Problem 22 - Calculus Early Transcendentals

Question

Sketch a graph of the rational function and label the coordinates of the stationary points and inflection points. Show the horizontal, vertical, oblique, and curvilinear asymptotes and label them with their equations. Label point(s), if any, where the graph crosses an asymptote. Check your work with a graphing utility. $$ x-\frac{1}{x}-\frac{1}{x^{2}} $$

Step-by-Step Solution

Answer

Graph the function, labeling its vertical asymptote at \( x = 0 \), oblique asymptote at \( y = x \), and stationary/inflection points.

Step 1: Identify the Function Components

We begin by analyzing the function \( y = x - \frac{1}{x} - \frac{1}{x^2} \). The terms of the function suggest the presence of vertical and horizontal asymptotes due to the rational components \( \frac{1}{x} \) and \( \frac{1}{x^2} \).

Step 2: Determine Asymptotes

**Vertical Asymptotes:**The vertical asymptotes occur where the denominator is zero. Both \( \frac{1}{x} \) and \( \frac{1}{x^2} \) suggest that their denominators become zero at \( x = 0 \). Thus, there is a vertical asymptote at \( x = 0 \).**Horizontal Asymptote:**As \( x \to \infty \) or \( x \to -\infty \), the \( -\frac{1}{x} \) and \( -\frac{1}{x^2} \) terms approach zero, meaning the function approaches \( y = x \). However, referring strictly to horizontal behavior at infinity neglects the linear term \( x \) itself, suggesting no horizontal asymptote in the traditional sense, but an oblique (slant) asymptote at \( y = x \).

Step 3: Calculate Stationary Points

To find stationary points, differentiate the function and set the derivative equal to zero:\[ y' = 1 + \frac{1}{x^2} + \frac{2}{x^3}\]Set \( y' = 0 \):\[ 1 + \frac{1}{x^2} + \frac{2}{x^3} = 0\]Multiply through by \( x^3 \) to clear fractions:\[ x^3 + x + 2 = 0\]Solve this cubic equation to find the stationary points. The solutions give us the x-coordinates of the stationary points.

Step 4: Determine Inflection Points

Inflection points occur where the second derivative changes sign. Determine the second derivative:\[ y'' = -\frac{2}{x^3} - \frac{6}{x^4}\]Set \( y'' = 0 \) and solve for \( x \):\[ -\frac{2}{x^3} - \frac{6}{x^4} = 0\]Solve \( -2x - 6 = 0 \) after multiplying by \( x^4 \) for simplicity:\[ x = -3\]Check the sign change around \( x = -3 \) to confirm an inflection point at this coordinate.

Step 5: Plot the Graph and Verify

Using the calculated stationary and inflection points, along with asymptotes, sketch the graph. The key elements to label are:- Vertical asymptote at \( x = 0 \).- Oblique asymptote at \( y = x \).- Stationary point solution coordinates from the derivative.- Inflection point at \( x = -3 \).Use a graphing utility to confirm the accuracy of the graph.

Key Concepts

Asymptotes

In rational functions, asymptotes are lines that the graph approaches but never touches or intersects, except possibly at specific points. Understanding asymptotes helps in sketching the general shape of the function. - **Vertical Asymptotes** occur where the function is undefined, primarily due to division by zero. For the given function \( y = x - \frac{1}{x} - \frac{1}{x^2} \), the vertical asymptote is found at \( x = 0 \) because both \( \frac{1}{x} \) and \( \frac{1}{x^2} \) become undefined when \( x = 0 \). This creates a boundary that the graph approaches but never crosses at this particular location.- **Horizontal Asymptotes** typically occur when \( x \) approaches infinity or negative infinity. However, in our function's case, the presence of \( x \) suggests an **oblique asymptote** instead. As \( x \to \infty \) or \( x \to -\infty \), the function approaches \( y = x \), forming a slant asymptote along this line. This indicates the graph's general direction but not a horizontal limit.- Sometimes, asymptotes can be crossed. Familiarize yourself with any exceptions by analyzing the function's behavior around known coordinates.

Stationary Points

Stationary points on a graph are where the slope of the tangent is zero. They indicate where the function changes direction, meaning the graph has a peak, trough, or point of inflection. These points are vital in understanding the shape and turning points of curves. To find the stationary points for the function \( y = x - \frac{1}{x} - \frac{1}{x^2} \), we first find the derivative:\[ y' = 1 + \frac{1}{x^2} + \frac{2}{x^3} \]Setting \( y' = 0 \) gives us:\[ 1 + \frac{1}{x^2} + \frac{2}{x^3} = 0 \]By clearing the fractions through multiplication by \( x^3 \), we derive:\[ x^3 + x + 2 = 0 \]Solving this cubic equation will give the \( x \)-coordinates of stationary points. Each solution represents where the graph shifts from increasing to decreasing and vice versa, offering insights into potential local maxima or minima.

Inflection Points

Inflection points occur where the function's concavity changes, indicating transitions between concave up (shaped like a cup) and concave down (shaped like a cap). These points are crucial for understanding more subtle changes in the graph's shape. For the given function \( y = x - \frac{1}{x} - \frac{1}{x^2} \), we find the second derivative:\[ y'' = -\frac{2}{x^3} - \frac{6}{x^4} \]Setting \( y'' = 0 \):\[ -\frac{2}{x^3} - \frac{6}{x^4} = 0 \]Simplifying by multiplying through by \( x^4 \), we solve:\[ -2x - 6 = 0 \]This simplifies to \( x = -3 \), indicating a potential inflection point. To confirm, check if the second derivative changes sign around \( x = -3 \). It tells us where the graph changes its curvature - from bending upwards to downwards or vice versa. Recognizing inflection points enriches your understanding of the continuity and flow of the graph.