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To determine the absolute extrema of the function \( f(x) = 3x^3 e^{-x} \) on the interval \([-1, 6]\), we need to evaluate \(f(x)\) at its critical points within this interval, as well as at the endpoints \( x = -1 \) and \( x = 6 \).

### Step 1: Find the derivative \( f'(x) \)
Using the product rule for derivatives, where \( f(x) = u(x)v(x) \) with \( u(x) = 3x^3 \) and \( v(x) = e^{-x} \):

\[
f'(x) = u'(x)v(x) + u(x)v'(x)
\]

1. \( u(x) = 3x^3 \Rightarrow u'(x) = 9x^2 \)
2. \( v(x) = e^{-x} \Rightarrow v'(x) = -e^{-x} \)

Thus:

\[
f'(x) = 9x^2 e^{-x} - 3x^3 e^{-x} = 3x^2 e^{-x}(3 - x)
\]

### Step 2: Set \( f'(x) = 0 \) to find critical points
We solve the equation:

\[
3x^2 e^{-x}(3 - x) = 0
\]

This equation is zero if any of its factors are zero:

1. \( 3x^2 = 0 \Rightarrow x = 0 \)
2. \( e^{-x} \neq 0 \) for all real \( x \) (exponential function is never zero).
3. \( 3 - x = 0 \Rightarrow x = 3 \)

So, the critical points are \( x = 0 \) and \( x = 3 \).

### Step 3: Evaluate \( f(x) \) at the critical points and endpoints
Now we evaluate \( f(x) \) at \( x = -1 \), \( x = 0 \), \( x = 3 \), and \( x = 6 \):

1. **At \( x = -1 \):**
   \[
   f(-1) = 3(-1)^3 e^{-(-1)} = -3e
   \]

2. **At \( x = 0 \):**
   \[
   f(0) = 3(0)^3 e^{-0} = 0
   \]

3. **At \( x = 3 \):**
   \[
   f(3) = 3(3)^3 e^{-3} = 3 \cdot 27 \cdot e^{-3} = 81e^{-3}
   \]

4. **At \( x = 6 \):**
   \[
   f(6) = 3(6)^3 e^{-6} = 3 \cdot 216 \cdot e^{-6} = 648e^{-6}
   \]

### Step 4: Determine the absolute maximum and minimum
Comparing the values:

- \( f(-1) = -3e \)
- \( f(0) = 0 \)
- \( f(3) = 81e^{-3} \)
- \( f(6) = 648e^{-6} \)

Since \( 0 > -3e \), \( 81e^{-3} > 648e^{-6} \), and \( 81e^{-3} > 0 \), we conclude:

- **Absolute maximum** occurs at \( x = 3 \) with value \( 81e^{-3} \).
- **Absolute minimum** occurs at \( x = -1 \) with value \( -3e \).

### Final Answer
- **Absolute maximum**: \( f(3) = 81e^{-3} \)
- **Absolute minimum**: \( f(-1) = -3e \)

Would you like further details on any specific part of this solution?

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### Related Questions:
1. How would the solution change if the interval were unbounded?
2. What if we changed the function to \( f(x) = 3x^3 e^{x} \)?
3. How does the exponential decay factor \( e^{-x} \) affect the extrema of \( f(x) \)?
4. What if the interval was restricted to only positive values, like \([0, 6]\)?
5. How do we interpret the behavior of \( f(x) \) as \( x \to \infty \)?

### Tip:
When dealing with products of polynomial and exponential functions, look for critical points by setting derivatives to zero, but don't forget to check boundary points in closed intervals!

Determine the location and value of the absolute extreme values of f on the given interval, if they exist. f(x) = 3x^3 e^{-x}; [-1, 6]

Learn how to find the absolute maximum and minimum values of the function f(x) = 3x^3 e^{-x} on the interval [-1, 6]. This solution involves finding critical points using derivatives, evaluating the function at endpoints, and comparing values for extrema. Ideal for college calculus students.

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