To find the absolute minimum and maximum values of the function $f(x) = x e^{-x^2/128}$ on the interval $[-5, 16]$, we need to evaluate the function at critical points within the interval and at the endpoints.

Here is the step-by-step approach:

1. Identify the critical points: We need to find where $f'(x) = 0$ or $f'(x)$ is undefined. To do this, we'll first compute the derivative of $f(x)$.

   $f(x) = x e^{-x^2/128}$

   Using the product rule and chain rule: $f'(x) = e^{-x^2/128} - \frac{x^2}{64} e^{-x^2/128} = e^{-x^2/128} \left(1 - \frac{x^2}{64}\right)$

   Setting $f'(x) = 0$: $e^{-x^2/128} \left(1 - \frac{x^2}{64}\right) = 0$

   Since $e^{-x^2/128} \neq 0$, we have: $1 - \frac{x^2}{64} = 0$ Solving for $x$, we find: $x^2 = 64 \Rightarrow x = \pm 8$

   So, the critical points are $x = -8$ and $x = 8$.

2. Evaluate the function at the endpoints and critical points: We evaluate $f(x)$ at $x = -5$, $x = 16$, $x = -8$, and $x = 8$:

   • $f(-5) = -5 e^{-(-5)^2/128} = -5 e^{-25/128}$
   • $f(16) = 16 e^{-(16)^2/128} = 16 e^{-256/128} = 16 e^{-2}$
   • $f(-8) = -8 e^{-(-8)^2/128} = -8 e^{-64/128} = -8 e^{-0.5}$
   • $f(8) = 8 e^{-(8)^2/128} = 8 e^{-0.5}$

3. Compare the values: We approximate each of these values to find the absolute minimum and maximum:

   • $f(-5) \approx -5 \cdot e^{-0.195} \approx -5 \cdot 0.822 \approx -4.11$
   • $f(16) \approx 16 \cdot e^{-2} \approx 16 \cdot 0.1353 \approx 2.1648$
   • $f(-8) \approx -8 \cdot e^{-0.5} \approx -8 \cdot 0.6065 \approx -4.852$
   • $f(8) \approx 8 \cdot e^{-0.5} \approx 8 \cdot 0.6065 \approx 4.852$

4. Conclusion:

   • The absolute minimum value on the interval is approximately $f(-8) = -4.852$.
   • The absolute maximum value on the interval is approximately $f(8) = 4.852$.

So, the answers are:

• Absolute minimum value: $-4.852$
• Absolute maximum value: $4.852$

Would you like further details or have any questions?

Here are some related questions for further understanding:

1. How is the product rule used when differentiating functions like $f(x) = x e^{-x^2/128}$?
2. Why do we evaluate the function at the endpoints as well as the critical points?
3. How do we interpret the exponential term $e^{-x^2/128}$ in terms of its effect on $f(x)$?
4. What are some other techniques to find the absolute extrema of a function on a closed interval?
5. How would the solution change if the interval or function were different?

Tip: Always evaluate a function at critical points and endpoints on a closed interval to determine absolute extrema.