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The First Derivative Test is a powerful tool to determine whether critical points are local minima, maxima, or neither. To employ the test, we first need to find the critical points by setting the first derivative of the function to zero. For the function given, \(f(x) = x^2 e^{-2x}\), the critical points are \(x = 0\) and \(x = 1\).

Here's how the First Derivative Test works:

• Calculate the first derivative, \(f'(x)\).
• Identify intervals between the critical points.
• Check the sign of \(f'(x)\) within each interval.

When \(x < 0\), \(f'(x)\) is negative, indicating a decreasing function. Between \(0 < x < 1\), \(f'(x)\) turns positive, suggesting an increasing function. After \(x > 1\), \(f'(x)\) becomes negative again, which means the function decreases.

These sign changes show that at \(x = 0\), the function changes from decreasing to increasing, implying a local minimum. At \(x = 1\), it changes from increasing to decreasing, indicating a local maximum.

The Product Rule is essential when differentiating functions that are products of two or more terms. It is particularly useful when dealing with functions like \(f(x) = x^2 e^{-2x}\), where you have both a polynomial and an exponential function.

The Product Rule states:
If you have a function \(g(x) = u(x) v(x)\), then the derivative is \(g'(x) = u'(x)v(x) + u(x)v'(x)\).

Applying this to our function, where \(u(x) = x^2\) and \(v(x) = e^{-2x}\):

• Find \(u'(x) = 2x\)
• Find \(v'(x) = -2e^{-2x}\)
• Plug these into the Product Rule: \(f'(x) = 2x e^{-2x} + x^2(-2e^{-2x})\)

After simplifying, you get \(f'(x) = 2x e^{-2x} - 2x^2 e^{-2x}\). By factoring out the common term \(2x e^{-2x}\), you simplify it to \(f'(x) = 2x e^{-2x} (1-x)\), making it much easier to analyze the critical points.

Extrema Analysis involves finding and classifying the minimum and maximum values of a function. For a function like \(f(x) = x^2 e^{-2x}\), identifying these extremas is key to understanding its behavior over its domain.

There are two main steps in extrema analysis:

• Locate critical points by setting \(f'(x) = 0\).
• Classify these points using the First Derivative Test or any other suitable method.

For \(f(x) = x^2 e^{-2x}\), we've found the critical points \(x = 0\) (local minimum) and \(x = 1\) (local maximum). Each point marks a significant change in direction for the function.

Ultimately, performing a thorough extrema analysis enables deeper understanding of how the function behaves graphically and dynamically, helping not just in calculus problems, but in practical applications where understanding peaks and valleys is crucial.