Problem 48 - Calculus Early Transcendentals

Question

Use any method to find the relative extrema of the function \(f .\) $$ f(x)=\left(x e^{x}\right)^{2} $$

Step-by-Step Solution

Answer

The relative maximum is at \(x=-1\) and the relative minimum is at \(x=0\).

Step 1: Find the First Derivative

To find relative extrema, we need the first derivative of the function. Given \(f(x) = (x e^x)^2\), we'll differentiate it. Let \(u = x e^x\). Therefore, \(f(x) = u^2\). Using the chain rule, \(f'(x) = 2u \cdot u'\). Now find \(u'\). Since \(u = x e^x\), \(u' = e^x + x e^x\). Thus, the derivative is \(f'(x) = 2(x e^x)(e^x + x e^x)\).

Step 2: Simplify the First Derivative

The first derivative is \(f'(x) = 2(x e^x)(e^x + x e^x)\). Simplifying further, \(f'(x) = 2 x e^{x}(e^x) + 2 x^2 e^{2x}\). Combine the terms: \(f'(x) = 2x e^{2x} + 2x^2 e^{2x} = 2xe^{2x}(1 + x)\).

Step 3: Find Critical Points

Set the first derivative to zero to find critical points: \(2xe^{2x}(1 + x) = 0\). This implies either \(2x = 0\) or \(e^{2x} = 0\) or \((1 + x) = 0\). Since \(e^{2x} eq 0\), consider \(x = 0\) and \(x = -1\). Thus, the critical points are \(x = 0\) and \(x = -1\).

Step 4: Determine Relative Extrema Using the Second Derivative Test

The second derivative \(f''(x)\) tests the concavity at the critical points. For \(f'(x) = 2xe^{2x}(1 + x)\), differentiate to find \(f''(x)\). Applying the product rule: \(f''(x) = 2e^{2x}(1+x) + 2xe^{2x} + 4xe^{2x}(1+x)\). Simplify, then evaluate \(f''(x)\) at the critical points. Substituting \(x=0\) into \(f''(x)\), check its sign (positive means local min, negative means local max). Repeat for \(x=-1\). Find \(x=0\) gives \(f''(0) > 0\), indicating a local minimum. \(x=-1\) gives \(f''(-1) < 0\), indicating a local maximum.

Key Concepts

First Derivative

To find the relative extrema of a function, we first need to determine its first derivative. The first derivative, often represented as \( f'(x) \), helps us understand how the function's slope behaves. In this case, our function is \( f(x) = (x e^x)^2 \). To find the derivative, we apply the chain rule, which is useful in differentiating composite functions. We let \( u = x e^x \) so that our function becomes \( u^2 \). The chain rule tells us that \( f'(x) = 2u \cdot u' \).

• Differentiate \( u \): Since \( u = x e^x \), we need the derivative \( u' = e^x + x e^x \).
• Multiply and Simplify: Substitute \( u' \) back to get \( f'(x) = 2(x e^x)(e^x + x e^x) \). On further simplification, this becomes \( f'(x) = 2x e^{2x}(1 + x) \).

Finding the first derivative is crucial because it lays down the path to locate critical points, which are essential in determining the relative extrema of the function.

Critical Points

Critical points occur where the first derivative \( f'(x) \) is zero or undefined. Solving for critical points allows us to determine where the function might have a relative maximum or minimum. Let's take the simplified first derivative \( f'(x) = 2x e^{2x}(1 + x) \) and set it equal to zero: \ \( 2x e^{2x}(1 + x) = 0 \)

• The exponential function \( e^{2x} \) is never zero, so we focus on the other factors.
• Setting \( 2x = 0 \) gives \( x = 0 \).
• Setting \( 1 + x = 0 \) yields \( x = -1 \).

Therefore, the critical points of this function are \( x = 0 \) and \( x = -1 \). Identifying these points is crucial because they are potential locations where the function could change direction, indicating a local maximum or minimum.

Second Derivative Test

The second derivative test helps us further analyze critical points to determine if they are indeed relative maximums or minimums. After finding the critical points, we look at the second derivative \( f''(x) \) to see the concavity of the function at those points. A positive second derivative indicates a local minimum, while a negative one indicates a local maximum.

• Calculate the Second Derivative: From the first derivative \( f'(x) = 2x e^{2x}(1 + x) \), apply the product rule to differentiate further and find \( f''(x) \).
• Evaluate the Second Derivative at Critical Points: \ - For \( x = 0 \), compute \( f''(0) \). If \( f''(0) > 0 \), then there's a local minimum. \ - For \( x = -1 \), compute \( f''(-1) \). If \( f''(-1) < 0 \), then there's a local maximum.

Thus, evaluating the second derivative at these critical points is key to determining the nature of these points – whether they represent a peak, a valley, or a point of inflection. Here, at \( x = 0 \), \( f''(0) \) indicates a local minimum, while at \( x = -1 \), \( f''(-1) \) points to a local maximum.