Problem 15 - Calculus with Concepts in Calculus

Question

Find all critical numbers of the given function. $$ f(x)=x^{2} e^{x} $$

Step-by-Step Solution

Answer

The critical numbers of the function are \( x = 0 \) and \( x = -2 \).

Step 1: Understand Critical Numbers

Critical numbers of a function are the x-values where the derivative is zero or undefined. For a differentiable function, these points indicate where the function's rate of change is zero, potentially identifying local maxima, minima, or inflection points.

Step 2: Find the Derivative

To find the critical numbers, start by differentiating the given function. The function is given as \( f(x) = x^2 e^x \). Use the product rule for differentiation: if \( u = x^2 \) and \( v = e^x \), then \( f'(x) = u'v + uv' \). Here, \( u' = 2x \) and \( v' = e^x \), so the derivative is:\[f'(x) = (x^2)' e^x + x^2 (e^x)' = 2x e^x + x^2 e^x\]

Step 3: Factor the Derivative

The derivative \( f'(x) = 2x e^x + x^2 e^x \) can be factored to make it easier to solve. Notice that there is a common factor of \( e^x \), so:\[f'(x) = e^x (2x + x^2)\]

Step 4: Set Derivative to Zero

Set the factored derivative equal to zero to find critical numbers:\[e^x (x^2 + 2x) = 0\]Since \( e^x \) is never zero, solve for where \( x^2 + 2x = 0 \).

Step 5: Solve the Quadratic Equation

Solve the equation \( x^2 + 2x = 0 \) by factoring:\[x(x + 2) = 0\]This gives solutions:\[x = 0 \quad \text{or} \quad x = -2\]

Step 6: Verify Critical Numbers

The solutions \( x = 0 \) and \( x = -2 \) are critical numbers, as they satisfy the condition \( f'(x) = 0 \). There are no other factors or terms that would need checking (such as points of discontinuity), confirming these are all the critical numbers.

Key Concepts

Derivative

When working with functions in calculus, the concept of a derivative is essential. A derivative represents the rate at which a function is changing at any given point. For the function \( f(x) = x^2 e^x \), finding the derivative involves determining how \( f(x) \) changes with respect to \( x \). The derivative is denoted as \( f'(x) \).To differentiate \( f(x) \), we look at small changes in \( x \) and measure the corresponding changes in \( f(x) \). The derivative not only helps to find critical numbers but also identifies where the slope of the tangent to the function is zero or undefined. Understanding these points can inform us about the behavior of the function, such as where it might reach a local maximum or minimum.

Product Rule

The product rule is a method used to differentiate a function that is the product of two other functions. For the function \( f(x) = x^2 e^x \), consider it as the product of \( u = x^2 \) and \( v = e^x \). The product rule states: if \( f(x) = u(x) v(x) \), then the derivative \( f'(x) \) is given by:

• \( f'(x) = u'(x) v(x) + u(x) v'(x) \)

In this exercise, \( u'(x) = 2x \) and \( v'(x) = e^x \). Therefore, applying the product rule results in:

• \( f'(x) = 2x e^x + x^2 e^x \)

Applying the product rule helps to correctly find the derivative of complex functions. This is essential for locating critical numbers and analyzing the function's properties.

Quadratic Equation

A quadratic equation takes the form \( ax^2 + bx + c = 0 \). Solving quadratic equations can be done through various methods, such as factoring or using the quadratic formula. In the context of the derivative \( f'(x) = e^x (x^2 + 2x) \), we set the expression \( x^2 + 2x = 0 \) to zero to find critical numbers.Since this is a simple quadratic equation with no constant term, factoring is a straightforward method to solve it. These critical points tell us where the derivative, and hence the slope of the function, is zero.After factoring, \( x^2 + 2x = x(x + 2) \), the solutions are obtained easily by setting each factor to zero. This yields the solutions \( x = 0 \) and \( x = -2 \). These solutions represent the x-values where the critical numbers of the original function lie.

Factorization

Factorization is a useful algebraic technique that simplifies expressions, making it easier to solve equations, particularly quadratic ones. When finding critical numbers, we often use factorization to set the derivative of a function equal to zero. For the derivative \( f'(x) = 2x e^x + x^2 e^x \), note that \( e^x \) can be factored out:

• \( f'(x) = e^x (2x + x^2) \)

The beauty of factorization is that it reduces the equation into smaller, more manageable parts. In this case, we are left with \( e^x \) and \( x(x + 2) \). Since \( e^x \) is never zero, we focus on the quadratic part, \( x(x + 2) \), to solve for zero. Factorization, thus, helps us efficiently find potential critical points such as \( x = 0 \) and \( x = -2 \). These points are crucial as they inform us about the function's potential extremities.