Problem 50 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Local minimum at \( x = 1 \); no absolute min/max or inflection points.

Step 1: Find the Derivative

To identify any extreme points, we need to find the derivative of the function. Given the function \( y = \frac{e^x}{x} \), apply the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \), where \( u = e^x \) and \( v = x \). This gives us: \[ y' = \frac{e^x \cdot x - e^x \cdot 1}{x^2} = \frac{e^x (x - 1)}{x^2}. \]

Step 2: Solve for Critical Points

Critical points occur where the derivative is zero or undefined. Set \( y' = 0 \): \[ \frac{e^x (x - 1)}{x^2} = 0. \] The exponential \( e^x \) is never zero, so set \( x - 1 = 0 \). Thus, \( x = 1 \) is a critical point. The derivative is undefined at \( x = 0 \).

Step 3: Second Derivative Test for Concavity and Inflection Points

Find the second derivative to determine concavity and potential inflection points. Differentiate \( y' \): \[ y'' = \frac{d}{dx}\left(\frac{e^x (x - 1)}{x^2}\right). \] Applying the quotient and product rules, we eventually get \( y'' = \frac{e^x (x^2 - 2x + 2)}{x^3}. \) Set \( y'' = 0 \): \( x^2 - 2x + 2 = 0 \). The discriminant \( b^2 - 4ac = 4 - 8 = -4 \), so no real solutions exist, indicating no inflection points.

Step 4: Analyze Behavior and Extreme Points

Analyze \( y' \) around the critical point \( x = 1 \). Test intervals like \( (0, 1) \) and \( (1, \infty) \). For \( 0 < x < 1 \), \( y' < 0 \). For \( x > 1 \), \( y' > 0 \). Thus, \( x = 1 \) is a local minimum of \( y \). The function tends towards infinity as \( x \to \infty \) and negative infinity as \( x \downarrow 0 \), thus there is no absolute minimum or maximum.

Step 5: Graphing the Function

The function has a vertical asymptote at \( x = 0 \) and a local minimum at \( (1, e) \). The function is decreasing on \( (0, 1) \) and increasing on \( (1, \infty) \). These behavior insights can be used to sketch the graph showing the shape and key features.

Key Concepts

Derivative

Derivatives are a fundamental concept in calculus, acting as the backbone for understanding the behavior of functions. Computing the derivative of a function allows us to determine how the function's output changes with respect to its input. For the function \( y = \frac{e^x}{x} \), we need to apply the quotient rule to find its derivative. The quotient rule is particularly handy for functions that are ratios of two other functions. It is given by: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \), where \( u = e^x \) and \( v = x \). By following this rule, we obtain the derivative:

• \( y' = \frac{e^x (x - 1)}{x^2} \)

This equation gives us valuable information about the rate of change of \( y \) with respect to \( x \). Understanding this rate of change helps in identifying critical points and analyzing the behavior of the function.

Critical Points

Critical points are locations on the graph of a function where its derivative is generally zero or undefined. These points are important since they can often indicate local maxima, minima, or points of inflection. For the function \( y = \frac{e^x}{x} \), by setting \( y' = 0 \), we can find if there are any critical points:

• \( \frac{e^x (x - 1)}{x^2} = 0 \)

Since \( e^x \) cannot be zero, we solve for \( x - 1 = 0 \), giving us a critical point at \( x = 1 \). Additionally, the derivative is undefined at \( x = 0 \), indicating a potential for non-regular behavior at this point. These critical points show where significant changes occur in the function's behavior.

Concavity

Concavity reveals whether a function curves upwards or downwards at a particular point, giving us more details about its graphical shape. To determine concavity, the second derivative test is used. For the function \( y = \frac{e^x}{x} \), we found:

• \( y'' = \frac{e^x (x^2 - 2x + 2)}{x^3} \)

Solving \( y'' = 0 \) helps us find potential inflection points. However, for our function, \( x^2 - 2x + 2 = 0 \) has no real solutions. Thus, there aren't any points of inflection. This means concavity does not change across the domain, simplifying the analysis of the function's curve.

Graphing Functions

Graphing functions is a powerful way to visualize their behavior and understand different characteristics, such as critical points and asymptotes. For the function \( y = \frac{e^x}{x} \):

• There is a vertical asymptote at \( x = 0 \) because the function is undefined there.
• The function has a local minimum at the point \( (1, e) \).
• The function decreases for \( x \) values within \((0, 1)\) and increases for \( x \) values greater than 1.

By plotting these characteristics, one can create a sketch of the function's graph. Observing these features can help us predict the function's behavior at different intervals and clearly see any potential extreme values or asymptotic behavior.