Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The line \(x=1\) is a vertical asymptote of the function \(f(x)=\frac{x^{2}-7 x+6}{x^{2}-1}.\) b. The line \(x=-1\) is a vertical asymptote of the function \(f(x)=\frac{x^{2}-7 x+6}{x^{2}-1}.\) c. If \(g\) has a vertical asymptote at \(x=1\) and \(\lim _{x \rightarrow 1^{+}} g(x)=\infty\) then \(\lim _{x \rightarrow 1^{-}} g(x)=\infty.\)

Short answer

a. The line \(x=1\) is a vertical asymptote of \(f(x)\). b. The line \(x=-1\) is a vertical asymptote of \(f(x)\). c. If there is a vertical asymptote at \(x=1\) and \(\lim_{x\rightarrow 1^+} g(x) = \infty\), then \(\lim_{x\rightarrow 1^-} g(x) = - \infty\). Answer: Only statement b is true.

Step-by-step solution

Analyze Function \(f(x)\)

Let's first analyze the function \(f(x)=\frac{x^{2}-7x+6}{x^{2}-1}.\) We can factor the numerator and denominator: \(f(x)=\frac{(x-6)(x-1)}{(x+1)(x-1)}\). Notice that there is a common factor of \((x-1)\) in both the numerator and the denominator, thus it can be canceled, resulting in a simplified function: \(f(x)=\frac{x-6}{x+1}\).

Statement a

We are given that the line \(x=1\) is a vertical asymptote of \(f(x)\). Let's examine the behavior of the simplified function around this point: \(f(x)=\frac{x-6}{x+1}\) has a vertical asymptote at \(x=-1\) because the denominator goes to zero while the numerator does not. However, there is no vertical asymptote at \(x=1\) because the function is defined at this point. So, statement A is false.

Statement b

We are given that the line \(x=-1\) is a vertical asymptote of \(f(x)\). As mentioned earlier, \(f(x)=\frac{x-6}{x+1}\) has a vertical asymptote at \(x=-1\). So, statement B is true.

Statement c

To determine the truth of statement C, we need to see if the limit as \(x\) approaches \(1\) from the left is also infinity when there is a vertical asymptote at \(x=1\) and the limit as \(x\) approaches \(1\) from the right is infinity. The statement assumes the limit from the right is infinity: \(\lim _{x \rightarrow 1^{+}} g(x)=\infty.\) It is not necessarily true that \(\lim _{x \rightarrow 1^{-}} g(x)=\infty,\) as the behavior of the function to the left of the asymptote can be different. Consider a function \(h(x)=\frac{1}{(x-1)^2}\). It has a vertical asymptote at \(x=1\). However, the limit from the left is also \(\lim _{x \rightarrow 1^{-}} (h(x))=\infty\), not \(-\infty\). So, statement C is false. In conclusion, statement A is false, statement B is true, and statement C is false.

Key concepts

Limits

In mathematics, limits help us understand the behavior of a function as it approaches a particular point. For vertical asymptotes, limits become crucial. They show how the function behaves near points where the denominator approaches zero. This is where the function tends to grow infinitely large (positive or negative), demonstrating a vertical asymptote.

When examining the statement about the limit \(\lim _{x \rightarrow 1^{+}} g(x)=\infty\) in our original exercise, it reflects that as \(x\) approaches 1 from the positive side, the function grows infinitely. Limits from both directions must be checked to conclude the overall behavior near asymptotes. If the values trend to infinity from both sides, then asymptotic behavior is consistent, and an asymptote exists.

For instance, \(\lim _{x \rightarrow 1^{-}} g(x)\) can be different from \(\lim _{x \rightarrow 1^{+}} g(x)=\infty\), illustrating how left-hand and right-hand behavior isn’t always the same. Understanding these concepts is essential to accurately identify asymptotic properties.

Function Simplification

Function simplification is a vital step to identify properties like vertical asymptotes clearly. By reducing a function to its simplest form, we can make sense of its behavior more easily. In our exercise, the function \(f(x)=\frac{x^{2}-7x+6}{x^{2}-1}\) is complicated in its original form.

To simplify, both the numerator and denominator are factored. After factoring, we get \(f(x)=\frac{(x-6)(x-1)}{(x+1)(x-1)}\). Observing that \((x-1)\) is common in both, it can be canceled out. The simpler form, \(\frac{x-6}{x+1}\), subsequently provides a clearer view to analyze further characteristics like vertical asymptotes.

Without simplification, visibility can be muddied by complex factors. Simplified functions aid in correctly identifying points where behavior like asymptotes occurs, as they remove extraneous complexities.

Rational Functions

Rational functions are quotients of polynomials and have particular characteristics like asymptotes. Analyzing rational functions helps determine where functions are undefined, which is key in identifying vertical asymptotes.

The denominator of a rational function often serves as a basis for finding vertical asymptotes. If the denominator becomes zero when the numerator doesn’t, a vertical asymptote exists. For instance, in our key problem, \(\frac{x-6}{x+1}\) reveals a vertical asymptote at \(x=-1\), since making \(x+1=0\) does not cancel any factor in the numerator.

Conversely, if both the numerator and denominator share a zero, such as \(x=1\) in our unreduced function, an asymptote does not occur, as this point can be resolved out by simplification. Understanding these principles helps unravel where key characteristics of these functions lie and how they behave graphically.