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Chapter 1: Problem 16

Finding Vertical Asymptotes In Exercises \(13-28\) , find the vertical asymptotes (if any) of the graph of the function. $$ f(x)=\frac{3 x}{x^{2}+9} $$

Short Answer

Expert verified
The function \(f(x)=\frac{3x}{x^{2}+9}\) does not have any vertical asymptotes.

Step by step solution

01

Identify the Function and its Denominator

The function given is \(f(x)=\frac{3x}{x^{2}+9}\). Identify the denominator of this function, which is \(x^{2}+9\).
02

Set the Denominator Equal to Zero

To find potential vertical asymptotes, set the denominator of the function equal to zero and solve for \(x\). In this case, this turns into a problem like this: \(x^{2}+9 = 0\).
03

Solve for x

After subtracting 9 from both sides, we end up with \(x^{2} = -9\). However, the square of any real number cannot be negative, so there are no real solutions to this equation.
04

Identify Vertical Asymptotes

Since there are no real solutions for \(x\), there are no vertical asymptotes in the graph of the function \(f(x)=\frac{3x}{x^{2}+9}\).

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Key Concepts

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

Asymptotes of Functions
Asymptotes are lines that a graph approaches as the inputs or outputs grow without bound. They are not always part of the graph of the function itself, but they serve an important role in understanding the behavior of functions as they extend towards infinity.

There are three types of asymptotes:
  • Vertical asymptotes occur when the function approaches a certain x-value but does not actually touch or cross that vertical line. These are found in locations where the function is undefined.
  • Horizontal asymptotes describe the behavior of a graph as it moves far to the left or the right along the x-axis, indicating the value that the function is approaching.
  • Oblique asymptotes occur when the graph approaches a line that isn't horizontal or vertical. This happens when the degree of the numerator is exactly one more than the degree of the denominator in a rational function.

For the given function \(f(x)=\frac{3x}{x^2+9}\), we aim to find vertical asymptotes by looking for the x-values that make the denominator equal to zero. However, since the equation \(x^2+9=0\) lacks a real solution, the function does not have vertical asymptotes. This means the graph of this function will not approach any vertical line as x grows larger or smaller.
Rational Functions
Rational functions are fractions that involve polynomials in both the numerator and the denominator. They can often be complex, with interesting behaviors as variables increase or decrease.

A rational function like \(f(x)=\frac{3x}{x^2+9}\) is defined for all x-values except those that make the denominator zero. These undefined values are where we look for vertical asymptotes. To further explore rational functions:
  • When the degree of the numerator is less than the degree of the denominator, as in our example, there will be a horizontal asymptote at y = 0.
  • If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote will be the ratio of the leading coefficients.
  • Whenever the degree of the numerator is greater than the degree of the denominator by one, we may find an oblique asymptote.

In the exploration for vertical asymptotes of rational functions, it is crucial to factor both the numerator and the denominator whenever possible to cancel out common factors. This will provide the simplest form of the function and reveal the true asymptotes.
Solving Equations
Solving equations is a fundamental skill in mathematics. It involves finding the values of the variables that make the equation true. Every equation represents a balance, and our goal is to isolate the variable of interest.

To solve for x in an equation, we perform operations that simplify the equation while maintaining balance. The solution can be a single value, multiple values, or in some cases, no real values as in the example equation \(x^2+9=0\).

Here are the steps to solving a simple quadratic equation:
  • Move all terms to one side of the equal sign to set the equation to zero.
  • Factor the quadratic equation, if possible, to find its roots.
  • If factoring is not possible or impractical, use the quadratic formula to find the roots.

In our example, the equation \(x^2+9=0\) does not have a solution among the real numbers because the square of a real number cannot be negative. Complex numbers would be required to solve such an equation. Hence, within the realm of real numbers, this particular step in solving the equation confirms the absence of vertical asymptotes for the rational function \(f(x)=\frac{3x}{x^2+9}\).

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Most popular questions from this chapter

Déjà Vu At \(8 : 00\) A.M. on Saturday, a man begins running up the side of a mountain to his weekend campsite (see figure). On Sunday morning at \(8 : 00\) A.M. he runs back down the mountain. It takes him 20 minutes to run up, but only 10 minutes to run down. At some point on the way down, he realizes that he passed the same place at exactly the same time on Saturday. Prove that he is correct. [Hint: Let \(s(t)\) and \(r(t)\) be the position functions for the runs up and down, and apply the Intermediate Value Theorem to the function \(f(t)=s(t)-r(t) . ]\)

Continuity of a Function Show that the function $$ f(x)=\left\\{\begin{array}{ll}{0,} & {\text { if } x \text { is rational }} \\\ {k x,} & {\text { if } x \text { is irrational }}\end{array}\right. $$ is continuous only at \(x=0 .\) (Assume that \(k\) is any nonzero real number.)

Proof (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.

Finding a Limit In Exercises \(7-26\) , find the limit (if it exists). If it does not exist, explain why. $$ \lim _{x \rightarrow 10^{+}} \frac{|x-10|}{x-10} $$

Removable and Nonremovable Discontinuities In Exercises \(35-60,\) find the \(x\) -values (if any) at which \(f\) is not continuous. Which of the discontinuities are removable? \ $$ f(x)=\left\\{\begin{array}{ll}{\csc \frac{\pi x}{6},} & {|x-3| \leq 2} \\\ {2,} & {|x-3|>2}\end{array}\right. $$
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