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Critical points are vital in calculus because they help us identify where the behavior of a function changes. These are the points where the function has either a potential maximum or minimum, or where the function's slope is flat, which means the first derivative is zero. To find critical points, you start by calculating the derivative of the function and set it equal to zero. So, for our function \[ f(x) = x^2 e^{-x} \], we find \[ f'(x) = e^{-x} (2x - x^2) \].We set the derivative to zero and solve:\[ e^{-x} (2x - x^2) = 0. \]Since \( e^{-x} \) can never be zero, we focus on \( 2x - x^2 = 0 \). Factoring gives us \( x(2-x) = 0 \), yielding the critical points \( x = 0 \) and \( x = 2 \). These points are essential as they mark possible shifts from increasing to decreasing behavior and vice versa.

Understanding whether a function is increasing or decreasing in a certain interval involves looking at the sign of the first derivative.- If \( f'(x) > 0 \), the function is increasing.- If \( f'(x) < 0 \), the function is decreasing.For the function \( f(x) \), we tested intervals around the critical points:1. **Interval \((-\infty, 0)\):** Choose a test point like \( x = -1 \), we find\[ f'(-1) = e^{1}(2(-1) - (-1)^2) < 0, \] indicating the function is decreasing.2. **Interval \((0, 2)\):** With \( x = 1 \),\[ f'(1) = e^{-1}(2(1) - 1) > 0, \] the function is increasing.3. **Interval \((2, \infty)\):** For \( x = 3 \),\[ f'(3) = e^{-3}(6 - 9) < 0, \] again, it's decreasing.These intervals help us understand how the function moves between growing larger and smaller across the \( x \) axis.

Finding local maxima and minima is an application of critical points in further identifying specific values where the function attains high or low points within a given range. Once critical points are found, the First Derivative Test helps determine whether each critical point is a local maximum, minimum, or neither. You need to check the change in the sign of the derivative around these points:- **Local Minimum at \( x = 0 \):** The derivative \( f'(x) \) changes from negative to positive as we pass through \( x = 0 \), thus it's a local minimum.- **Local Maximum at \( x = 2 \):** Here, \( f'(x) \) changes from positive to negative at \( x = 2 \), meaning there's a local maximum.Local extrema are significant because they provide insight into the function's "turning points," helping us understand where the function output reaches critical values within intervals without considering boundaries.

The Product Rule is crucial when you're dealing with functions that are the product of two or more functions. It's used to find the derivative of \( f(x) = u(x) \, v(x) \), where both \( u \) and \( v \) are functions of \( x \). The rule states:\[ (uv)' = u'v + uv'. \]In the context of our function \( f(x) = x^2 e^{-x} \), let\( u = x^2 \)and\( v = e^{-x} \).We have:- \( u' = 2x \)- \( v' = -e^{-x} \)Plugging these into the product rule,\[ f'(x) = (2x)(e^{-x}) + (x^2)(-e^{-x}) = 2x e^{-x} - x^2 e^{-x}. \]Using the product rule allows us to accurately calculate the derivative of functions that fundamentally rely on multiplication of simpler variable-based expressions.

The Chain Rule is essential for differentiating composite functions. Suppose you have a function \( y = f(g(x)) \). The Chain Rule helps in finding the derivative \( f'(x) \). It states:\[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x). \]In our exercise, the function \( e^{-x} \) is essentially a composition if we perceive the exponent \(-x\) as the inner function \( g(x) \).Utilizing the chain rule in this context:- The outer function \( f(u) = e^u \)- Resulting in \( f'(u) = e^u \)where \( u = -x \), and then,- The inner derivative is \( g'(x) = -1 \).Thus,\[ \frac{d}{dx}[e^{-x}] = e^{-x} \cdot (-1) = -e^{-x}. \]The Chain Rule simplifies the differentiation of nested functions, playing a pivotal role in breaking down and simplifying complex formulas for analysis.