase:** \( \left(\frac{1}{e}, \infty\right) \)
- **Local Minimum:** \( -\frac{1}{e} \) at \( x = \frac{1}{e} \)

Do you need further clarification on any step?

---

Here are some related questions for deeper understanding:
1. How would you find intervals of concavity for the function \( f(x) = x \ln(x) \)?
2. Can you determine if the local minimum at \( x = \frac{1}{e} \) is also a global minimum?
3. How would the behavior of \( f(x) = x \ln(x) \) change if we considered negative values of \( x \)?
4. What happens to the function \( f(x) = x \ln(x) \) as \( x \) approaches 0 from the right?
5. How would you calculate the derivative of a similar function, such as \( g(x) = x^2 \ln(x) \)?

**Tip:** Always check the domain of the function before starting any calculus operations to ensure your intervals are valid.

Find the open intervals on which the function is increasing and those on which it is decreasing for f(x) = x * ln(x). Identify the function's local extreme values, if any, and where they occur.

Explore how to find the increasing and decreasing intervals of the function f(x) = x * ln(x), as well as its local minimum values. This problem involves calculus concepts such as differentiation, critical points, and interval analysis, suitable for college-level calculus students.

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