To find the Maclaurin series for the function $f(x) = 9x e^{-4x}$, we start by recalling that a Maclaurin series expands a function around $x = 0$. Let's break it down step-by-step:

1. Separate the Function: We have $f(x) = 9x \cdot e^{-4x}$. We can treat $9x$ and $e^{-4x}$ as two parts and expand $e^{-4x}$ into its own Maclaurin series.

2. Maclaurin Series for $e^{-4x}$: The exponential function $e^u$ has a Maclaurin series: $e^u = \sum_{n=0}^{\infty} \frac{u^n}{n!}$ Here, we substitute $u = -4x$, giving: $e^{-4x} = \sum_{n=0}^{\infty} \frac{(-4x)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-4)^n x^n}{n!}$

3. Multiply by $9x$: Now, we multiply this series by $9x$: $f(x) = 9x \cdot \sum_{n=0}^{\infty} \frac{(-4)^n x^n}{n!} = \sum_{n=0}^{\infty} \frac{9 \cdot (-4)^n x^{n+1}}{n!}$

4. Rewrite in Summation Form: To put this in a standard summation form, let’s rewrite it as: $f(x) = \sum_{n=0}^{\infty} \frac{9 \cdot (-4)^n}{n!} x^{n+1}$

This is the Maclaurin series for $f(x) = 9x e^{-4x}$. If there is an error in your answer, it could be due to the structure of the summation or terms not aligning with your solution's requirements.

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Related Questions:

1. How do you find the Maclaurin series of more complex functions involving $e^x$ and trigonometric terms?
2. What is the general form of the Maclaurin series for a function $f(x) = g(x) \cdot h(x)$?
3. How does multiplying by a polynomial affect the convergence of a Maclaurin series?
4. What are the applications of Maclaurin series in physics or engineering?
5. How can you verify the convergence of a Maclaurin series for different functions?

Tip:

Always remember to separate each part of a function when calculating Maclaurin series for products, and use known series expansions to simplify calculations.