Problem 54 - University Calculus: Early Transcendentals

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=x e^{-x}$$

Step-by-Step Solution

Answer

Local max at (1, e^{-1}), inflection point at (2, 2e^{-2}).

Step 1: Find the First Derivative

To identify the critical points, we first need to find the derivative of the function. Given the function \( y = x e^{-x} \), use the product rule to find the derivative: \( y' = (1) e^{-x} + x (-e^{-x}) = e^{-x} - x e^{-x} \), which simplifies to \( y' = e^{-x}(1 - x) \).

Step 2: Set the First Derivative to Zero

The critical points occur where the first derivative is zero or undefined. Set \( y' = e^{-x}(1-x) = 0 \). Since \( e^{-x} \) is never zero, we solve \( 1 - x = 0 \), giving the critical point \( x = 1 \).

Step 3: Second Derivative Test for Local Extrema

Use the second derivative test to determine the nature of the critical point. Find the second derivative: \( y'' = \frac{d}{dx} [e^{-x} - x e^{-x}] \). Using the product rule, \( y'' = -e^{-x} - [e^{-x} - x e^{-x}] \) which simplifies after combining like terms to \( y'' = -2e^{-x} + x e^{-x} \). Evaluate \( y''(1) \): \( y''(1) = -2e^{-1} + 1e^{-1} = -e^{-1} \), which is less than zero, indicating a local maximum at \( x=1 \).

Step 4: Identify Inflection Points

An inflection point occurs where the second derivative changes sign. Find where \( y'' = 0 \): \(-2e^{-x} + x e^{-x} = 0 \), which simplifies to \( x = 2 \) after canceling \( e^{-x} \). Check if the second derivative changes sign around \( x = 2 \):- For \( x < 2 \), \( y'' < 0 \)- For \( x > 2 \), \( y'' > 0 \)This change in sign confirms an inflection point at \( x = 2 \).

Step 5: Absolute Extrema and Graphing

Since \( x e^{-x} \) approaches zero as \( x \to \infty \) and is zero at \( x = 0 \), evaluate the function at these points and the local max we found:- \( y(0) = 0 \) (endpoint)- \( y(1) = e^{-1} \) (local maximum)Plot the function with these important points:- Local maximum at \( (1, e^{-1}) \)- Inflection point at \( (2, 2e^{-2}) \)

Key Concepts

First Derivative

The first derivative of a function is a crucial tool in calculus that helps identify critical points, where a function can have local maxima or minima. For a given function, such as \( y = x e^{-x} \), the first derivative provides insights into the behavior of the function. To find it, we use differentiation techniques like the product rule, which is necessary when a function is a product of two simpler functions.
In our example, we have:

• The first term from the derivative of \( x \) is \( e^{-x} \)
• The second term is from the derivative of \( e^{-x} \), combined with \( x \), resulting in \( -x e^{-x} \)

Therefore, the first derivative is \( y' = e^{-x} - x e^{-x} \), which simplifies to \( y' = e^{-x}(1 - x) \).
This form helps in easily finding the critical points by setting the derivative to zero. Notice, since \( e^{-x} \) cannot be zero, we focus on \( 1-x=0 \), which gives \( x=1 \) as a critical point.

Second Derivative

The second derivative of a function provides more in-depth insight into the function's concavity and the nature of any local extrema found. It is the derivative of the first derivative, giving information on how the slope of the original function is changing. For our function \( y = x e^{-x} \), the second derivative tells us about the concavity and helps in determining the presence of inflection points.
To find it, apply the product rule again to \( y' = e^{-x}(1-x) \), yielding:

• The derivative of \( e^{-x} \) is \( -e^{-x} \)
• Applied to \( (1-x) \), you get \(-2e^{-x} + x e^{-x} \)

This simplifies to the second derivative as \( y'' = -2e^{-x} + x e^{-x} \).
The value of the second derivative at critical points helps determine their nature: if \( y''(x) < 0 \) at a critical point, it is a local maximum; if \( y''(x) > 0 \), it is a local minimum.

Local Maximum

A local maximum occurs at a point where a function reaches a peak relative to its immediate surroundings. For the function \( y = x e^{-x} \), determining local maxima involves using both the first and second derivatives.
First, identify the critical points by setting the first derivative \( y' = 0 \), giving \( x = 1 \). Then, apply the second derivative test to classify this point:

• Calculate \( y''(1) = -e^{-1} \)
• Since \( y''(1) < 0 \), it confirms a local maximum at \( x = 1 \)

Thus, the function has a local maximum at \((1, e^{-1})\). This is where the graph of the function peaks before descending again, a key concept in understanding the function's behavior.

Inflection Point

An inflection point is where a function changes its concavity. This change signifies a shift from a curve that is concave up (bowl-shaped) to concave down (hill-shaped), or vice versa. For our function, we find the inflection points by looking where the second derivative changes its sign.
Solve \( y'' = -2e^{-x} + x e^{-x} = 0 \), simplifying it to \( x = 2 \), helps us locate the inflection point.

• Check around \( x=2 \) to see how the sign of \( y'' \) changes.
• For \( x < 2 \), \( y'' < 0 \)
• For \( x > 2 \), \( y'' > 0 \)

This change in sign confirms the inflection point at \( x = 2 \). At this point, the graph transitions from concavity downwards to upwards, marking a significant point where the behavior of the function changes.