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To solve this, let’s start with the Maclaurin series for \( \cos(x) \):

\[
\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots
\]

Since we need the series for \( f(x) = x^3 \cos(4x) \), let’s substitute \( 4x \) into the Maclaurin series for \( \cos(x) \):

\[
\cos(4x) = 1 - \frac{(4x)^2}{2!} + \frac{(4x)^4}{4!} - \frac{(4x)^6}{6!} + \cdots
\]

Simplifying each term in the series for \( \cos(4x) \):

\[
\cos(4x) = 1 - \frac{16x^2}{2} + \frac{256x^4}{24} - \cdots
\]
\[
\cos(4x) = 1 - 8x^2 + \frac{32x^4}{3} - \cdots
\]

Now, we multiply this by \( x^3 \) to obtain the series for \( f(x) = x^3 \cos(4x) \):

\[
f(x) = x^3 \left(1 - 8x^2 + \frac{32x^4}{3} - \cdots \right)
\]

Expanding this:

\[
f(x) = x^3 - 8x^5 + \frac{32x^7}{3} + \cdots
\]

The first three non-zero terms of the Maclaurin series for \( f(x) = x^3 \cos(4x) \) are:

\[
x^3, -8x^5, \text{ and } \frac{32x^7}{3}
\]

Would you like further clarification on any step?

Here are some related questions to explore:

1. What is the Maclaurin series for other trigonometric functions like \( \sin(x) \)?
2. How would the result change if we had \( f(x) = x^3 \sin(4x) \) instead?
3. How is the Maclaurin series useful in approximating values near \( x = 0 \)?
4. What role does the factorial in each term’s denominator play in the convergence of the series?
5. Can we find higher-order terms in the Maclaurin series for more accurate approximations?

**Tip:** Maclaurin series are particularly useful for approximating functions near \( x = 0 \) and often simplify complex expressions involving powers and trigonometric functions.

Using the Maclaurin series for \( \cos(x) \), find the first three non-zero terms of the Maclaurin series for \( f(x) = x^3 \cos(4x) \).

Discover how to use the Maclaurin series for \( \cos(x) \) to find the first three non-zero terms of \( f(x) = x^3 \cos(4x) \). This detailed solution includes step-by-step simplification of the series expansion, suitable for students in Grades 11-12.

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