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To solve this problem, let’s analyze the given function \( f(x) = x - 3 \ln(x) \) over the interval \( \frac{1}{3} \leq x \leq 10 \).

### Steps to Solve:

1. **Find the First Derivative**: We need \( f'(x) \) to locate the critical points within the interval.
   - The derivative of \( f(x) = x - 3 \ln(x) \) is:
     \[
     f'(x) = 1 - \frac{3}{x}
     \]
   - Set \( f'(x) = 0 \):
     \[
     1 - \frac{3}{x} = 0 \Rightarrow x = 3
     \]
   - Thus, \( x = 3 \) is a critical point.

2. **Evaluate \( f(x) \) at the Endpoints and the Critical Point**:
   - Calculate \( f(x) \) at \( x = \frac{1}{3} \), \( x = 3 \), and \( x = 10 \):
     - \( f\left(\frac{1}{3}\right) = \frac{1}{3} - 3 \ln\left(\frac{1}{3}\right) \)
     - \( f(3) = 3 - 3 \ln(3) \)
     - \( f(10) = 10 - 3 \ln(10) \)

3. **Compare Values**:
   - Compute each expression to find the global maximum and minimum over the interval:
     - \( f\left(\frac{1}{3}\right) \approx 4.011 \)
     - \( f(3) \approx 0.901 \)
     - \( f(10) \approx 4.001 \)

From these calculations, we can determine:
- **Global Maximum**: The function reaches its maximum value of approximately \( 4.011 \) at \( x = \frac{1}{3} \).
- **Global Minimum**: The function reaches its minimum value of approximately \( 0.901 \) at \( x = 3 \).

It appears that:
- The maximum value provided in the answer, \( 4.011 \), is correct, but it should occur at \( x = \frac{1}{3} \) rather than \( x = 10 \).
- The minimum value \( 0.901 \) at \( x = 3 \) is already correctly identified.

Would you like further details on these steps, or do you have any specific questions about this problem?

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Here are some related questions to consider:
1. How would you find critical points of a different function?
2. What is the significance of the first derivative test in determining maxima and minima?
3. How do endpoint evaluations influence finding absolute extrema?
4. Could you use the second derivative test to confirm the type of extremum at \( x = 3 \)?
5. How might this method change for functions defined on unbounded intervals?

**Tip:** When looking for global maxima or minima on a closed interval, always evaluate the function at critical points and endpoints.

Consider the function f(x) = x - 3 ln(x), with 1/3 ≤ x ≤ 10. Find the global maximum and minimum values and their corresponding x values.

Learn how to find the global maximum and minimum of the function f(x) = x - 3 ln(x) within the interval [1/3, 10]. This solution covers calculus concepts, including the first derivative test and endpoint evaluations, and is ideal for college-level calculus students.

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