To find the inflection point of the function $f(x) = 4xe^{-9x}$, we need to analyze the second derivative of the function. Inflection points occur where the concavity of the function changes, which happens when the second derivative changes sign. Here are the steps to find it:

1. Compute the First Derivative, $f'(x)$:

   Using the product rule: $f'(x) = \frac{d}{dx}(4x) \cdot e^{-9x} + 4x \cdot \frac{d}{dx}(e^{-9x})$ $= 4 \cdot e^{-9x} - 36x \cdot e^{-9x}$ $= e^{-9x}(4 - 36x)$

2. Compute the Second Derivative, $f''(x)$:

   Applying the product rule to $f'(x) = e^{-9x}(4 - 36x)$: $f''(x) = \frac{d}{dx}(e^{-9x}) \cdot (4 - 36x) + e^{-9x} \cdot \frac{d}{dx}(4 - 36x)$ $= (-9e^{-9x})(4 - 36x) + e^{-9x} \cdot (-36)$ $= e^{-9x}(-36 + 324x - 36)$ $= e^{-9x}(-72 + 324x)$ $= 36e^{-9x}(9x - 2)$

3. Set the Second Derivative Equal to Zero: $36e^{-9x}(9x - 2) = 0$

   Since $e^{-9x} \neq 0$ for any real $x$, we solve: $9x - 2 = 0$ $x = \frac{2}{9}$

4. Confirm the Inflection Point:

   To confirm, we check if the sign of $f''(x)$ changes around $x = \frac{2}{9}$. Since it does, this is an inflection point.

Conclusion:

The inflection point of $f(x) = 4xe^{-9x}$ occurs at $x = \frac{2}{9}$.

Would you like further details on any step or have any questions?

Here are some related questions you might consider:

1. How do inflection points relate to the graph's concavity?
2. What is the product rule, and how does it apply to derivatives?
3. Why does setting the second derivative to zero help find inflection points?
4. How does exponential decay affect the behavior of $f(x) = 4xe^{-9x}$?
5. How can you confirm if a point is an inflection point by examining concavity?

Tip: Always check if the sign of the second derivative changes around the inflection point to confirm the concavity shift.