To find the absolute maximum and minimum values of the function $f(x) = x - 2\ln(x)$ on the interval $\left[\frac{1}{2}, 6\right]$, follow these steps:

1. Find the critical points of $f(x)$:

   First, find the derivative $f'(x)$ and set it equal to zero to find the critical points. $f(x) = x - 2 \ln(x)$ $f'(x) = 1 - \frac{2}{x}$ Set the derivative equal to zero: $1 - \frac{2}{x} = 0$ Solve for $x$: $\frac{2}{x} = 1$ $x = 2$

2. Evaluate $f(x)$ at the critical points and endpoints of the interval:

   The critical point is $x = 2$. The endpoints of the interval are $x = \frac{1}{2}$ and $x = 6$.

   $f\left(\frac{1}{2}\right) = \frac{1}{2} - 2 \ln\left(\frac{1}{2}\right) = \frac{1}{2} - 2(-\ln(2)) = \frac{1}{2} + 2 \ln(2)$ $f(2) = 2 - 2 \ln(2)$ $f(6) = 6 - 2 \ln(6)$

3. Compare the values to find the absolute maximum and minimum:

   Calculate the values: $f\left(\frac{1}{2}\right) = \frac{1}{2} + 2 \ln(2) \approx 0.5 + 2 \cdot 0.693 = 0.5 + 1.386 = 1.886$ $f(2) = 2 - 2 \ln(2) \approx 2 - 2 \cdot 0.693 = 2 - 1.386 = 0.614$ $f(6) = 6 - 2 \ln(6) \approx 6 - 2 \cdot 1.792 = 6 - 3.584 = 2.416$

   Therefore: $f\left(\frac{1}{2}\right) \approx 1.886$ $f(2) \approx 0.614$ $f(6) \approx 2.416$

   So, the absolute minimum value is $f(2) \approx 0.614$ and the absolute maximum value is $f(6) \approx 2.416$.

To summarize:

• Absolute minimum value: $\approx 0.614$ at $x = 2$
• Absolute maximum value: $\approx 2.416$ at $x = 6$

Would you like further details or have any questions?

Here are 5 additional questions you might ask:

1. How do you find the derivative of a function?
2. What is the second derivative test for concavity?
3. How do you solve logarithmic equations?
4. What are the properties of logarithms?
5. Can you explain the concept of critical points in calculus?

Tip: When solving for critical points, always check the function's endpoints within the given interval to ensure you find the absolute maximum and minimum values.