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We are given the function \( f(x) = x - 3 \ln(x) \) over the interval \( \frac{1}{3} \leq x \leq 8 \). The task is to find the absolute maximum and minimum values of the function within this interval and the corresponding \( x \)-values.

### Steps:

1. **Find the derivative of \( f(x) \):**
   \[
   f'(x) = \frac{d}{dx} \left( x - 3 \ln(x) \right) = 1 - \frac{3}{x}
   \]

2. **Set the derivative equal to zero to find critical points:**
   \[
   f'(x) = 0 \quad \Rightarrow \quad 1 - \frac{3}{x} = 0 \quad \Rightarrow \quad x = 3
   \]
   So, \( x = 3 \) is a critical point.

3. **Evaluate the function at the critical point and the endpoints of the interval:**

   - At \( x = \frac{1}{3} \):
     \[
     f\left( \frac{1}{3} \right) = \frac{1}{3} - 3 \ln\left( \frac{1}{3} \right) = \frac{1}{3} - 3 \ln(1) + 3 \ln(3) = \frac{1}{3} + 3 \ln(3)
     \]
     \[
     f\left( \frac{1}{3} \right) \approx \frac{1}{3} + 3 \times 1.0986 = \frac{1}{3} + 3.2958 \approx 3.6291
     \]

   - At \( x = 8 \):
     \[
     f(8) = 8 - 3 \ln(8) = 8 - 3 \times 2.0794 = 8 - 6.2382 = 1.7618
     \]

   - At \( x = 3 \) (the critical point):
     \[
     f(3) = 3 - 3 \ln(3) = 3 - 3 \times 1.0986 = 3 - 3.2958 \approx -0.2958
     \]

4. **Conclusion:**
   - The absolute maximum value is approximately **3.6291** at \( x = \frac{1}{3} \).
   - The absolute minimum value is approximately **-0.2958** at \( x = 3 \).

Would you like a more detailed explanation or help with another aspect of the problem?

### 5 Questions:
1. What is the significance of finding critical points in an interval?
2. Can you explain why we need to evaluate the endpoints of the interval?
3. How does the derivative help in finding local extrema?
4. What happens if a function has no critical points within the given interval?
5. How do we know if the values at the critical points are maxima or minima?

### Tip:
Always check both the critical points and the boundary values when finding absolute extrema, as the highest or lowest values can occur at the edges of the interval.

Consider the function f(x) = x - 3ln(x), 1/3 ≤ x ≤ 8. Find the absolute maximum and minimum values and the corresponding x-values.

Solve the problem of finding the absolute maximum and minimum values of the function f(x) = x - 3ln(x) on the interval 1/3 ≤ x ≤ 8. The solution involves calculus concepts like derivatives, critical points, and the Extreme Value Theorem. Suitable for advanced high school or college calculus students.

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