The given function is $f(x) = e^{-x} - e^{-5x}$ over the interval $[0, 1]$. To find the absolute maximum and minimum values, we follow these steps:

Step 1: Find the derivative of the function

Differentiate $f(x)$: $f'(x) = -e^{-x} + 5e^{-5x}$ This derivative helps find the critical points.

Step 2: Set the derivative equal to zero to find critical points

We solve the equation: $-e^{-x} + 5e^{-5x} = 0$ $e^{-x} = 5e^{-5x}$ Divide both sides by $e^{-5x}$: $e^{4x} = 5$ Taking the natural logarithm of both sides: $4x = \ln(5)$ $x = \frac{\ln(5)}{4}$

Step 3: Evaluate $f(x)$ at the critical points and endpoints

Now evaluate the function at the critical point and the interval endpoints $x = 0$ and $x = 1$.

1. $f(0) = e^{0} - e^{0} = 1 - 1 = 0$
2. $f(1) = e^{-1} - e^{-5}$ $f(1) \approx 0.3679 - 0.0067 = 0.3612$
3. At $x = \frac{\ln(5)}{4}$: $f\left( \frac{\ln(5)}{4} \right) = e^{-\frac{\ln(5)}{4}} - e^{-5 \cdot \frac{\ln(5)}{4}}$ Simplifying this requires numerical evaluation: $f\left( \frac{\ln(5)}{4} \right) \approx 0.7248 - 0.064 = 0.6608$

Step 4: Conclusion

The absolute minimum value is already correctly identified as $0$ at $x = 0$.

For the absolute maximum value, the highest value occurs at $x = \frac{\ln(5)}{4}$, and $f\left( \frac{\ln(5)}{4} \right) \approx 0.6608$.

Let me know if you'd like further details on the steps!

Here are five related questions to extend your understanding:

1. How would the function change if the interval was extended beyond $[0, 1]$?
2. Can you explain why the critical point $x = \frac{\ln(5)}{4}$ lies between $0$ and $1$?
3. How does the behavior of exponential decay affect the shape of the function?
4. What is the significance of the endpoints in determining absolute extrema?
5. Can you generalize this method to find extrema for other exponential functions?

Tip: Always check both critical points and interval endpoints when finding absolute extrema of a function.