Problem 40 - 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 \(f .\) \(f(x)=\frac{e^{x}}{1-e^{x}}\)

Step-by-Step Solution

Answer

Vertical asymptote: \(x=0\). Horizontal asymptotes: \(y=-1\) and \(y=0\). Increasing on \((-\infty, 0)\), decreasing on \((0, \infty)\). No finite local extrema. Inflection point at \(x=\ln(\frac{1}{2})\).

Step 1: Find Vertical Asymptotes

A vertical asymptote occurs where the denominator of a function is zero but the numerator is not zero. For the function \(f(x)=\frac{e^{x}}{1-e^{x}}\), set the denominator \(1-e^x = 0\). Solving gives \(e^x = 1\) which implies \(x = 0\). Therefore, \(x = 0\) is a vertical asymptote.

Step 2: Find Horizontal Asymptotes

A horizontal asymptote is found by analyzing the behavior of \(f(x)\) as \(x\) approaches infinity or negative infinity. As \(x \to -\infty\), \(e^x \to 0\), so \(f(x) \to 0\). Therefore, \(y = 0\) is a horizontal asymptote. As \(x \to \infty\), \(f(x)\) approaches \(-1\); thus, \(y = -1\) is another horizontal asymptote.

Step 3: Determine Intervals of Increase or Decrease

To find the intervals where the function is increasing or decreasing, compute the first derivative \(f'(x)\). Applying the quotient rule to \(f(x) = \frac{e^{x}}{1-e^{x}}\), we find \(f'(x) = \frac{e^x (1-e^x) - e^x(e^x)}{(1-e^x)^2} = \frac{e^x}{(1-e^x)^2}\). The critical points occur where \(f'(x) = 0\), but since the numerator \(e^x\) is always positive, the function has no critical points. Evaluate the sign of \(f'(x)\): it's positive for \(x < 0\) and negative for \(x > 0\), indicating an increase on \((-\infty, 0)\) and a decrease on \((0, \infty)\).

Step 4: Identify Local Maximum and Minimum Values

Since \(f(x)\) is increasing on \((-\infty, 0)\) and decreasing on \((0, \infty)\), there is a local maximum at \(x = 0\). Calculate \(f(0) = \frac{1}{1-1} = \text{undefined}\) because of the vertical asymptote, so there is no finite local extremum.

Step 5: Determine Intervals of Concavity and Inflection Points

Find the second derivative \(f''(x)\) from \(f'(x)\). Using the quotient rule, \(f''(x) = \frac{e^x(1-e^x)^2 - 2e^x(1-e^2x)}{(1-e^x)^4}\). Simplifying, find \(f''(x) = -\frac{e^x(1+2e^x)}{(1-e^x)^3}\). The sign change in \(f''(x)\) indicates inflection points. The concavity is determined by the sign of \(f''(x)\): \(f(x)\) is concave up where \(f''(x) > 0\) and concave down where \(f''(x) < 0\). Here, \(f(x)\) changes concavity at \(x = ln(\frac{1}{2})\), indicating an inflection point.

Step 6: Use Information to Sketch Graph

Considering the vertical asymptote at \(x = 0\), horizontal asymptotes at \(y = 0\) and \(y = -1\), intervals of increase on \((-\infty, 0)\) and decrease on \((0, \infty)\), and the inflection point at \(x = ln(\frac{1}{2})\), you can sketch the graph. Begin from the left, approaching \(y = -1\) from below, increasing past \(x = ln(\frac{1}{2})\), and sharply decreasing near \(x = 0\). Continue the decrease into \((0, \infty)\), getting closer to \(y = -1\).

Key Concepts

Vertical and Horizontal Asymptotes

When analyzing a function's graph, identifying asymptotes helps predict its end behavior. Vertical asymptotes occur where the function is undefined because the denominator is zero while the numerator remains non-zero. For the function \(f(x)=\frac{e^{x}}{1-e^{x}}\), set the denominator \(1-e^x = 0\). Solving this equation gives \(e^x = 1\), leading to \(x = 0\). Thus, \(x = 0\) is a vertical asymptote.
Horizontal asymptotes describe the behavior of \(f(x)\) as \(x\) approaches infinity or negative infinity. For this function, as \(x \to -\infty\), \(e^x \to 0\), so \(f(x) \to 0\), showing that \(y = 0\) is a horizontal asymptote. As \(x \to \infty\), \(f(x)\) approaches \(-1\), hence \(y = -1\) is also a horizontal asymptote. These lines give insights into how the graph stretches out towards infinity.

Intervals of Increase and Decrease

To determine where a function is increasing or decreasing, you'll need the first derivative, \(f'(x)\). The given function is \(f(x) = \frac{e^{x}}{1-e^{x}}\). Applying the quotient rule, we get
\[f'(x) = \frac{e^x}{(1-e^x)^2}.\]
Since the numerator \(e^x\) is always positive, \(f'(x)\) never equals zero, implying no critical points. Assessing \(f'(x)\) for its sign reveals it is positive when \(x < 0\) and negative for \(x > 0\). Consequently, \(f(x)\) is increasing on \(( -\infty, 0)\) and decreasing on \((0, \infty)\). Recognizing these intervals helps understand the movement of the graph.

• Increasing: \(( -\infty, 0)\)
• Decreasing: \((0, \infty)\)

Local Maximum and Minimum Values

Discovering local maxima and minima involves checking where the function changes direction. Given that \(f(x)\) is increasing before \(x = 0\) and decreasing afterward, we typically expect a local maximum at \(x = 0\). However, as calculated \(f(0)\) is undefined due to the vertical asymptote at \(x = 0\), therefore, no finite local extremum exists. Understanding this behavior ensures clarity in predicting points where the graph cannot have peaks or valleys.

Concavity and Inflection Points

Concavity refers to whether a function appears "bowl-shaped" or "cap-shaped." This feature is determined through the second derivative \(f''(x)\). For \(f(x) = \frac{e^{x}}{1-e^{x}}\), calculating the second derivative gives:
\[f''(x) = -\frac{e^x(1+2e^x)}{(1-e^x)^3}.\]
Evaluating \(f''(x)\) reveals changes in sign, which indicate inflection points, where the graph changes concavity. For \(f(x)\), such a transformation in concavity occurs at \(x = \ln(\frac{1}{2})\).
The graph is concave up where \(f''(x) > 0\) and concave down where \(f''(x) < 0\). Understanding these intervals enhances the sketch precision, revealing some underlying graph behavior unseen by the first derivative alone.

Graph Sketching

Once you analyze a function's various characteristics, you're equipped to sketch its graph accurately. Begin with the vertical asymptote at \(x = 0\) and horizontal asymptotes at \(y = 0\) and \(y = -1\). Take note of the intervals of increase \(( -\infty, 0)\) and decrease \((0, \infty)\). Ensure you mark the inflection point at \(x = \ln(\frac{1}{2})\) and the sign changes in concavity to build a consistent sketch.
The graph approaches the horizontal asymptote \(y = -1\) from below while increasing past \(x = \ln(\frac{1}{2})\) and sharply decreasing as it nears \(x = 0\). It continues to decline into \((0, \infty)\), closely aligning to \(y = -1\). These visual cues provide an insightful reflection of all the properties analyzed, aiding stronger comprehension of the graph's dynamics.