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Chapter 2: Problem 33

Locate the discontinuities of the function and illustrate by graphing. $$ y=\frac{1}{1+e^{1 / x}} $$

Short Answer

Expert verified
The function has a jump discontinuity at \( x = 0 \), where it is not defined. As \( x \to 0^+ \), \( y \to 0 \); as \( x \to 0^- \), \( y \to 1 \).

Step by step solution

01

Understand the Function

The function provided is \( y = \frac{1}{1+e^{1/x}} \). This is a composite function involving an exponential function \( e^{1/x} \). Our objective is to identify points of discontinuity, if any.
02

Analyze the Denominator

Discontinuities occur when the denominator is zero or undefined. The expression in the denominator is \( 1 + e^{1/x} \). Exponential functions \( e^{1/x} \) are never zero, and \( 1 + e^{1/x} \) is always positive, implying the denominator is never zero.
03

Identify Undefined Points for \( 1/x \)

The expression \( 1/x \) is undefined when \( x = 0 \). Therefore, the function is not defined at \( x = 0 \). This is a point to explore for potential discontinuity.
04

Explore Behavior Near \( x = 0 \)

Analyze the behavior of the function as \( x \to 0^+ \) and \( x \to 0^- \). As \( x \to 0^+ \), \( \frac{1}{x} \to +\infty \) making \( e^{1/x} \to +\infty \), so \( y \approx \frac{1}{1+\infty} \to 0 \). As \( x \to 0^- \), \( \frac{1}{x} \to -\infty \), making \( e^{1/x} \to 0 \), so \( y \approx \frac{1}{1+0} = 1 \). Therefore, there's a jump discontinuity at \( x = 0 \).
05

Graph the Function

Graphically, as \( x \) approaches 0 from the positive side, the output tends towards 0, while from the negative side, it tends toward 1. This jump, indicating a discontinuity, can be shown by plotting the function values approaching either side of \( x = 0 \). Join the graph on both sides to see this jump.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Composite Functions
A composite function is a function that is formed when one function is applied to the results of another. In the given problem, our function is \( y = \frac{1}{1+e^{1/x}} \). This function involves two main components:
  • An exponential function \( e^{1/x} \)
  • The algebraic fraction \( \frac{1}{1+e^{1/x}} \)
These two pieces create a composite function because they work together to form the overall function. Understanding composite functions helps us know how the input passes through these multiple layers of operations, which affects how we graph the function and find any discontinuities.
Exponential Functions
Exponential functions have the general form \( f(x) = e^x \), where \( e \) is a mathematical constant approximately equal to 2.71828. In this function, the exponential part is \( e^{1/x} \). Here, the exponent itself is not a fixed number but a function of \( x \) (i.e., \( 1/x \)). This introduces a more complex behavior because it changes dynamically as \( x \) changes.
This dynamic nature affects the values of \( e^{1/x} \):
  • When \( x \to 0^+ \), \( 1/x \to +\infty \), and \( e^{1/x} \to +\infty \).
  • When \( x \to 0^- \), \( 1/x \to -\infty \), and \( e^{1/x} \to 0 \).
Understanding exponential functions' behavior is crucial, as it helps determine the regions where the function approaches certain values, leading to discontinuities.
Jump Discontinuity
A jump discontinuity occurs when there is a sudden 'jump' in the function values at a particular point. In this case, as we approach \( x = 0 \) from both directions, we observe:
  • For \( x \to 0^+ \), the function \( y \to 0 \)
  • For \( x \to 0^- \), the function \( y \to 1 \)
There is a clear and sudden change in the function value from 0 to 1 at \( x = 0 \), which can't be bridged by a continuous curve. Thus, \( x = 0 \) marks a jump discontinuity. Recognizing this helps us paint an accurate picture of how the function behaves near this point and affects the graph.
Graphing Functions
Graphing functions is a visual way to understand how they behave. For our function \( y = \frac{1}{1+e^{1/x}} \), the graph helps showcase the jump discontinuity at \( x = 0 \).
When graphing:
  • Plot points for \( x > 0 \) closely approaching 0; the function value approaches 0.
  • Plot points for \( x < 0 \) closely approaching 0; the function value approaches 1.
The graph will show a gap at \( x = 0 \), showing that there's no value that bridges \( y = 0 \) and \( y = 1 \). This gap represents the jump discontinuity. Graphing provides a concrete means of visualizing how a function behaves and where discontinuities like jumps occur.

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Most popular questions from this chapter

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