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.