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.