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