Problem 40
Question
Find all vertical and horizontal asymptotes of the graph of \(f\). You may wish to use a graphics calculator to assist you. $$ f(x)=\frac{\ln \left(1+e^{x}\right)}{\ln \left(2+e^{3 x}\right)} $$
Step-by-Step Solution
Verified Answer
No vertical asymptotes; horizontal asymptotes are at \( y = \frac{1}{3} \) and \( y = 0 \).
1Step 1: Determine vertical asymptotes
To find vertical asymptotes, we need to determine where the function is undefined. Vertical asymptotes occur when the denominator is zero (as long as the numerator is not also zero). So, solve the equation \( \ln(2 + e^{3x}) = 0 \). This requires \( 2 + e^{3x} = 1 \), leading to \( e^{3x} = -1 \). Since the exponential function \( e^{3x} \) is always positive, there is no real solution, and therefore no vertical asymptotes exist for this function.
2Step 2: Determine horizontal asymptotes
Horizontal asymptotes are found by analyzing the behavior of the function as \( x \to \infty \) and \( x \to -\infty \). For large \( x \), \( e^x \) and \( e^{3x} \) dominate the expressions \( 1+e^x \) and \( 2+e^{3x} \). Thus \( \ln(1+e^x) \approx x \) and \( \ln(2+e^{3x}) \approx 3x \), so \( \lim_{{x \to \infty}} f(x) = \frac{x}{3x} = \frac{1}{3} \). As \( x \to -\infty \), \( e^x \to 0 \) and \( e^{3x} \to 0 \), therefore \( \ln(1+e^x) \approx \ln(1) = 0 \) and \( \ln(2+e^{3x}) \approx \ln(2) \), so \( \lim_{{x \to -\infty}} f(x) = \frac{0}{\ln(2)} = 0 \). Thus, the horizontal asymptotes are \( y = \frac{1}{3} \) and \( y = 0 \).
Key Concepts
Vertical AsymptotesHorizontal AsymptotesLimits
Vertical Asymptotes
Vertical asymptotes are a key feature in graph analysis and occur when a function approaches infinity as the input gets closer to a certain value. This typically happens when the denominator of a rational function equals zero and the numerator does not. In simpler terms, the function becomes undefined and shoots off to infinity, creating a vertical asymptote.
For our function, we examine the denominator \( \ln(2 + e^{3x}) \) to find vertical asymptotes. We set it to zero: \( \ln(2 + e^{3x}) = 0 \), meaning \( 2 + e^{3x} = 1 \). Solving \( e^{3x} = -1 \) leads to no solution since \( e^{3x} \) is always positive. Hence, there are no vertical asymptotes in this function. It's a good example of an instance where you might initially assume there could be vertical asymptotes but later conclude otherwise based on algebraic analysis.
For our function, we examine the denominator \( \ln(2 + e^{3x}) \) to find vertical asymptotes. We set it to zero: \( \ln(2 + e^{3x}) = 0 \), meaning \( 2 + e^{3x} = 1 \). Solving \( e^{3x} = -1 \) leads to no solution since \( e^{3x} \) is always positive. Hence, there are no vertical asymptotes in this function. It's a good example of an instance where you might initially assume there could be vertical asymptotes but later conclude otherwise based on algebraic analysis.
Horizontal Asymptotes
Horizontal asymptotes describe the behavior of a function as the input grows very large or very small. Specifically, they are horizontal lines that the graph of the function approaches as \( x \to \infty \) or \( x \to -\infty \).
To find horizontal asymptotes for our function \( f(x) = \frac{\ln(1 + e^x)}{\ln(2 + e^{3x})} \), we evaluate the limits at infinity. As \( x \to \infty \), both numerator and denominator are dominated by exponential terms, respectively leading to \( \ln(1 + e^x) \approx x \) and \( \ln(2 + e^{3x}) \approx 3x \). Thus \( \lim_{{x \to \infty}} f(x) = \frac{x}{3x} = \frac{1}{3} \). At \( x \to -\infty \), exponential terms \( e^x \) and \( e^{3x} \) approach zero, so the expression simplifies to \( \ln(1) = 0 \) and \( \ln(2) \). Therefore, \( \lim_{{x \to -\infty}} f(x) = \frac{0}{\ln(2)} = 0 \). Consequently, the graph has horizontal asymptotes at \( y = \frac{1}{3} \) and \( y = 0 \). These asymptotes give a clear picture of how the function behaves at extreme values of \( x \).
To find horizontal asymptotes for our function \( f(x) = \frac{\ln(1 + e^x)}{\ln(2 + e^{3x})} \), we evaluate the limits at infinity. As \( x \to \infty \), both numerator and denominator are dominated by exponential terms, respectively leading to \( \ln(1 + e^x) \approx x \) and \( \ln(2 + e^{3x}) \approx 3x \). Thus \( \lim_{{x \to \infty}} f(x) = \frac{x}{3x} = \frac{1}{3} \). At \( x \to -\infty \), exponential terms \( e^x \) and \( e^{3x} \) approach zero, so the expression simplifies to \( \ln(1) = 0 \) and \( \ln(2) \). Therefore, \( \lim_{{x \to -\infty}} f(x) = \frac{0}{\ln(2)} = 0 \). Consequently, the graph has horizontal asymptotes at \( y = \frac{1}{3} \) and \( y = 0 \). These asymptotes give a clear picture of how the function behaves at extreme values of \( x \).
Limits
Limits are foundational in calculus, describing how a function behaves as the input approaches a certain value. Understanding limits helps in determining asymptotic behavior, which reveals the end behavior of a graph.
For the given function, limits are crucial in identifying horizontal asymptotes. When finding \( \lim_{{x \to \infty}} \) and \( \lim_{{x \to -\infty}} \), the function simplifies due to the dominance of exponential expressions. The approach involves approximating the logarithms based on their dominating exponential terms, and understanding that properties of limits allow us to predict that the values \( \frac{1}{3} \) and \( 0 \) represent the horizontal asymptotes, corresponding to the equation \( \frac{x}{3x} \) as \( x \to \infty \) and constant \( \ln(2) \) for \( x \to -\infty \). This clear explanation of limits not only aids in solving the problem at hand, but also reinforces how limits govern the predictive behavior of function graphs.
For the given function, limits are crucial in identifying horizontal asymptotes. When finding \( \lim_{{x \to \infty}} \) and \( \lim_{{x \to -\infty}} \), the function simplifies due to the dominance of exponential expressions. The approach involves approximating the logarithms based on their dominating exponential terms, and understanding that properties of limits allow us to predict that the values \( \frac{1}{3} \) and \( 0 \) represent the horizontal asymptotes, corresponding to the equation \( \frac{x}{3x} \) as \( x \to \infty \) and constant \( \ln(2) \) for \( x \to -\infty \). This clear explanation of limits not only aids in solving the problem at hand, but also reinforces how limits govern the predictive behavior of function graphs.
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