Problem 49 - University Calculus: Early Transcendentals

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=x e^{1 / x}$$

Step-by-Step Solution

Answer

Local minimum at \( (1, e) \), inflection point at \( (2, 2e^{1/2}) \).

Step 1: Find the First Derivative

To locate the extrema of the function, begin by computing the first derivative of the function. The function is given as \( y = x e^{1/x} \). Applying the product rule, we have:\( y' = e^{1/x} + x(-1/x^2) e^{1/x} = e^{1/x} - \frac{1}{x} e^{1/x} = e^{1/x} \left( 1 - \frac{1}{x} \right) \).

Step 2: Solve for Critical Points

Set the derivative found in Step 1 equal to zero to find critical points:\( e^{1/x} \left( 1 - \frac{1}{x} \right) = 0 \).Since \( e^{1/x} \) is never zero, focus on the other factor: \( 1 - \frac{1}{x} = 0 \).Solve for \( x \):\( 1 = \frac{1}{x} \), which gives \( x = 1 \).

Step 3: Identify Inflection Points using the Second Derivative

Find the second derivative to locate inflection points. The first derivative is \( y' = e^{1/x} \left( 1 - \frac{1}{x} \right) \). Differentiate it again using the product rule:\( y'' = -\frac{1}{x^2} e^{1/x} \left( 1 - \frac{1}{x} \right) + e^{1/x} \frac{1}{x^3} \).Simplifying gives:\( y'' = e^{1/x} \left( \frac{2}{x^3} - \frac{1}{x^2} \right) \).Set \( y'' = 0 \) to find inflection points:\( \frac{2}{x^3} - \frac{1}{x^2} = 0 \) which simplifies to \( 2 = x \), so \( x = 2 \).

Step 4: Determine Local and Absolute Extrema

Using the critical value \( x = 1 \) found in Step 2, evaluate the second derivative to use the second derivative test:\( y''(1) = e^{1/1} \left( \frac{2}{1^3} - \frac{1}{1^2} \right) = e \times 1 = e > 0 \).This means \( x = 1 \) is a local minimum.

Step 5: Evaluate Function at Key Points

Evaluate the function at \( x = 1 \) (local minimum) and \( x = 2 \) (potential inflection point):- At \( x = 1 \), \( y(1) = 1 \cdot e^{1/1} = e \).- At \( x = 2 \), \( y(2) = 2 \cdot e^{1/2} \), which is a calculation value and serves to compare points.

Step 6: Graph the Function

Plot the function \( y = x e^{1/x} \). Mark key points: local minimum at \( (1, e) \) and inflection point at \( (2, 2 \cdot e^{1/2}) \). The graph should show a local minima at \( x = 1 \) and a change in concavity at \( x = 2 \).

Key Concepts

Critical Points

To find critical points for a function, we need to determine where the derivative of the function is zero or undefined. These points are where the function’s slope shifts from positive to negative or vice versa, which can indicate local maxima, minima, or saddle points.
In the given function, \( y = x e^{1/x} \), we calculated the first derivative as \( y' = e^{1/x} \left( 1 - \frac{1}{x} \right) \). To find where this derivative equals zero, we set the expression inside the parentheses to zero, resulting in \( 1 - \frac{1}{x} = 0 \).
Solving, we find \( x = 1 \). This is a critical point because it’s where the slope changes. Understanding these points is crucial as they provide vital information on the behavior of a graph.

Inflection Points

Inflection points are locations on a graph where the function changes its concavity, meaning it switches from being concave up to concave down, or vice versa. This is found by analyzing the second derivative of the function.
In our exercise, the second derivative is computed as \( y'' = e^{1/x} \left( \frac{2}{x^3} - \frac{1}{x^2} \right) \). To locate inflection points, we solve \( y'' = 0 \), leading to the equation \( \frac{2}{x^3} - \frac{1}{x^2} = 0 \). Solving yields \( x = 2 \).
At \( x = 2 \), the function changes concavity, indicating an inflection point. These points are instrumental in understanding the shape and behavior of the function’s graph.

Graphing Functions

Graphing functions involves plotting points that reveal the function's behavior, such as critical points and inflection points. For the function \( y = x e^{1/x} \), we analyzed both kinds of points to construct the graph.
The critical point at \( x = 1 \) is a local minimum, as evidenced by \( y''(1) \) being positive. At this point, the function \( y(1) = e \). The inflection point at \( x = 2 \) marks where the function changes its concavity, not necessarily an extremum, but crucial for shaping the graph.
These insights help draw a more accurate graph, showing how the function behaves over its domain. By marking these points on the graph, you're able to predict the function's tendencies.

Derivatives

Derivatives provide information about the rate of change of a function. The first derivative indicates the slope of the tangent line to the function at a given point, which helps in identifying critical points.
The application of product or chain rules is often needed, as seen in \( y = x e^{1/x} \), leading to \( y' = e^{1/x} \left( 1 - \frac{1}{x} \right) \). This tells us where the slope is zero or non-existent, indicating a change in the graph's direction.
The second derivative, \( y'' \), informs us about the function's concavity and helps find inflection points. A positive second derivative means the function is concave up, while a negative one implies concave down. Thus, derivatives are essential tools in calculus for analyzing and sketching graphs.