Problem 73 - Algebra and Trigonometry

Question

Draw the graph of the function in a suitable viewing rec- tangle, and use it to find the domain, the asymptotes, and the local maximum and minimum values. $$ y=\frac{\ln x}{x} $$

Step-by-Step Solution

Answer

Domain: \((0, \infty)\); Asymptote: horizontal y=0; Local max at \(x=e, y=\frac{1}{e}\).

Step 1: Determine the Domain

The function involved is given by \( y = \frac{\ln x}{x} \). To determine the domain, notice that the natural logarithm function \( \ln x \) is defined for \( x > 0 \). Additionally, since \( x \) is in the denominator, \( x \) cannot be zero. Therefore, the domain of \( y \) is all positive real numbers, \( (0, \infty) \).

Step 2: Identify Vertical and Horizontal Asymptotes

In this step, we look for any asymptotes. For vertical asymptotes, none exist because the natural logarithm \( \ln x \) is defined over the entire domain. For horizontal asymptotes, as \( x \to \infty \), \( \ln x \to \infty \), but \( \frac{1}{x} \to 0 \), making \( y = \frac{\ln x}{x} \to 0 \). Therefore, the x-axis \( y = 0 \) is a horizontal asymptote.

Step 3: Find Critical Points for Local Maximum and Minimum Values

To find the critical points, we will take the derivative of \( y = \frac{\ln x}{x} \). Using the quotient rule, the derivative is \( y' = \frac{(1 - \ln x)}{x^2} \). Set \( y' = 0 \) to find the critical points: \( 1 - \ln x = 0 \) implies \( \ln x = 1 \), so \( x = e \). This is our critical point.

Step 4: Determine Local Maxima and Minima

Using the second derivative test or evaluating the sign change of \( y' \, \) it's determined that around the critical point \( x=e \), the sign of \( y' \, \) changes from positive to negative, indicating a local maximum at \( x=e \). Substitute \( x=e \) in the original equation to find \( y(e) = \frac{1}{e} \) as the local maximum value.

Step 5: Construct the Graph

Using the domain \( (0, \infty) \), vertical asymptotes, and the local maximum, sketch a graph. The graph should rise to the point \((e, \frac{1}{e}) \) and then decrease asymptotically toward the x-axis as it moves further right. Ensure the behavior around the local maximum and asymptote are accurately depicted.

Key Concepts

Domain of a Function

Understanding the domain of a function is crucial for graphing and analyzing it correctly. The domain refers to all possible input values (i.e., values of the variable) for which the function is defined.
For the function given, \( y = \frac{\ln x}{x} \), the domain is determined by two conditions:

• The natural logarithm, \( \ln x \), which only accepts positive values, implying \( x > 0 \).
• Since \( x \) is in the denominator, it cannot be zero.

Hence, the domain includes all positive real numbers, expressed as \((0, \infty)\). Remember: always consider restrictions from both the numerator and denominator to find the domain.

Asymptotes

Asymptotes are important features of a graph where the function approaches a line, but never actually touches it. In our case, we look for both vertical and horizontal asymptotes.

• Vertical Asymptotes: There are no vertical asymptotes because \( \ln x \) is defined across the entire domain \((0, \infty)\).
• Horizontal Asymptotes: As \( x \to \infty \), \( \ln x \to \infty \) and \( \frac{1}{x} \to 0 \). This means the function \( y = \frac{\ln x}{x} \) approaches zero, resulting in a horizontal asymptote at \( y = 0 \).

The horizontal asymptote (the x-axis) reflects the curve's behavior at extreme values of \( x \).

Local Maximum and Minimum

Local maxima and minima are points where a function reaches a peak or valley, respectively, within a small neighborhood. These points are crucial for understanding the function's behavior.
To find them, first identify critical points by setting the first derivative of the function to zero.

• For \( y = \frac{\ln x}{x} \), the critical point arises at \( x = e \) since this is where \( y' = \frac{(1 - \ln x)}{x^2} = 0 \).

The critical point at \( x = e \) is a local maximum because the derivative changes from positive to negative at this point. Evaluating the function at \( x=e \), we find the local maximum value \( y(e) = \frac{1}{e} \). No local minima exist in this context.

Derivative and Critical Points

The derivative is a tool that helps find the slope of a function at any point and is essential in identifying critical points for maxima and minima. Using the quotient rule, the derivative of \( y = \frac{\ln x}{x} \) is:
\[ y' = \frac{1 - \ln x}{x^2} \] To find critical points, set \( y' = 0 \):

• Solving \( 1 - \ln x = 0 \) gives \( \ln x = 1 \), hence \( x = e \).

Thus, the only critical point is at \( x = e \). By analyzing the sign change of \( y' \), we deduce that there is a local maximum at this critical point. Remember, critical points can indicate where a function might change its increasing or decreasing behavior.