Chapter 4: Problem 58
Graph the function and comment on vertical and horizontal asymptotes. $$y=2^{1 / x}$$
Short Answer
Expert verified
The function has a vertical asymptote at \( x = 0 \) and a horizontal asymptote at \( y = 1 \).
Step by step solution
01
Understand Function Behavior
The function given is \( y = 2^{1/x} \). Notice that this form suggests exponential growth or decay depending on the sign and magnitude of \( 1/x \). Analyzing \( 1/x \) is crucial to understanding the behavior of \( y \).
02
Identify Vertical Asymptotes
Vertical asymptotes occur where the function becomes undefined or tends toward infinity. Since the function \( y = 2^{1/x} \) becomes undefined at \( x = 0 \) (as you cannot divide by zero), there is a vertical asymptote at \( x = 0 \).
03
Determine Horizontal Asymptotes
Horizontal asymptotes are identified by evaluating the limits as \( x \to \infty \) and \( x \to -\infty \). As \( x \to \infty \), \( 1/x \to 0 \), so \( y \to 2^0 = 1 \). As \( x \to -\infty \), \( 1/x \to 0 \) again implies \( y \to 1 \). Therefore, there is a horizontal asymptote at \( y = 1 \).
04
Sketch the Graph
Draw the axes, and note the vertical asymptote at \( x = 0 \) and the horizontal asymptote at \( y = 1 \). As \( x \to 0^+ \), \( 2^{1/x}\) rapidly increases without bound, and as \( x \to 0^- \), \( 2^{1/x}\) approaches zero. As \( x \to \infty \), the function approaches the horizontal asymptote \( y = 1 \) from below, and as \( x \to -\infty \), it approaches \( y = 1 \) from above.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vertical Asymptotes
In the world of graphing, vertical asymptotes are like invisible walls that a function can never cross or reach. When examining the function \( y = 2^{1/x} \), we need to find where the function behaves in such a manner. A vertical asymptote typically appears when the function becomes undefined.
For the function \( y = 2^{1/x} \), this happens at \( x = 0 \), because dividing by zero is undefined in mathematics. At this point, the function essentially "blows up," meaning it tends towards infinity as \( x \) gets closer and closer to 0 from either side.
Keep in mind:
For the function \( y = 2^{1/x} \), this happens at \( x = 0 \), because dividing by zero is undefined in mathematics. At this point, the function essentially "blows up," meaning it tends towards infinity as \( x \) gets closer and closer to 0 from either side.
Keep in mind:
- Vertical asymptotes mark the location where the function spikes to infinity.
- They divide the graph into distinct sections where behavior changes dramatically.
Horizontal Asymptotes
Horizontal asymptotes help us understand the behavior of a function as \( x \) goes to infinity—or gets very, very large in magnitude. Unlike vertical asymptotes, horizontal ones represent a leveling-off behavior. In simpler terms, no matter how big or small \( x \) gets, the function's value approaches but never actually reaches a certain number.
For the function \( y = 2^{1/x} \), we check the behavior as \( x \) approaches positive and negative infinity. As \( x \to \infty \), the expression \( 1/x \to 0 \), resulting in \( y \to 2^0 = 1 \). Similarly, when \( x \to -\infty \), the same kind of leveling-off occurs, again bringing the value of \( y \) very close to 1.
It's important to keep in mind:
For the function \( y = 2^{1/x} \), we check the behavior as \( x \) approaches positive and negative infinity. As \( x \to \infty \), the expression \( 1/x \to 0 \), resulting in \( y \to 2^0 = 1 \). Similarly, when \( x \to -\infty \), the same kind of leveling-off occurs, again bringing the value of \( y \) very close to 1.
It's important to keep in mind:
- Horizontal asymptotes allow us to predict the long-term behavior of a function.
- They tell us where the function "settles down" both values of \( x \) get extremely large or small.
Behavior of Exponential Functions
Exponential functions are fascinating because they can grow or decay at astonishing rates. In \( y = 2^{1/x} \), the function's behavior changes depending on the sign and magnitude of \( 1/x \).
To better understand the behavior:
To better understand the behavior:
- As \( x \to 0^+ \) (approaching from the right), \( 2^{1/x} \) grows towards infinity, indicating rapid exponential growth.
- On the other hand, as \( x \to 0^- \) (approaching from the left), the function diminishes towards zero, indicating rapid decay.
- When \( x \) moves towards either positive or negative infinity, the function tends to stabilize and hover around the horizontal asymptote \( y = 1 \).