Problem 52
Question
Find the horizontal and vertical asymptotes of each curve. If you have a graphing device, check your work by graphing the curve and estimating the asymptotes. \( y = \dfrac{2e^x}{e^x - 5} \)
Step-by-Step Solution
Verified Answer
The horizontal asymptote is at \( y = 2 \), and the vertical asymptote is at \( x = \ln 5 \).
1Step 1: Identify Horizontal Asymptote
To determine the horizontal asymptote for the function \( y = \frac{2e^x}{e^x - 5} \), consider the behavior of the function as \( x \to \infty \) and \( x \to -\infty \). For large \( x \), the terms \( e^x \) dominate in the numerator and the denominator. Thus, the expression simplifies to \( \frac{2e^x}{e^x} = 2 \). Therefore, the horizontal asymptote is \( y = 2 \).
2Step 2: Check Vertical Asymptotes
Vertical asymptotes occur where the denominator of a function goes to zero, leading to undefined values of \( y \). Set the denominator \( e^x - 5 = 0 \) to find \( x \). Solving \( e^x = 5 \), we take the natural logarithm of both sides: \( x = \ln 5 \). Thus, there is a vertical asymptote at \( x = \ln 5 \).
Key Concepts
Horizontal AsymptotesVertical AsymptotesNatural LogarithmExponential Function
Horizontal Asymptotes
Horizontal asymptotes are lines that the graph of a function approaches as the input value (often represented by \( x \)) goes to positive or negative infinity. These asymptotes give us an idea of the end behavior of the function.
In the function \( y = \frac{2e^x}{e^x - 5} \), as \( x \) becomes very large (\( x \to \infty \)), the terms involving \( e^x \) dominate both the numerator and the denominator. The function simplifies to \( \frac{2e^x}{e^x} = 2 \), indicating a horizontal asymptote at \( y = 2 \).
As \( x \to -\infty \), \( e^x \to 0 \) and the effect is negligible compared to the constant \( -5 \), confirming that the horizontal asymptote remains \( y = 2 \).
Horizontal asymptotes describe the behavior of functions as \( x \) goes to extreme values and do not usually affect the graph's shape in its middle portions.
In the function \( y = \frac{2e^x}{e^x - 5} \), as \( x \) becomes very large (\( x \to \infty \)), the terms involving \( e^x \) dominate both the numerator and the denominator. The function simplifies to \( \frac{2e^x}{e^x} = 2 \), indicating a horizontal asymptote at \( y = 2 \).
As \( x \to -\infty \), \( e^x \to 0 \) and the effect is negligible compared to the constant \( -5 \), confirming that the horizontal asymptote remains \( y = 2 \).
Horizontal asymptotes describe the behavior of functions as \( x \) goes to extreme values and do not usually affect the graph's shape in its middle portions.
Vertical Asymptotes
Vertical asymptotes occur where a function's value increases or decreases without bound. This happens when the denominator of a function approaches zero, making the value of the function undefined.
For the function \( y = \frac{2e^x}{e^x - 5} \), a vertical asymptote is found when the denominator \( e^x - 5 = 0 \).
For the function \( y = \frac{2e^x}{e^x - 5} \), a vertical asymptote is found when the denominator \( e^x - 5 = 0 \).
- Set \( e^x = 5 \) to identify such points, where the function is undefined.
- Taking the natural logarithm of both sides, we get \( x = \ln 5 \).
Natural Logarithm
The natural logarithm, denoted as \( \ln \), is the logarithm to the base \( e \), where \( e \approx 2.718 \). It is particularly useful because it simplifies expressions involving exponential growth or decay, showing the relationship between the rate of change and the current state.
In our function, to find the vertical asymptote, we used the natural logarithm to solve \( e^x = 5 \) for \( x \):
In our function, to find the vertical asymptote, we used the natural logarithm to solve \( e^x = 5 \) for \( x \):
- \( \ln e^x = x \ln e \)
- Since \( \ln e = 1 \), this simplifies to \( x = \ln 5 \).
Exponential Function
The exponential function, widely utilized in mathematics, physics, and many scientific disciplines, involves the constant \( e \) raised to the power \( x \). It is expressed as \( e^x \) and represents continuous growth or decay.
In the function \( y = \frac{2e^x}{e^x - 5} \), the exponential factor \( e^x \) greatly influences both the numerator and denominator, determining the asymptotes:
In the function \( y = \frac{2e^x}{e^x - 5} \), the exponential factor \( e^x \) greatly influences both the numerator and denominator, determining the asymptotes:
- As \( x \) increases, \( e^x \) becomes very large, dominating the behavior of the function and leading to a horizontal asymptote at \( y = 2 \).
- When \( e^x = 5 \), it affects the denominator, causing a vertical asymptote at \( x = \ln 5 \).
Other exercises in this chapter
Problem 51
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