Problem 35 - Essential Calculus Early Transcendentals

Question

\(23-36=\) Find the critical numbers of the function. $$f(x)=x^{2} e^{-3 x}$$

Step-by-Step Solution

Answer

The critical numbers are 0 and \( \frac{2}{3} \).

Step 1: Find the Derivative

To find the critical numbers of the function, we need to find its derivative. The function is given as \( f(x) = x^2 e^{-3x} \). We apply the product rule: \( (u\cdot v)' = u'v + uv' \), with \( u = x^2 \) and \( v = e^{-3x} \). So, \( u' = 2x \) and \( v' = -3e^{-3x} \). Therefore, \[ f'(x) = (2x)e^{-3x} + (x^2)(-3e^{-3x}) \] This simplifies to: \[ f'(x) = 2xe^{-3x} - 3x^2e^{-3x} \] Which can be factored as: \[ f'(x) = e^{-3x}(2x - 3x^2) \].

Step 2: Set the Derivative Equal to Zero

To find the critical numbers, we need to solve the equation \( f'(x) = 0 \). Setting the derivative to zero, we have:\[ 0 = e^{-3x}(2x - 3x^2) \].The exponential component, \( e^{-3x} \), is never zero. Thus, we focus on solving:\[ 2x - 3x^2 = 0 \].

Step 3: Solve for x

Factor the quadratic equation \( 2x - 3x^2 = 0 \).Factor out an \( x \):\[ x(2 - 3x) = 0 \].Set each factor equal to zero:1. \( x = 0 \)2. \( 2 - 3x = 0 \). Solve for \( x \):\[ 3x = 2 \ x = \frac{2}{3} \].Thus, the critical numbers are \( x = 0 \) and \( x = \frac{2}{3} \).

Step 4: Verify the Critical Numbers are Within the Domain

The function \( f(x) = x^2 e^{-3x} \) is defined for all real numbers, so both critical numbers \( x = 0 \) and \( x = \frac{2}{3} \) lie within the domain of the function.

Key Concepts

Understanding the Product Rule

When dealing with derivatives of products of functions, the Product Rule becomes a handy tool. If you have two functions, say \( u(x) \) and \( v(x) \), their product is \( u(x) \cdot v(x) \). The Product Rule tells you how to differentiate this product. Instead of simply differentiating each function separately, you use this formula:

• \( (u \, v)' = u' \, v + u \, v' \)

In our problem, the function \( f(x) = x^2 e^{-3x} \) is a product where \( u = x^2 \) and \( v = e^{-3x} \). By applying the Product Rule, we derive each part:

• The derivative of \( u = x^2 \) is \( u' = 2x \).
• The derivative of \( v = e^{-3x} \) is \( v' = -3e^{-3x} \).

Applying the Product Rule combines these parts into a single derivative that helps identify critical points by setting it to zero.

Calculating the Derivative

To find the derivative of a function, we look for the rate at which it changes. Consider our function \( f(x) = x^2 e^{-3x} \). The goal is to calculate \( f'(x) \), the derivative. With the Product Rule, we've constructed \( f'(x) = 2xe^{-3x} - 3x^2e^{-3x} \). After simplifying, you factor out the common term:

• \( f'(x) = e^{-3x}(2x - 3x^2) \)

This expression simplifies the derivative into a form that is usable for further calculations. The derivative tells us how the function's value changes as \( x \) changes and sets the stage for finding critical numbers, where this rate of change is zero.

Solving a Quadratic Equation

A quadratic equation appears in our problem through the expression \( 2x - 3x^2 = 0 \). To solve it, we apply basic algebraic strategies. Since it's quadratic, it typically can be solved by factoring or using the quadratic formula. Here, we factor:

• Factor out \( x \): \( x(2 - 3x) = 0 \)

This leads to two simpler equations:

• \( x = 0 \)
• \( 2 - 3x = 0 \), which simplifies to \( x = \frac{2}{3} \)

These solutions, \( x = 0 \) and \( x = \frac{2}{3} \), are the critical numbers. They represent points where the function's rate of change is zero, indicating possible maxima, minima, or points of inflection.

Exploring the Domain of a Function

Before we conclude that any solution is valid, it’s important to check whether it falls within the domain of the function. The domain defines all possible inputs \( x \) for which the function \( f(x) \) is defined. Here, \( f(x) = x^2 e^{-3x} \) is valid for all real numbers. The exponential part \( e^{-3x} \) doesn’t impose any restrictions like prohibited dividers or imaginary components.

• Therefore, the critical numbers \( x = 0 \) and \( x = \frac{2}{3} \) are valid since they lie completely in the domain.

This step ensures that all solutions are applicable, as a critical number must exist within a function's domain to be meaningful. These criteria confirm that our critical numbers are indeed points of interest for further analysis of the function's behavior.