Problem 43 - Essential Calculus Early Transcendentals

Question

(a) Find the vertical and horizontal asymptotes. (b) Find the intervals of increase or decrease. (c) Find the local maximum and minimum values. (d) Find the intervals of concavity and the inflection points. (e) Use the information from parts ( d ) to sketch the graph of \(f .\) \(f(x)=\ln (1-\ln x)\)

Step-by-Step Solution

Answer

The function has a vertical asymptote at \(x=0\), decreases on \((0,e)\), with an inflection point at \(x=1\). No horizontal asymptotes or extremums exist.

Step 1: Find the Domain

The function \( f(x) = \ln(1-\ln x) \) is defined where both \( \ln x \) and \( 1-\ln x \) are defined. \( \ln x \) is defined for \( x > 0 \), and \( 1-\ln x > 0 \) implies \( \ln x < 1 \). Therefore, the domain is \( 0 < x < e \).

Step 2: Find Vertical Asymptotes

Vertical asymptotes occur where the function is undefined or tends to infinity. In the domain \( 0 < x < e \), the function becomes undefined as \( x \to 0^+ \). Thus, \( x = 0 \) is a vertical asymptote.

Step 3: Find Horizontal Asymptotes

To find horizontal asymptotes, evaluate the limit as \( x \to e^- \). This makes \( \ln x \to 1^- \), causing \( 1-\ln x \to 0^+ \) and therefore \( \ln(1-\ln x) \to -\infty \). There are no horizontal asymptotes.

Step 4: First Derivative Test for Increase/Decrease

Find the derivative: \( f'(x) = \frac{-1}{(1-\ln x)x} \). Set \( f'(x) = 0 \) does not yield any real solution, but we analyze sign changes: \( f'(x) < 0 \) for all \( x \) in the domain. Thus, the function is decreasing on \( (0, e) \).

Step 5: Find Local Maxima and Minima

Since \( f'(x) < 0 \) throughout \( (0, e) \), the function always decreases and has no local maxima or minima.

Step 6: Second Derivative Test for Concavity

Find the second derivative: \( f''(x) = \frac{1 + \ln x}{(1-\ln x)^2 x^2} \). The second derivative changes sign at \( x = 1 \), indicating this is an inflection point. \( f''(x) > 0 \) when \( 0 < x < 1 \) (concave up) and \( f''(x) < 0 \) when \( 1 < x < e \) (concave down).

Step 7: Sketch the Graph Using Found Information

Plot the vertical asymptote at \( x=0 \), mark intervals decreasing (entire domain) and concavity changes (inflection at \( x=1 \)). Because the function decreases and changes concavity, it transitions from concave up to concave down without extremums in the interval.

Key Concepts

Asymptotes

In calculus, asymptotes are lines that a graph of a function approaches but never actually reaches. These lines can be vertical, horizontal, or even slanted. For the function \( f(x) = \ln(1-\ln x) \), the step-by-step solution highlights the presence of a vertical asymptote at \( x = 0 \). This occurs because as \( x \) approaches zero, the natural logarithm \( \ln x \) approaches negative infinity, making the function \( \ln(1-\ln x) \) undefined at 0. It is crucial to remember that vertical asymptotes indicate points where the function does not exist due to division by zero or other undefined behavior, leading the graph to shoot upwards or downwards infinitely in those regions.

Horizontal asymptotes describe the behavior of a function as \( x \) approaches infinity or another limit. Here, the solution shows there are no horizontal asymptotes for this function as \( x \to e^- \), the value of \( f(x) \) approaches \(-\infty\) significantly, indicating the absence of stabilization at a finite height.

Increase/Decrease Intervals

Identifying intervals of increase or decrease involves examining the first derivative of the function, which helps determine where the function is rising or falling. For \( f(x) = \ln(1-\ln x) \), the derivative is \( f'(x) = \frac{-1}{(1-\ln x)x} \). While setting \( f'(x) = 0 \) in this case provides no real solutions, the sign of the first derivative reveals the function's behavior.

Analyzing the problem, we find that \( f'(x) < 0 \) for all values in the domain \( (0, e) \), indicating that the function is continuously decreasing across this range. This continuous decrease suggests the absence of turning points, meaning the function also lacks local maxima or minima. Such information is fundamental in understanding the function's overall shape and behavior.

Concavity and Inflection Points

Concavity refers to the direction in which a curve bends. It is determined by the second derivative \( f''(x) \). To find where the function is concave up (bends upwards) or concave down (bends downwards), we look at the sign of \( f''(x) \). For \( f(x) = \ln(1-\ln x) \), the second derivative \( f''(x) = \frac{1 + \ln x}{(1-\ln x)^2 x^2} \) changes sign around \( x = 1 \).

This change of sign indicates an inflection point at \( x = 1 \), characterized by the curve changing from concave up on the interval \( 0 < x < 1 \) to concave down as \( x \) progresses from \( 1 < x < e \). Inflection points highlight where the curve of the function switches concavity, providing essential insight into the graph's geometric properties.

Graph Sketching

Graph sketching involves using calculated properties to visualize a function's behavior graphically. For \( f(x) = \ln(1-\ln x) \), you follow these steps using the information gathered:

• Draw the vertical asymptote at \( x=0 \), where the function becomes undefined.
• Mark intervals of decrease across the domain \( (0, e) \) due to the negative first derivative.
• Identify the inflection point at \( x = 1 \), where the character of the curve changes.
• Illustrate the function being concave up for \( 0 < x < 1 \).
• Depict a concave down shape for \( 1 < x < e \).

These aspects aid in constructing an intuitive picture of how the function behaves, offering insights into its progression and tendencies. Understanding these core elements ensures a thorough comprehension of the function's influence and helps immensely in calculus studies.