Chapter 2: Problem 97
Use analytical methods to identify all the asymptotes of \(f(x)=\frac{\ln x^{6}}{\ln x^{3}-1} .\) Plot a graph of the function with a graphing utility and then sketch a graph by hand, correcting any errors in the computer- generated graph.
Short Answer
Expert verified
The function \(f(x) = \frac{\ln x^6}{\ln x^3 - 1}\) has a vertical asymptote at \(x=e^{\frac{1}{3}}\) and a horizontal asymptote at \(y=2\).
Step by step solution
01
Simplify the function
Simplify the function \(f(x) = \frac{\ln x^6}{\ln x^3 - 1}\) using the property of logarithms \(\ln a^b = b\ln a\). The function now becomes:
$$
f(x) = \frac{6\ln x}{3\ln x - 1}.
$$
02
Find the vertical asymptotes
The denominator of the function is \(3\ln x - 1\). Set the denominator equal to zero and solve for \(x\):
$$
3\ln x - 1 = 0 \Rightarrow \ln x = \frac{1}{3} \Rightarrow x = e^{\frac{1}{3}}.
$$
Thus, there is a vertical asymptote at \(x = e^{\frac{1}{3}}\).
03
Find the horizontal asymptote
To find the horizontal asymptote, investigate the end behavior of the function as \(x\) approaches infinity and negative infinity:
$$
\lim_{x\to \infty} \frac{6\ln x}{3\ln x - 1} \text{ and } \lim_{x\to -\infty} \frac{6\ln x}{3\ln x - 1}.
$$
Since the natural logarithm only exists for positive \(x\) values, we don't need to evaluate the limit when \(x\) goes to negative infinity. As for the other limit, the powers of the logarithm are equal, so we can apply L'Hopital's rule:
$$
\lim_{x\to \infty} \frac{6\ln x}{3\ln x - 1} = \lim_{x\to \infty} \frac{6\cdot\frac{1}{x}}{3\cdot\frac{1}{x}} = \lim_{x\to \infty} \frac{6}{3} = 2.
$$
Hence, there is a horizontal asymptote at \(y=2\).
04
Plot the function
Use a graphing utility to plot the function \(f(x) = \frac{6\ln x}{3\ln x - 1}\). Observe the graph and verify that the found asymptotes are correct.
05
Sketch the graph by hand and correct errors
Sketch the graph of \(f(x) = \frac{6\ln x}{3\ln x - 1}\) by hand, including the asymptotes found in the previous steps. Compare the hand-drawn graph with the computer-generated one and correct any inaccuracies.
In conclusion, the function \(f(x) = \frac{\ln x^6}{\ln x^3 - 1}\) has a vertical asymptote at \(x=e^{\frac{1}{3}}\) and a horizontal asymptote at \(y=2\). Plotting the function reveals the asymptotes converge, and the hand sketch confirms this.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vertical Asymptote
A vertical asymptote in a function occurs when the function approaches infinity or negative infinity as the input reaches a particular value. It signifies a point where the function is undefined. To identify vertical asymptotes analytically, we typically look at the values of the input, usually denoted as \(x\), that cause the denominator of a rational function to equal zero.
If you consider the equation \(f(x) = \frac{6\ln x}{3\ln x - 1}\), finding the vertical asymptote involves solving \(3\ln x - 1 = 0\). In our example, this gives us \(\ln x = \frac{1}{3}\). By exponentiating both sides to solve for \(x\), we find that \(x = e^{\frac{1}{3}}\).
If you consider the equation \(f(x) = \frac{6\ln x}{3\ln x - 1}\), finding the vertical asymptote involves solving \(3\ln x - 1 = 0\). In our example, this gives us \(\ln x = \frac{1}{3}\). By exponentiating both sides to solve for \(x\), we find that \(x = e^{\frac{1}{3}}\).
- Set the denominator \(= 0\).
- Solve for \(x\).
- If \(x = a\), then \(x = a\) is the vertical asymptote.
Horizontal Asymptote
Horizontal asymptotes describe the behavior of a function as it heads towards infinity. They indicate the value that a function approaches as the input becomes extremely large or small. In the function \(f(x) = \frac{6\ln x}{3\ln x - 1}\), we observe the terms in the numerator and denominator of the rational function as \(x\) approaches infinity.
For our case, as \(x\to \infty\), both the numerator and the denominator become dominated by terms involving \(\ln x\). When these terms have matching growth rates, their coefficients determine the horizontal asymptote: \(y = \frac{6}{3} = 2\).
For our case, as \(x\to \infty\), both the numerator and the denominator become dominated by terms involving \(\ln x\). When these terms have matching growth rates, their coefficients determine the horizontal asymptote: \(y = \frac{6}{3} = 2\).
- Evaluate \(\lim_{x\to\infty} f(x)\).
- Balance dominant terms in the numerator and denominator when they share growth rates.
- Coefficient ratio of dominant terms gives the horizontal asymptote.
L'Hopital's Rule
L'Hopital's Rule is a powerful mathematical tool used to evaluate limits of indeterminate forms. These indeterminate forms frequently occur as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). Applying this rule allows the calculation of limits that aren't straightforward by their initial forms. It simplifies complex rational functions by using derivatives.
In this exercise, after simplifying \(f(x)\), we see forming \(\lim_{x\to \infty} \frac{6\ln x}{3\ln x - 1}\) into \(\frac{\infty}{\infty}\) by considering each function's growth. Applying L'Hopital's rule lets us differentiate the numerator and the denominator:
In this exercise, after simplifying \(f(x)\), we see forming \(\lim_{x\to \infty} \frac{6\ln x}{3\ln x - 1}\) into \(\frac{\infty}{\infty}\) by considering each function's growth. Applying L'Hopital's rule lets us differentiate the numerator and the denominator:
- Differentiate \(6\ln x\) to get \(\frac{6}{x}\)
- Differentiate \(3\ln x - 1\) to get \(\frac{3}{x}\)
Graphing Utility
Graphing utilities are essential tools for visual learners, offering a visual representation of functions. They help confirm calculations performed by hand and identify key features of graphs such as asymptotes, intercepts, and behavior at infinity. Using graphing utilities allows us to observe these traits even before plotting manually.
For the function \(f(x) = \frac{6\ln x}{3\ln x - 1}\), a graphing utility provides a quick way to confirm the vertical and horizontal asymptotes. Checkpoints such as \(x = e^{\frac{1}{3}}\) for vertical asymptotes and \(y = 2\) for horizontal ones are validated visually.
For the function \(f(x) = \frac{6\ln x}{3\ln x - 1}\), a graphing utility provides a quick way to confirm the vertical and horizontal asymptotes. Checkpoints such as \(x = e^{\frac{1}{3}}\) for vertical asymptotes and \(y = 2\) for horizontal ones are validated visually.
- Input the function into the graphing tool.
- Adjust window settings to focus on areas of interest.
- Compare your calculated outcomes with the graph's features.
Natural Logarithm
The natural logarithm, denoted as \(\ln x\), is a logarithm with the base \(e\), where \(e\) is an irrational constant approximately equal to 2.71828. Natural logarithms are commonly used in mathematical equations involving growth rates and compound interest because of the unique properties of the exponential function.
- Property: \(\ln e^x = x\)
- Simplification tip: \(\ln a^b = b\ln a\)