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In calculus, the derivative of a function represents how the function's output value changes with respect to its input value. Think of it as the rate of change or the slope of the function at any given point. In mathematical terms, if you have a function \( f(x) \), the derivative is written as \( f'(x) \). This new function \( f'(x) \) tells you how rapidly \( f(x) \) increases or decreases as \( x \) changes.

For the function \( f(x) = x^2 e^{-3x} \), we first need to find its derivative to locate critical numbers. These are points on the graph where the slope of the tangent (the line that just "touches" the curve) is zero or where the derivative does not exist. This is crucial because critical points can reveal maximums, minimums, or points of inflection in the function.

By using calculus rules like the product rule (which we'll explore next) and differentiating each part carefully, we arrive at the expression for \( f'(x) \). The goal is then to solve it or identify where it's undefined to find your critical numbers.

The product rule is a differentiation rule used when you need to find the derivative of a product of two functions. If you have two functions, \( u(x) \) and \( v(x) \), that you want to multiply, their derivative is not simply the product of their derivatives. Instead, the formula is:

• \( (uv)' = u'v + uv' \)

This might look a little confusing at first, but break it down step by step.

In our exercise, we have \( u(x) = x^2 \) and \( v(x) = e^{-3x} \). So first, we need to differentiate each separately:

• \( u'(x) = 2x \)
• \( v'(x) = -3e^{-3x} \)

With these, you plug them into the product rule to get: \( f'(x) = 2x e^{-3x} + x^2(-3)e^{-3x} \).

The result is a combination of both terms: \( (2x - 3x^2)e^{-3x} \). Be careful with signs and multiplication as these can alter your final derivative, leading to incorrect critical numbers if not handled correctly.

Exponential functions are a key part of calculus and are often expressed in the form \( e^{x} \) where \( e \) is approximately equal to 2.71828. They have unique properties that make them rise or fall very steeply. The derivative of an exponential function has a distinct quality because the derivative of \( e^{x} \) itself is also \( e^{x} \). This can be very helpful in calculus, as it simplifies the process.

In the function \( f(x) = x^2 e^{-3x} \), the term \( e^{-3x} \) plays a crucial role. When you differentiate it, you encounter another essential part of calculus: the chain rule. This makes the derivative \( -3e^{-3x} \)—you multiply by the derivative of the exponent, which is \(-3 \) in this case.

Understanding how exponential functions behave and how to differentiate them are foundational skills. These skills help solve problems involving critical points, allowing us to analyze and understand a function's behavior more thoroughly.