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I'll go through each problem one by one to find the required solutions.

### Problem 1
Given:
\[
f(x) = xe^{-kx}
\]
We are told that \( x = 4 \) is a critical point, meaning that \( f'(4) = 0 \).

1. Differentiate \( f(x) \) with respect to \( x \):
   \[
   f(x) = xe^{-kx}
   \]
   Using the product rule:
   \[
   f'(x) = e^{-kx} + x \cdot (-k) e^{-kx} = e^{-kx}(1 - kx)
   \]
2. Set \( f'(4) = 0 \):
   \[
   e^{-4k}(1 - 4k) = 0
   \]
   Since \( e^{-4k} \neq 0 \), we have:
   \[
   1 - 4k = 0
   \]
   Solving for \( k \):
   \[
   k = \frac{1}{4}
   \]

**Answer for Problem 1:** \( k = \frac{1}{4} \).

---

### Problem 2
Given:
\[
f(x) = x - 5 \ln(x)
\]
over the interval \( \frac{1}{5} \leq x \leq 11 \).

To find the absolute maximum and minimum, we need to evaluate \( f(x) \) at critical points within this interval and at the endpoints \( x = \frac{1}{5} \) and \( x = 11 \).

1. Differentiate \( f(x) \) with respect to \( x \):
   \[
   f'(x) = 1 - \frac{5}{x}
   \]
2. Set \( f'(x) = 0 \) to find critical points:
   \[
   1 - \frac{5}{x} = 0
   \]
   \[
   x = 5
   \]
3. Calculate \( f(x) \) at \( x = \frac{1}{5} \), \( x = 5 \), and \( x = 11 \):
   - \( f\left(\frac{1}{5}\right) = \frac{1}{5} - 5 \ln\left(\frac{1}{5}\right) = \frac{1}{5} + 5 \ln(5) \)
   - \( f(5) = 5 - 5 \ln(5) \)
   - \( f(11) = 11 - 5 \ln(11) \)

4. Compare these values to determine the maximum and minimum.

After evaluating, we get:
   - **Absolute Maximum Value:** \( f(x) \) at \( x = \frac{1}{5} \)
   - **Absolute Minimum Value:** \( f(x) \) at \( x = 11 \)

---

### Problem 4
Given:
\[
f(x) = 1kx^3 - (k^2 + 2)x + 8k
\]
We are told that \( f(x) \) has a local minimum at \( x = 1 \), so we’ll apply the second derivative test.



Get final answers for the following math problems:
1. Suppose that x = 4 is a critical point of f(x) = x e^(-kx). Find the value of k.
2. Consider the function f(x) = x - 5 ln(x), 1/5 ≤ x ≤ 11. Determine the absolute maximum and minimum values and the points where they occur.
4. If the function f(x) = 1kx^3 - (k^2 + 2)x + 8k has a local minimum at x = 1, find all values of k.

Find the value of k for critical points in exponential functions, determine absolute maxima and minima for logarithmic functions, and apply the second derivative test for local minima. These calculus problems demonstrate key concepts and methods suitable for Grades 11-12 and college-level students.

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