Problem 95 - Essential Calculus

Question

Writing Consider the function \(f(x)=\frac{2}{1+e^{1 / x}}\) (a) Use a graphing utility to graph \(f\). (b) Write a short paragraph explaining why the graph has a horizontal asymptote at \(y=1\) and why the function has a nonremovable discontinuity at \(x=0\).

Step-by-Step Solution

Answer

The function \(f(x)=\frac{2}{1+e^{1/x}}\) has a horizontal asymptote at \(y=2\) rather than at \(y=1\) as \(x\) approaches positive or negative infinity. The function exhibits a nonremovable discontinuity at \(x=0\) because the function is undefined at this point, creating a hole in the graph.

Step 1: Graphing the function

To graph the function \(f(x)=\frac{2}{1+e^{1/x}}\), use a graphing calculator or online graphing tool. Alternatively, one can sketch the graph by plotting some key points and noting the behavior of the function as \(x\to 0^+\), \(x\to 0^-\), \(x\to +\infty\), and \(x\to -\infty\).

Step 2: Identifying the horizontal asymptote

The graph has a horizontal asymptote at \(y=1\) because the function tends to \(1\) as \(x\) tends to positive or negative infinity. Mathematically, this is because the term \(e^{1/x}\) tends to \(0\) as \(x\) tends to positive or negative infinity, thus making the entire function to tend to \(2/(1+0) = 2/1 = 2\). But since the entire expression could only reach a maximum of 2, the line \(y=2\) becomes a ceiling that the function cannot cross, forming a horizontal asymptote at \(y=2\).

Step 3: Explaining the nonremovable discontinuity

The function has a nonremovable discontinuity at \(x=0\) because the term \(e^{1/x}\) is undefined at \(x=0\). This means that the function \(f(x)\) does not exist at \(x=0\), creating a gap or a hole in the graph at this point, a characteristic of nonremovable discontinuity.

Key Concepts

Graphing Functions

Graphing functions is an essential skill in calculus, as it helps us understand how functions behave visually. When confronted with the function \( f(x)=\frac{2}{1+e^{1 / x}} \), it's useful to use a graphing tool to illustrate its behavior. Graphing utilities or calculators are perfect for plotting a vast number of points simultaneously, allowing us to see the overall shape.

• Observe the behavior as \( x \) approaches certain points, such as \( 0^+ \), \( 0^- \), \( \infty \), and \( -\infty \).
• In this function, as \( x \to \infty \), the term \( e^{1/x} \) approaches zero, meaning the function's value nears 2.
• Similarly, as \( x \to -\infty \), \( e^{1/x} \) still approaches zero, keeping the function's value at 2.

By plotting key points and examining these limits, you gain insight into the function’s overall behavior and how asymptotes might form.

Nonremovable Discontinuity

A nonremovable discontinuity occurs when a function cannot be made continuous by simply redefining a point. In the function \( f(x)=\frac{2}{1+e^{1 / x}} \), there is a nonremovable discontinuity at \( x=0 \). This happens because \( e^{1/x} \) becomes undefined at \( x=0 \).

• The undefined term means that you cannot calculate a value for \( f(x) \) at \( x=0 \), leading to a break or hole in the graph.
• Nonremovable discontinuities create gaps that cannot be "filled" by simply redefining the function at that point. Instead, these are intrinsic properties of the function.

Understanding this concept is crucial when examining graphs that showcase unpredictable behavior around specific points.

Horizontal Asymptotes

Horizontal asymptotes are lines that a function approaches as \( x \) approaches infinity or negative infinity. They showcase the end behavior of a function. For this function, \( f(x)=\frac{2}{1+e^{1 / x}} \), a horizontal asymptote forms at \( y=2 \).

• As \( x \to \infty \), \( e^{1/x} \) tends to zero, thereby simplifying the expression to \( f(x) \to 2 \).
• The horizontal asymptote at \( y=2 \) represents a line that the function will approach but never touch or cross as \( x \) becomes extremely large or small.

This asymptotic behavior helps predict the general direction and limit of the function as it stretches infinitely.