We are tasked with determining the absolute extreme values of the function $f(x) = 4x^3 e^{-x}$ over the interval $[-1, 5]$. To solve this, let's go through the steps to find the absolute maximum and minimum.

Step 1: Identify Critical Points

We need to find the derivative of $f(x) = 4x^3 e^{-x}$ and set it equal to zero to find critical points.

The function is a product, so we apply the product rule for differentiation:

$f'(x) = \frac{d}{dx}(4x^3) \cdot e^{-x} + 4x^3 \cdot \frac{d}{dx}(e^{-x})$

$f'(x) = 12x^2 e^{-x} - 4x^3 e^{-x}$

Factoring out $4x^2 e^{-x}$:

$f'(x) = 4x^2 e^{-x}(3 - x)$

Now, set the derivative equal to zero to find critical points:

$4x^2 e^{-x}(3 - x) = 0$

Since $e^{-x} \neq 0$ for all $x$, we can set the factors equal to zero:

$4x^2 = 0 \quad \text{or} \quad 3 - x = 0$

From $4x^2 = 0$, we get $x = 0$.
From $3 - x = 0$, we get $x = 3$.

Step 2: Evaluate at Critical Points and Endpoints

We now evaluate $f(x)$ at the critical points $x = 0$ and $x = 3$, as well as at the endpoints of the interval $x = -1$ and $x = 5$.

$f(0) = 4(0)^3 e^0 = 0$ $f(3) = 4(3)^3 e^{-3} = 4 \cdot 27 e^{-3} = 108 e^{-3} \approx 1.96$ $f(-1) = 4(-1)^3 e^{-(-1)} = -4 e^1 \approx -10.87$ $f(5) = 4(5)^3 e^{-5} = 4 \cdot 125 e^{-5} \approx 0.34$

Step 3: Compare Values

• $f(-1) \approx -10.87$
• $f(0) = 0$
• $f(3) \approx 1.96$
• $f(5) \approx 0.34$

Thus, the absolute maximum is approximately $1.96$ at $x = 3$, and there is no absolute minimum since the lowest value, $f(-1)$, is a negative number but not lower than $-\infty$.

Conclusion:

The correct answer is B: The absolute maximum is $108 e^{-3}$ at $x = 3$, but there is no absolute minimum.

Would you like more details or have any questions about this solution?

Here are 5 follow-up questions:

1. How is the product rule applied when differentiating?
2. Why does $e^{-x}$ never equal zero?
3. What is the significance of checking endpoints in optimization problems?
4. Can there be an absolute minimum if a function approaches negative infinity?
5. How can we estimate $e^{-3}$ more accurately?

Tip: Always evaluate a function at both the critical points and endpoints when determining absolute extrema on a closed interval.