Problem 80 - Calculus Single Variable

Question

A function \(f\) with domain either \(I=(-\infty, \infty)\) or \(I=(0, \infty)\) is given. Sketch the graph of \(f\). (The set \(C\) of critical points of \(f\) and the set \(I\) of inflection points of \(f\) are provided in some cases.) Use l'Hôpital's Rule to determine the horizontal asymptote of the graph. If \(I=(0, \infty),\) use l'Hôpital's Rule to determine \(\lim _{x \rightarrow 0^{+}} f(x)\). \(f(x)=\ln (x) / x ; I=(0, \infty)\)

Step-by-Step Solution

Answer

The graph of \(f(x) = \frac{\ln(x)}{x}\) has a local maximum at \(x = e\), an inflection point at \(x = e^2\), a horizontal asymptote at \(y = 0\), and as \(x \to 0^+\), \(f(x) \to -\infty\).

Step 1: Determine the Derivatives

Calculate the first and second derivatives of the function to find critical and inflection points. The first derivative is \(f'(x) = \frac{1 - \ln(x)}{x^2}\), and the second derivative is \(f''(x) = \frac{\ln(x) - 2}{x^3}\).

Step 2: Find Critical Points

Set the first derivative equal to zero: \(\frac{1 - \ln(x)}{x^2} = 0\). Solving gives \(\ln(x) = 1\), so \(x = e\). Thus, the critical point is \(x = e\).

Step 3: Determine if Critical Points are Maximum or Minimum

Use the second derivative test. \(f''(e) = \frac{1 - 2}{e^3} = -\frac{1}{e^3} < 0\), indicating a local maximum at \(x = e\).

Step 4: Find Inflection Points

Find where the second derivative changes sign: \(\frac{\ln(x) - 2}{x^3} = 0\). Solving gives \(\ln(x) = 2\), so \(x = e^2\). There is an inflection point at \(x = e^2\).

Step 5: Determine Horizontal Asymptote Using L'Hôpital's Rule

Evaluate \(\lim_{x \to \infty} \frac{\ln(x)}{x}\) by L'Hôpital's Rule: differentiate the numerator and denominator, then take the limit as \(x \to \infty\). The limit is \(\lim_{x \to \infty} \frac{1/x}{1} = \lim_{x \to \infty} \frac{1}{x} = 0\). Thus, the horizontal asymptote is \(y = 0\).

Step 6: Determine Limit as \(x \to 0^+\)

Evaluate \(\lim_{x \to 0^+} \frac{\ln(x)}{x}\) by L'Hôpital's Rule. Differentiate the numerator and denominator: \(\lim_{x \to 0^+} \frac{1/x}{1} = \lim_{x \to 0^+} \frac{1}{x} = -\infty\). Thus, as \(x\) approaches 0 from the right, \(f(x)\) approaches \(-\infty\).

Step 7: Sketch the Graph

Using the information from previous steps, sketch the graph. The function has a local maximum at \(x = e\), approaches zero as \(x\) goes to \(+\infty\), and goes to \(-\infty\) as \(x\) approaches 0 from the right. Include the inflection point at \(x = e^2\).

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus used to find limits that initially result in indeterminate forms such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). By differentiating the numerator and the denominator separately, you can simplify and evaluate the limit effectively.
For the function \(f(x) = \frac{\ln(x)}{x}\), both \(\lim_{x \to \infty}\) and \(\lim_{x \to 0^{+}}\) times initially result in indeterminate forms. Applying L'Hôpital's Rule involves:

• Differentiating the numerator to get \(\frac{1}{x}\).
• Differentiating the denominator, which remains 1.

This transforms the limit to \(\lim_{x \to \infty} \frac{1/x}{1} = \lim_{x \to \infty} \frac{1}{x} = 0\) for \(x \to \infty\), indicating a horizontal asymptote at \(y=0\). For \(x \to 0^+\), it becomes \(\lim_{x \to 0^{+}} \frac{1/x}{1} = -\infty\), illustrating the function's behavior towards \(-\infty\). This helps depict how the function behaves at its extremes.

Derivatives

Derivatives are fundamental in calculus used to analyze the rate of change of a function, uncovering its critical and inflection points.
By taking the first derivative of \(f(x) = \frac{\ln(x)}{x}\), we find \(f'(x) = \frac{1 - \ln(x)}{x^2}\). This function provides critical insights:

• Setting \(f'(x) \) to zero helps to find critical points, indicating where the function could potentially have a local maximum or minimum.
• Solving \(\frac{1 - \ln(x)}{x^2} = 0\), we find \(\ln(x) = 1\), leading to \(x = e\), a critical point.

The second derivative, \(f''(x) = \frac{\ln(x) - 2}{x^3}\), determines the convexity of the function:

• If \(f''(x) > 0\), the function is concave up, suggesting a local minimum.
• If \(f''(x) < 0\), the function is concave down, indicating a local maximum.

At \(x = e\), \(f''(e) < 0\), showing a local maximum. Derivatives effectively outline the function's behavior and shape.

Critical Points

Critical points are where the first derivative of a function is zero or undefined. These points are key to determining local maxima and minima.
For the function \(f(x) = \frac{\ln(x)}{x}\), the critical point is found by setting the first derivative, \(f'(x) = \frac{1 - \ln(x)}{x^2}\), to zero.

• Solve \(1 - \ln(x) = 0\) to find \(x = e\).

At \(x = e\), the function changes direction, making it a significant point for examining the graph's local behavior.
Using the second derivative \(f''(x) = \frac{\ln(x) - 2}{x^3}\), we test whether it's a maximum or minimum:

• \(f''(e) = -\frac{1}{e^3} < 0\), indicating a local maximum at \(x = e\).

Understanding this point helps to sketch or interpret the graph of the function more accurately.

Inflection Points

Inflection points are where the concavity of a function changes; they are found by setting the second derivative to zero and solving for \(x\).
For \(f(x) = \frac{\ln(x)}{x}\), the second derivative is \(f''(x) = \frac{\ln(x) - 2}{x^3}\).
By solving \(\frac{\ln(x) - 2}{x^3} = 0\), we obtain:

• \(\ln(x) = 2\).
• Thus, \(x = e^2\).

At \(x = e^2\), \(f''(x)\) changes sign, confirming an inflection point. This point is crucial for sketching the function as it indicates a change in curvature, giving the graph a distinct look.