Let's solve this problem step by step.

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.