To find the absolute maximum and minimum values of the function $f(x) = e^{-x} - e^{-9x}$ on the interval $[0, 1]$, we will follow these steps:

Step 1: Find the critical points

1. Compute the derivative of $f(x)$: $f'(x) = -e^{-x} \cdot (-1) + 9e^{-9x} \cdot (-1) = -e^{-x} + 9e^{-9x}.$ Simplifying, we get: $f'(x) = -e^{-x} + 9e^{-9x}.$

2. Set $f'(x) = 0$ to find critical points: $-e^{-x} + 9e^{-9x} = 0 \implies e^{-x} = 9e^{-9x}.$

3. Solve for $x$: Take the natural logarithm of both sides: $-x = -9x + \ln(9).$ Simplify to get: $8x = \ln(9) \implies x = \frac{\ln(9)}{8}.$

   Since $\ln(9) \approx 2.197$, we find: $x \approx \frac{2.197}{8} \approx 0.2746.$

Step 2: Evaluate $f$ at critical points and endpoints

We need to find $f(x)$ at $x = 0$, $x = 1$, and $x \approx 0.2746$.

1. Evaluate $f(0)$: $f(0) = e^{-0} - e^{-9 \cdot 0} = 1 - 1 = 0.$

2. Evaluate $f(1)$: $f(1) = e^{-1} - e^{-9} \approx 0.3679 - 0.0001234 \approx 0.3678.$

3. Evaluate $f\left(\frac{\ln(9)}{8}\right)$: $f\left(\frac{\ln(9)}{8}\right) = e^{-\frac{\ln(9)}{8}} - e^{-9 \cdot \frac{\ln(9)}{8}}.$ Simplifying this, we get: $f\left(\frac{\ln(9)}{8}\right) = 9^{-\frac{1}{8}} - 9^{-\frac{9}{8}}.$ Using $9^{-\frac{1}{8}} \approx 0.7937$ and $9^{-\frac{9}{8}} \approx 0.1122$: $f\left(\frac{\ln(9)}{8}\right) \approx 0.7937 - 0.1122 = 0.6815.$

Step 3: Identify the absolute maximum and minimum values

From the evaluations above:

• $f(0) = 0$
• $f(1) \approx 0.3678$
• $f\left(\frac{\ln(9)}{8}\right) \approx 0.6815$

Conclusion:

• Absolute maximum: $f\left(\frac{\ln(9)}{8}\right) \approx 0.6815$ at $x \approx 0.2746$.
• Absolute minimum: $f(0) = 0$ at $x = 0$.

Would you like more details or have any questions on this process?

Related Questions:

1. How do you find critical points for a function?
2. Why do we evaluate $f(x)$ at endpoints when finding absolute extrema?
3. What is the significance of the derivative being zero at a point?
4. How can logarithms help in solving equations like $e^{-x} = 9e^{-9x}$?
5. What does $e^{-x}$ represent, and why does it decay as $x$ increases?

Tip:

For functions involving exponentials like $e^{-x}$, always consider taking the natural logarithm when equating expressions. It can simplify calculations significantly!