Problem 37 - 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 \(f .\) \(f(x)=1+\frac{1}{x}-\frac{1}{x^{2}}\)

Step-by-Step Solution

Answer

Vertical asymptote: \(x=0\); horizontal asymptote: \(y=1\); local max at \(x=2, f(2)=\frac{5}{4}\); inflection at \(x=3\).

Step 1: Finding Vertical Asymptotes

To find the vertical asymptotes, we look for the values of \(x\) that make the denominator zero. The function is \(f(x) = 1 + \frac{1}{x} - \frac{1}{x^2}\). The denominator is \(x^2\), so setting it to zero gives \(x^2 = 0\), which means \(x = 0\). Therefore, there is a vertical asymptote at \(x = 0\).

Step 2: Finding Horizontal Asymptotes

The horizontal asymptote is found by examining the behavior of \(f(x)\) as \(x\) approaches infinity. As \(x\to \infty\), the terms \(\frac{1}{x}\) and \(\frac{1}{x^2}\) approach 0, so \(f(x)\) approaches 1. Thus, the horizontal asymptote is \(y = 1\).

Step 3: Finding Intervals of Increase or Decrease

First, find the derivative \(f'(x) = -\frac{1}{x^2} + \frac{2}{x^3}\), and set it equal to zero to find critical points: \(-\frac{1}{x^2} + \frac{2}{x^3} = 0\) simplifies to \(x = 2\). To determine intervals of increase or decrease, test intervals around \(x = 2\) and \(x = 0\) (vertical asymptote). Calculate \(f'(x)\) for \(x < 0\), \(0 < x < 2\), and \(x > 2\). We find that \(f(x)\) is decreasing for \(x < 0\), increasing on \(0 < x < 2\), and decreasing for \(x > 2\).

Step 4: Finding Local Maximum and Minimum Values

Evaluate \(f'(x)\) around the critical point \(x = 2\). Since \(f'(x)\) changes from positive to negative at \(x = 2\), \(f(x)\) has a local maximum at \(x = 2\). Calculate \(f(2) = 1 + \frac{1}{2} - \frac{1}{4} = \frac{5}{4}\), so the local maximum value is \(\frac{5}{4}\).

Step 5: Finding Intervals of Concavity and Inflection Points

Calculate the second derivative \(f''(x) = \frac{2}{x^3} - \frac{6}{x^4}\). Set \(f''(x) = 0\) to find possible inflection points: \(\frac{2}{x^3} - \frac{6}{x^4} = 0\) simplifies to \(x = 3\). Test intervals around \(x = 3\) and \(x = 0\). The function is concave up for \(x < 0\) and \(x > 3\), and concave down on \(0 < x < 3\). Thus, there's an inflection point at \(x = 3\).

Step 6: Graphing with Found Information

Using the asymptotes, intervals of increase/decrease, and concavity, sketch the graph of \(f(x)\). The graph approaches \(y = 1\) as \(x\) increases or decreases without bound and has a vertical asymptote at \(x = 0\). The local maximum at \(x = 2\) and inflection point at \(x = 3\) guide the curves of the graph.

Key Concepts

Vertical Asymptote

When graphing rational functions, vertical asymptotes are lines where the function takes on extremely large positive or negative values, often heading towards infinity. This typically happens when the denominator of a function becomes zero.
In the function \( f(x) = 1 + \frac{1}{x} - \frac{1}{x^2} \), the denominator, \( x^2 \), is zero when \( x = 0 \).
This indicates that there is a vertical asymptote at \( x = 0 \). As \( x \to 0 \), the function's value will grow without bounds, making the understanding of asymptotes crucial for sketching the behavior of the graph near this region.

Horizontal Asymptote

Horizontal asymptotes denote the behavior of a function as the independent variable, \( x \), tends towards large positive or negative values. Unlike vertical asymptotes, they don't depend on the points where a function is undefined, but rather on its 'end behavior.'
For \( f(x) = 1 + \frac{1}{x} - \frac{1}{x^2} \), as \( x \to \infty \) or \( x \to -\infty \), the fractions \( \frac{1}{x} \) and \( \frac{1}{x^2} \) tend towards zero.
Thus, the function approaches the constant value of 1, establishing \( y = 1 \) as a horizontal asymptote. Remember, horizontal asymptotes illustrate where the function might settle as the influence of compatible terms vanishes.

Intervals of Increase or Decrease

Identifying intervals of increase or decrease helps map the function’s movement upward or downward across different sections of the graph.
To discover these intervals for \( f(x) = 1 + \frac{1}{x} - \frac{1}{x^2} \), we find the derivative, which is \( f'(x) = -\frac{1}{x^2} + \frac{2}{x^3} \).
Solving \( f'(x) = 0 \) reveals critical points. For this function, \( x = 2 \) is key. Testing intervals around \( x = 2 \) and the vertical asymptote \( x = 0 \), our function increases between \( 0 < x < 2 \) and decreases on intervals \( x < 0 \) and \( x > 2 \). This pattern aids in pinpointing where \( f(x) \) gains or sheds value rapidly or gently, which is crucial in graph interpretation.

Concavity and Inflection Points

Concavity and inflection points inform us how the graph "bends." This is determined using the second derivative \( f''(x) = \frac{2}{x^3} - \frac{6}{x^4} \).
Setting \( f''(x) = 0 \) helps find potential inflection points where concavity changes. Here, \( x = 3 \) serves as a point of interest after simplification.
By examining values around this \( x \) and near the asymptote at \( x = 0 \), we see the concavity shifts:

• Concave up when \( x < 0 \) or \( x > 3 \).
• Concave down between \( 0 < x < 3 \).

Inflection points, like \( x = 3 \), mark transitions in this curvature, which is useful for drawing the graph's general shape.