Problem 82 - 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 \(\lim _{x \rightarrow 0^{+}} f(x)\). \(f(x)=\ln (x) /\left(1 / x+e^{x-1}\right) ; \quad I=(0, \infty) ; \quad C=\\{0.730 \ldots\\} ; I=\) \(\\{1.646 \ldots\\}\)

Step-by-Step Solution

Answer

The horizontal asymptote for the function is \(y=0\). As \(x\to 0^{+}\), \(f(x)\to 0\). Sketch the graph accordingly based on critical points and asymptotes.

Step 1: Determine the Horizontal Asymptote

Evaluate the limit of the function as \(x\) approaches infinity to determine the horizontal asymptote using l'Hôpital's Rule. We have:\[\lim_{x \to \infty} \frac{\ln(x)}{\frac{1}{x} + e^{x-1}}\]This results in an indeterminate form of \(\frac{\infty}{\infty}\). By applying l'Hôpital's Rule, differentiate the numerator and the denominator:\[\lim_{x \to \infty} \frac{\frac{d}{dx}[\ln(x)]}{\frac{d}{dx}[\frac{1}{x} + e^{x-1}]} = \lim_{x \to \infty} \frac{\frac{1}{x}}{-\frac{1}{x^2} + e^{x-1}} = \lim_{x \to \infty} \frac{1/x}{e^{x-1} - 1/x^2}\]As \(x\) approaches infinity, \(e^{x-1}\) grows much faster than \(1/x^2\) and hence dominates:\[\lim_{x \to \infty} \frac{1/x}{e^{x-1}} = 0.\] Thus, the horizontal asymptote is \(y=0\).

Step 2: Evaluate Limit as x approaches 0 from the Right

Determine the behavior of the function as \(x\) approaches 0 from the right. The limit is given by:\[\lim_{x \to 0^{+}} \frac{\ln(x)}{\frac{1}{x} + e^{x-1}}\]As \(x \to 0^+\), \(\ln(x) \to -\infty\) and \(\frac{1}{x} \to \infty\), resulting in the indeterminate form \(\frac{-\infty}{\infty}\). Applying l'Hôpital's Rule, we differentiate the numerator and the denominator:\[\lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2} + e^{x-1}} = \lim_{x \to 0^+} \frac{x}{1 - x^2 e^{x-1}}\]This expression simplifies further, since as \(x \to 0^+\), \(- x^2 e^{x-1}\) approaches zero. Thus, the limit approaches \(0\).

Step 3: Sketch the Graph

Using the critical points and inflection points given, along with the asymptotic behavior, sketch the graph:- At \(C = \{0.730\ldots\}\), there is a critical point, indicating that \(f(x)\) has a stationary (peak or trough) point.- At \(I = \{1.646\ldots\}\), there is an inflection point where the concavity of the function changes.- The horizontal asymptote as \(x\) approaches infinity is \(y=0\), and the limit as \(x\) approaches 0 from the right is also \(0\).- Consider these points and behaviors to sketch a curve that generally increases, reaches the critical point at \(0.730\ldots\), changes concavity at \(1.646\ldots\), and decays towards \(y=0\) as \(x\) approaches infinity.

Key Concepts

Graph of a function

Understanding the graph of a function can help visualize its behavior over different intervals. When you graph a function like \( f(x) = \ln(x) / \left( \frac{1}{x} + e^{x-1} \right) \), you get to see the interplay between key components such as critical points, inflection points, and asymptotic behavior.

When sketching the graph:

• Start by identifying intercepts and asymptotes. These will form the framework of your sketch.
• Consider the critical and inflection points, as they indicate where the function's slope changes or its concavity shifts.
• Notice how the function reacts when approaching infinity or zero from the positive or negative sides.

In our case, the function \( f(x) \) has a critical point at approximately 0.730 and an inflection point at 1.646. Additionally, it's important to consider asymptotic behavior as it helps us understand long-term behaviors.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus for finding limits that result in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).

It works by taking the derivative of the numerator and the denominator separately, which usually simplifies the function enough to evaluate the limit.

Here’s how you use it:

• Identify indeterminate forms: Confirm that as \( x \to a \), both the numerator and denominator approach zero or infinity.
• Differentiate: Take the derivative of the top and bottom functions separately.
• Evaluate the new limit: Substitute the value where \( x \to a \). If still indeterminate, apply L'Hôpital’s Rule again.

In our exercise, applying L'Hôpital’s Rule helped find the limits for \( x \to 0^+ \) and \( x \to \infty \), clarifying the function’s horizontal asymptote and behavior at the origin.

Critical points

Critical points of a function are where the function's derivative is zero or undefined, marking potential peaks, troughs, or plateau points.

These points are crucial as they help indicate where the function changes direction, representing maximum, minimum, or other interesting behavior.

To find critical points you:

• Differentiate the function: Calculate \( f'(x) \).
• Set the derivative equal to zero: Solve \( f'(x) = 0 \).
• Check endpoints or undefined areas: Consider these for additional critical points.

In our example, the critical point at \( x = 0.730 \ldots \) indicates where the function might have a turning point, be it a maximum, minimum, or something else. Analyzing these points helps in the overall sketch of the function to determine its general behavior and specific features.

Inflection points

Inflection points occur where a function changes concavity; that is, where it transitions from a concave up to a concave down or vice versa.

They are found by evaluating the second derivative:

• Differentiate twice: Find \( f''(x) \).
• Set the second derivative to zero: Solve \( f''(x) = 0 \) to find possible inflection points.
• Test intervals around these points: Confirm a change in concavity by evaluating signs of \( f''(x) \) on either side.

In our function, there's an inflection point at \( x = 1.646 \ldots \), implying a switch in how the function curves on the graph. By identifying inflection points, you gain insight into the nature of the function's slope and its increasing or decreasing acceleration.

Horizontal asymptote

A horizontal asymptote shows the value that a function approaches as \( x \) tends toward positive or negative infinity.

These asymptotes offer insight into the long-term behavior of functions:

• Approach infinity: Evaluate the function as \( x \to \infty \).
• Consider limits at extreme values: Sometimes expressed using L'Hôpital’s Rule to handle indeterminate forms.
• Look for constants: If the function tends to a constant number as \( x \to \infty \), that’s the horizontal asymptote.

In this example, through L'Hôpital’s Rule, we determined the horizontal asymptote is \( y = 0 \). As \( x \to \infty \), the function reaches a steady behavior represented by this horizontal line, providing a comprehensive understanding of its ultimate direction.