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### Given function: 
\[
f(x) = \frac{1}{x} \ln(x)
\]
is defined for \( x \in \mathbb{R}_{>0} \).

### Part (a): Find the extrema of \( f(x) \) and determine whether they are minima or maxima.

#### Step 1: Compute the first derivative \( f'(x) \).
Using the product and chain rules for derivatives:

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

We apply the product rule. Let \( u(x) = \frac{1}{x} \) and \( v(x) = \ln(x) \). Then,

\[
f'(x) = u'(x) v(x) + u(x) v'(x)
\]

where \( u'(x) = -\frac{1}{x^2} \) and \( v'(x) = \frac{1}{x} \).

Thus,

\[
f'(x) = \left( -\frac{1}{x^2} \right) \ln(x) + \frac{1}{x^2}
\]

Simplifying:

\[
f'(x) = \frac{1 - \ln(x)}{x^2}
\]

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

\[
\frac{1 - \ln(x)}{x^2} = 0
\]

This implies:

\[
1 - \ln(x) = 0 \quad \Rightarrow \quad \ln(x) = 1
\]

Solving for \( x \):

\[
x = e
\]

Thus, \( x = e \) is a critical point.

#### Step 3: Use the second derivative to determine the nature of the critical point.

Compute \( f''(x) \):

\[
f''(x) = \frac{d}{dx} \left( \frac{1 - \ln(x)}{x^2} \right)
\]

Using the quotient rule for derivatives:

\[
f''(x) = \frac{-2x(1 - \ln(x)) - x^2 \cdot \frac{1}{x}}{x^4} = \frac{-2(1 - \ln(x)) - 1}{x^3}
\]

Simplifying:

\[
f''(x) = \frac{-2 + 2\ln(x) - 1}{x^3} = \frac{-3 + 2\ln(x)}{x^3}
\]

At \( x = e \):

\[
f''(e) = \frac{-3 + 2\cdot 1}{e^3} = \frac{-1}{e^3}
\]

Since \( f''(e) < 0 \), \( x = e \) is a **local maximum**.

### Part (b): Investigate the monotonicity of \( f(x) \) in the interval \( (0, e) \).

#### Step 1: Analyze the sign of \( f'(x) \) on \( (0, e) \).

\[
f'(x) = \frac{1 - \ln(x)}{x^2}
\]

For \( 0 < x < e \), we know that \( \ln(x) < 1 \), so \( 1 - \ln(x) > 0 \). Hence, \( f'(x) > 0 \) in this interval, meaning that \( f(x) \) is **increasing** on \( (0, e) \).

#### Step 2: Analyze the monotonicity beyond \( e \).

For \( x > e \), \( \ln(x) > 1 \), so \( 1 - \ln(x) < 0 \). Thus, \( f'(x) < 0 \) for \( x > e \), meaning that \( f(x) \) is **decreasing** in this region.

### Conclusion:

- The function has a local maximum at \( x = e \).
- The function is increasing on \( (0, e) \) and decreasing on \( (e, \infty) \).

---

Would you like more details on any step? Here are five related questions to expand your understanding:

1. How is the product rule applied when differentiating functions like \( \frac{1}{x} \ln(x) \)?
2. Why does the second derivative test help determine whether a critical point is a minimum or maximum?
3. How would the behavior of the function change if the logarithmic base were different?
4. Can you explain the physical meaning of finding extrema in real-world contexts?
5. How does the function behave as \( x \to 0^+ \) and as \( x \to \infty \)?

**Tip**: When applying the quotient rule, simplify the expression as much as possible before differentiating again to avoid errors.

Gegeben sei die Funktion f : R_{>0} → R mit f(x) = 1/x ln(x). (a) Bestimmen Sie die Extremwerte von f und untersuchen Sie deren Art (Minima oder Maxima). (b) Untersuchen Sie f(x) im Intervall (0, e) auf Monotonie.

Step-by-step solution for finding the extrema and analyzing the monotonicity of the function f(x) = 1/x ln(x). Learn how to differentiate and apply the first and second derivative tests to determine the nature of critical points and behavior of the function in the interval (0, e). Suitable for undergraduate-level calculus students.

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