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Chapter 1: Problem 9

Find the vertical asymptotes (if any) of the graph of the function. \(f(x)=\frac{1}{x^{2}}\)

Short Answer

Expert verified
The vertical asymptote of the function \(f(x)=\frac{1}{x^{2}}\) is \(x=0\).

Step by step solution

01

Set the denominator equals to zero

To find vertical asymptote(s), first set the denominator of the function to zero. Here the denominator is \(x^{2}\). So we've got the equation, \(x^{2}=0\).
02

Solve the equation

Now, solve this equation for x. The square root of 0 is 0, so the solution to the equation \(x^{2}=0\) is \(x=0\).
03

Determine the vertical asymptote

The solution to this equation, \(x=0\), is the value where the function is undefined. This creates a vertical asymptote for the function. So, \(x=0\) is a vertical asymptote of the function \(f(x)=\frac{1}{x^{2}}\).

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Key Concepts

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

Rational Functions
Rational functions are a particular type of function that appear frequently in algebra and calculus. These functions are composed of a ratio of two polynomials. In simple terms, think of them as fractions where the numerator and denominator are both polynomial expressions.
For instance, in the function \(f(x) = \frac{1}{x^{2}}\), the numerator is the constant \(1\), and the denominator is \(x^{2}\), which is a polynomial. Understanding the structure of rational functions is vital because it helps us analyze their behavior and properties.
  • They can be written in the form \(\frac{p(x)}{q(x)}\) where \(p(x)\) and \(q(x)\) are polynomials.
  • They exhibit unique traits such as asymptotes, intercepts, and domains influenced by the degrees of their polynomials.
This composition impacts where the function might be undefined or where it takes on special characteristics like asymptotic behavior.
Denominator Equals Zero
An essential concept in finding the vertical asymptotes of a rational function lies in identifying the values of \(x\) that make the denominator equal zero. When the denominator of a fraction equals zero, the expression becomes undefined because division by zero is not possible.
Here's how you determine where the denominator is zero:
  • Take the denominator of the rational function.
  • Set it equal to zero.
  • Solve for \(x\).
In our example, for \(f(x) = \frac{1}{x^{2}}\), we set \(x^{2} = 0\). The solution to this equation is \(x = 0\). Hence, \(x = 0\) is the point where the function becomes undefined, giving the location of a potential vertical asymptote.
Finding Asymptotes
Asymptotes are lines that a graph approaches but never actually touches or crosses. In the context of rational functions, vertical asymptotes occur at values of \(x\) that make the function undefined, often due to a zero in the denominator.
When identifying vertical asymptotes, you follow these steps:
  • Determine where the denominator equals zero by solving \(q(x) = 0\).
  • Confirm that the numerator does not also become zero for the same \(x\) value (if it does, further analysis is needed as it might be a hole rather than an asymptote).
For \(f(x) = \frac{1}{x^{2}}\), solving \(x^{2} = 0\) gives \(x = 0\), which is indeed a vertical asymptote. It's a hallmark of rational functions; these vertical lines represent a boundary or limit where the function's behavior becomes extreme.

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Most popular questions from this chapter

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(f(x)=g(x)\) for all real numbers other than \(x=0\), and \(\lim _{x \rightarrow 0} f(x)=L, \quad\) then \(\quad \lim _{x \rightarrow 0} g(x)=L\)

(a) Let \(f_{1}(x)\) and \(f_{2}(x)\) be continuous on the closed interval \([a, b]\). If \(f_{1}(a)f_{2}(b)\), prove that there exists \(c\) between \(a\) and \(b\) such that \(f_{1}(c)=f_{2}(c)\). (b) Show that there exists \(c\) in \(\left[0, \frac{\pi}{2}\right]\) such that \(\cos x=x\). Use a graphing utility to approximate \(c\) to three decimal places.

Let \(f(x)=\left(\sqrt{x+c^{2}}-c\right) / x, c>0 .\) What is the domain of \(f ?\) How can you define \(f\) at \(x=0\) in order for \(f\) to be continuous there?

Find the \(x\) -values (if any) at which \(f\) is not continuous. Which of the discontinuities are removable? $$ f(x)=x^{2}-2 x+1 $$

(a) Prove that if \(\lim _{x \rightarrow c}|f(x)|=0\), then \(\lim _{x \rightarrow c} f(x)=0\). (Note: This is the converse of Exercise \(110 .\) ) (b) Prove that if \(\lim _{x \rightarrow c} f(x)=L\), then \(\lim _{x \rightarrow c}|f(x)|=|L|\). [ Hint: Use the inequality \(\|f(x)|-| L\| \leq|f(x)-L| .]\)
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