Problem 31 - Calculus of a Single Variable

Question

Vertical Asymptote or Removable Discontinuity. In Exercises \(29-32\) , determine whether the graph of the function has a vertical asymptote or a removable discontinuity at \(x=-1 .\) Graph the function using a graphing utility to confirm your answer. $$ f(x)=\frac{x^{2}+1}{x+1} $$

Step-by-Step Solution

Answer

The function \(f(x)=\frac{x^{2}+1}{x+1}\) has a removable discontinuity at x=-1.

Step 1: Analyzing the Function

The function is \(f(x)=\frac{x^{2}+1}{x+1}\). Analyzing the denominator, \(x+1=0\), we find that x=-1 is a critical point where the function is either undefined or has an asymptote.

Step 2: Simplify the function algebraically

Rewrite the numerator as \(x^2 -1 +2\). Therefore, the function becomes \(f(x)=\frac{(x-1)(x+1)+2}{x+1}\). Now we can notice that at x = -1, we get 0/0, which is indeterminate, and this can often indicate a removable discontinuity.

Step 3: Factor and Simplify Further

We can then cancel out the common factor in the numerator and denominator, which will leave us with f(x) = (x-1) + \(\frac{2}{x+1}\) after cancellation.

Step 4: Evaluate the Simplified Function at x = -1

Substituting x=-1 into the reduced function f(x) = (x-1) + \(\frac{2}{x+1}\), we get 0.

Step 5: Graph the Function

If you plot the graph of the function using a graphing utility, you will see that the graph is continuous at x=-1 and intersects the y-axis at y=0. This confirms that x=-1 is not a vertical asymptote; but a removable discontinuity.

Key Concepts

Vertical Asymptote

In calculus, a vertical asymptote occurs in a graph of a function when there is a point where the function grows indefinitely. In simpler words, the function has no finite value at this point, usually because it results in division by zero. For the function \(f(x)=\frac{x^{2}+1}{x+1}\), the denominator becomes zero at \(x=-1\). This means the function might have a vertical asymptote at this point.
However, not all points where the denominator is zero result in a vertical asymptote. To determine whether \(x=-1\) is indeed a vertical asymptote, you must simplify the function to check if a straightforward form persists. In this case, further examination showed that instead of a vertical asymptote, there is a removable discontinuity. This is a clue that the function may just need a bit of algebraic simplification.

Removable Discontinuity

A removable discontinuity occurs when a function is undefined at a particular point, but this undefined point can be resolved by simplifying the function. It is like a "hole" in the graph that can be "filled" by redefining the function at that point. For the function \(f(x)=\frac{x^{2}+1}{x+1}\), analyzing the point \(x=-1\) gives you the indeterminate form \(\frac{0}{0}\). This often signals a removable discontinuity.
By rewriting and simplifying the function, we found that we could cancel the common factor \((x+1)\) in both the numerator and the denominator. The revised function \(f(x) = (x-1) + \frac{2}{x+1}\) is continuous, except at \(x=-1\). Here, it becomes apparent that instead of a vertical asymptote, the function merely has a removable discontinuity, which can be handled easily.

Function Simplification

Function simplification is a key process in calculus that involves rewriting a complex function into a simpler form. This makes it easier to evaluate, analyze and understand the behavior of the function. For \(f(x)=\frac{x^{2}+1}{x+1}\), simplification involves factoring the numerator, allowing for cancellation with the common factors in the denominator.
The original expression \(x^2 + 1\) in the numerator can be rewritten using basic algebraic techniques. It simplifies to \((x-1)(x+1) + 2\). This simplification helps you identify that \(x+1\) is a common factor, which when removed leads you to the simplified form \(f(x) = (x-1) + \frac{2}{x+1}\). Simplification exposes the "true" behavior of the function, making features like removable discontinuities and finite limits more discernible.

Graphing Utility

A graphing utility is a tool or software that helps visualize mathematical functions. It plots points on a graph to provide a visual representation of the function's behavior. Using a graphing utility to plot \(f(x)=\frac{x^{2}+1}{x+1}\), especially around critical points like \(x=-1\), can confirm your algebraic conclusions.
Graphing utilities provide various helpful features:

• Visual representation of functions
• Identifying critical points and discontinuities
• Testing conclusions from algebraic simplifications

When you graph the function \(f(x)=\frac{x^{2}+1}{x+1}\), you will observe that the graph is smooth at \(x=-1\), indicating a removable discontinuity. It verifies that there is no abrupt break or infinite value at this point, as would happen with a vertical asymptote. Graphing utilities reinforce the simplicity and reliability of algebraic methods in understanding function behavior.