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calculate and get final answers for each part

Let's go through each problem one by one to find the answers.

### Problem 1
Given that \( x = 2 \) is a critical point of \( f(x) = x^3 e^{-kx} \), we are asked to find the value of \( k \).

1. A critical point occurs when \( f'(x) = 0 \).
2. First, find \( f'(x) \) using the product rule:
   \[
   f(x) = x^3 e^{-kx} \Rightarrow f'(x) = 3x^2 e^{-kx} + x^3 \cdot (-k) e^{-kx} = e^{-kx} (3x^2 - kx^3)
   \]
3. Set \( f'(x) = 0 \):
   \[
   e^{-kx} (3x^2 - kx^3) = 0
   \]
   Since \( e^{-kx} \neq 0 \), we need \( 3x^2 - kx^3 = 0 \).
4. Factor out \( x^2 \):
   \[
   x^2(3 - kx) = 0
   \]
5. At \( x = 2 \), \( 3 - k \cdot 2 = 0 \Rightarrow k = \frac{3}{2} \).

**Answer for Problem 1:** \( k = \frac{3}{2} \).

---

### Problem 2
Consider the function \( f(x) = x - 4 \ln(x) \) over the interval \( \frac{1}{4} \leq x \leq 10 \). We need to find the absolute maximum and minimum values.

1. Take the derivative:
   \[
   f'(x) = 1 - \frac{4}{x}
   \]
2. Set \( f'(x) = 0 \) to find critical points:
   \[
   1 - \frac{4}{x} = 0 \Rightarrow x = 4
   \]
3. Evaluate \( f(x) \) at \( x = \frac{1}{4} \), \( x = 4 \), and \( x = 10 \):
   - \( f\left(\frac{1}{4}\right) = \frac{1}{4} - 4 \ln\left(\frac{1}{4}\right) = \frac{1}{4} + 4 \ln(4) \)
   - \( f(4) = 4 - 4 \ln(4) \)
   - \( f(10) = 10 - 4 \ln(10) \)

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

**Final Answers for Problem 2:**
- Absolute maximum value is at \( x = 10 \).
- Absolute minimum value is at \( x = 4 \).

---

### Problem 4
Given \( f(x) = 1kx^3 - (k^2 + 2)x + 5k \), and knowing it has a local minimum at \( x = 1 \), find all values of \( k \).

1. Find \( f'(x) \):
   \[
   f'(x) = 3kx^2 - (k^2 + 2)
   \]
2. Set \( f'(1) = 0 \) for a local minimum at \( x = 1 \):
   \[
   3k - (k^2 + 2) = 0
   \]
3. Solve this equation for \( k \):

**Final Answer for Problem 

Calculate and get final answers for each part

Solve calculus problems involving critical points, absolute maximum and minimum values, and the Second Derivative Test. Learn how to calculate values for critical points in functions involving exponential terms and logarithmic expressions.

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