To analyze the function \( F(x) = 9x e^{-3x} \), we can explore its key properties such as the first derivative, the second derivative, and critical points.

### 1. First Derivative
To find the first derivative, \( F'(x) \), we use the product rule. The product rule states that if \( u(x) = 9x \) and \( v(x) = e^{-3x} \), then:

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

First, we find the derivatives of \( u(x) \) and \( v(x) \):

\[
u(x) = 9x \implies u'(x) = 9
\]
\[
v(x) = e^{-3x} \implies v'(x) = -3e^{-3x}
\]

Applying the product rule:

\[
F'(x) = 9 \cdot e^{-3x} + 9x \cdot (-3e^{-3x})
\]
\[
F'(x) = 9e^{-3x} - 27xe^{-3x}
\]
\[
F'(x) = e^{-3x}(9 - 27x)
\]

### 2. Critical Points
To find the critical points, we set \( F'(x) \) to zero and solve for \( x \):

\[
e^{-3x}(9 - 27x) = 0
\]

Since \( e^{-3x} \neq 0 \), we only need to solve:

\[
9 - 27x = 0
\]
\[
27x = 9
\]
\[
x = \frac{1}{3}
\]

### 3. Second Derivative
To determine the concavity and find possible points of inflection, we find the second derivative, \( F''(x) \).

Using the first derivative \( F'(x) = e^{-3x}(9 - 27x) \), we apply the product rule again:

\[
F''(x) = \frac{d}{dx} \left( e^{-3x}(9 - 27x) \right)
\]

Let \( u(x) = e^{-3x} \) and \( v(x) = 9 - 27x \). Then:

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

We already have:

\[
u(x) = e^{-3x} \implies u'(x) = -3e^{-3x}
\]
\[
v(x) = 9 - 27x \implies v'(x) = -27
\]

Applying the product rule:

\[
F''(x) = (-3e^{-3x})(9 - 27x) + (e^{-3x})(-27)
\]
\[
F''(x) = -3e^{-3x}(9 - 27x) - 27e^{-3x}
\]
\[
F''(x) = e^{-3x} \left[ -3(9 - 27x) - 27 \right]
\]
\[
F''(x) = e^{-3x} \left[ -27 + 81x - 27 \right]
\]
\[
F''(x) = e^{-3x} \cdot 81x
\]
\[
F''(x) = 81xe^{-3x}
\]

### Summary of Findings
- The first derivative: \( F'(x) = e^{-3x}(9 - 27x) \)
- The critical point: \( x = \frac{1}{3} \)
- The second derivative: \( F''(x) = 81xe^{-3x} \)

Would you like more details or have any questions?

Here are some related questions you might find interesting:

1. How do you determine the concavity of \( F(x) = 9x e^{-3x} \) using the second derivative?
2. What is the maximum value of \( F(x) = 9x e^{-3x} \) on the interval \( [0, 1] \)?
3. How do you find the inflection points of \( F(x) = 9x e^{-3x} \)?
4. What is the behavior of \( F(x) = 9x e^{-3x} \) as \( x \to \infty \)?
5. How do you solve the equation \( 9x e^{-3x} = 1 \) for \( x \)?
6. What are the roots of the equation \( 9x e^{-3x} = 0 \)?
7. How do you sketch the graph of \( F(x) = 9x e^{-3x} \)?
8. What are the applications of the function \( F(x) = 9x e^{-3x} \) in real-life scenarios?

**Tip**: When analyzing functions involving products of polynomials and exponentials, the product rule and chain rule are essential tools for finding derivatives and understanding the function's behavior.

Explore the function F(x) = 9x e^-3x with detailed analysis of its derivatives, critical points, and concavity. Learn how to apply the product rule and chain rule to find derivatives, determine critical points, and understand the concavity of exponential functions. Suitable for advanced high school students and calculus enthusiasts.

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