*f*(*x*)=8*x*⋅*e^-3x* on the interval [−2,3][−2,3].

Now that the function \( f(x) = 8x \cdot e^{-3x} \) has been provided, we can proceed with applying the Mean Value Theorem (MVT) on the interval \([-2, 3]\).

### Steps to Solve Using the Mean Value Theorem

1. **State the Mean Value Theorem**: 
   The MVT states that if a function \( f(x) \) is continuous on a closed interval \([a, b]\) and differentiable on the open interval \( (a, b) \), then there exists at least one point \( c \in (a, b) \) such that:
   \[
   f'(c) = \frac{f(b) - f(a)}{b - a}
   \]

2. **Check the conditions for the MVT**:
   - The function \( f(x) = 8x \cdot e^{-3x} \) is continuous and differentiable on any interval because it is a product of continuous and differentiable functions.

3. **Calculate \( f(a) \) and \( f(b) \)**:
   - \( a = -2 \), \( b = 3 \).
   - \( f(-2) = 8(-2) \cdot e^{-3(-2)} = -16e^6 \).
   - \( f(3) = 8(3) \cdot e^{-3(3)} = 24e^{-9} \).

4. **Calculate the slope of the secant line**:
   \[
   \frac{f(3) - f(-2)}{3 - (-2)} = \frac{24e^{-9} - (-16e^6)}{5} = \frac{24e^{-9} + 16e^6}{5}
   \]

5. **Find the derivative \( f'(x) \)**:
   - Apply the product rule:
     \[
     f'(x) = \frac{d}{dx} \left(8x \cdot e^{-3x}\right)
     \]
     Let \( u(x) = 8x \) and \( v(x) = e^{-3x} \).
     - \( u'(x) = 8 \)
     - \( v'(x) = -3e^{-3x} \)
     - Using the product rule \( f'(x) = u'(x)v(x) + u(x)v'(x) \):
       \[
       f'(x) = 8e^{-3x} + 8x(-3e^{-3x}) = 8e^{-3x} - 24xe^{-3x} = 8e^{-3x}(1 - 3x)
       \]

6. **Set \( f'(c) \) equal to the slope of the secant line**:
   \[
   8e^{-3c}(1 - 3c) = \frac{24e^{-9} + 16e^6}{5}
   \]

7. **Solve for \( c \)**:
   - Simplify the equation to find the value(s) of \( c \) in the interval \((-2, 3)\) that satisfy the equation. 

This equation is transcendental and may require numerical methods or iterative approximation to solve for \( c \).

Would you like to proceed with the numerical solution, or do you have any specific questions or need more details on a particular step?

### Related Questions:
1. How does the negative exponent in the function \( f(x) = 8x \cdot e^{-3x} \) affect the behavior of the function?
2. Can the value \( c \) obtained from the Mean Value Theorem ever be outside the interval \((a, b)\)?
3. What methods can be used to solve transcendental equations like the one in this problem?
4. How does the Mean Value Theorem relate to the concept of the average rate of change of a function?
5. What are some common real-world applications of the Mean Value Theorem?

**Tip:** When solving problems involving exponentials, it’s often helpful to consider the behavior of the function as \( x \) increases or decreases, especially when dealing with large positive or negative exponents.

Learn how to apply the Mean Value Theorem in calculus to solve the function f(x) = 8x * e^-3x on the interval [-2, 3]. This problem involves calculating derivatives, evaluating exponential functions, and applying the Mean Value Theorem to find the value of c. Suitable for college-level students studying calculus.

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