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?

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!