To find the interval on which the function $f(x) = \frac{x \cos(\pi x)}{e^{2x}}$ is decreasing, we need to analyze the derivative of $f(x)$ and determine where it is negative.

Step 1: Differentiate $f(x)$

Using the quotient rule: $f(x) = \frac{u(x)}{v(x)}, \quad \text{where } u(x) = x \cos(\pi x) \text{ and } v(x) = e^{2x}$ Then: $f'(x) = \frac{u'(x) v(x) - u(x) v'(x)}{[v(x)]^2}$

Differentiate $u(x) = x \cos(\pi x)$

Using the product rule: $u'(x) = \cos(\pi x) - x \pi \sin(\pi x)$

Differentiate $v(x) = e^{2x}$

$v'(x) = 2e^{2x}$

Substitute these into the quotient rule formula: $f'(x) = \frac{\left( \cos(\pi x) - x \pi \sin(\pi x) \right) e^{2x} - x \cos(\pi x) \cdot 2e^{2x}}{e^{4x}}$

Step 2: Simplify $f'(x)$

Factor out $e^{2x}$ in the numerator: $f'(x) = \frac{e^{2x} \left( \cos(\pi x) - x \pi \sin(\pi x) - 2x \cos(\pi x) \right)}{e^{4x}} = \frac{\cos(\pi x) - x \pi \sin(\pi x) - 2x \cos(\pi x)}{e^{2x}}$

Simplifying further in the numerator: $f'(x) = \frac{\cos(\pi x) (1 - 2x) - x \pi \sin(\pi x)}{e^{2x}}$

Step 3: Set $f'(x) < 0$

Since $e^{2x} > 0$ for all $x$, the sign of $f'(x)$ depends only on the numerator: $\cos(\pi x) (1 - 2x) - x \pi \sin(\pi x) < 0$

Finding the exact intervals where this inequality holds is challenging analytically, so numerical or graphical analysis may be helpful.

Let me know if you would like me to proceed with a numerical solution, graphing approach, or any further explanation.

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Follow-up questions to deepen understanding:

1. What is the significance of the quotient rule in derivative calculations?
2. How do trigonometric functions affect the intervals of increase or decrease?
3. Why does the exponential function $e^{2x}$ always stay positive?
4. What does the behavior of $\cos(\pi x)$ suggest about periodicity in $f(x)$?
5. How does factoring help in simplifying derivatives for solving inequalities?

Tip: When working with trigonometric functions in derivatives, watch for points where they reach their maximum or minimum values, as these often impact the sign of the derivative.