To solve the problem, we are tasked with finding the absolute maximum value of the function $f(x) = x \ln x$ over the interval $x \in [e^{-2}, e]$.

Solution Steps:

1. Find the derivative $f'(x)$ of $f(x) = x \ln x$: $f'(x) = \ln x + 1$

2. Find the critical points: Set the derivative equal to zero to find critical points: $\ln x + 1 = 0 \implies \ln x = -1 \implies x = e^{-1}$

   So, $x = e^{-1}$ is a critical point within the interval.

3. Evaluate the function at the endpoints and the critical point:

   • At $x = e^{-2}$: $f(e^{-2}) = e^{-2} \ln(e^{-2}) = e^{-2} \cdot (-2) = -2e^{-2}$

   • At $x = e^{-1}$: $f(e^{-1}) = e^{-1} \ln(e^{-1}) = e^{-1} \cdot (-1) = -e^{-1}$

   • At $x = e$: $f(e) = e \ln(e) = e \cdot 1 = e$

4. Determine the maximum value: Comparing the values:

   • $f(e^{-2}) = -2e^{-2}$
   • $f(e^{-1}) = -e^{-1}$
   • $f(e) = e$

   The maximum value is $f(e) = e$.

Answer:

The absolute maximum value of $f(x)$ over the interval is $\boxed{e}$.

Would you like a more detailed explanation or have any questions?

Here are 5 related questions to consider:

1. What is the importance of critical points in finding the maximum or minimum of a function?
2. How does the derivative $f'(x) = \ln x + 1$ help identify the behavior of the function $f(x) = x \ln x$?
3. Why do we need to evaluate the function at the endpoints of the interval?
4. How can the natural logarithm properties be used to solve problems involving exponential functions?
5. What would the result be if the interval was different, say $[1, e^2]$?

Tip: Remember, when finding maximum or minimum values of a function over a closed interval, always check both critical points and endpoints to ensure accuracy.