Chapter 4, Problem 26RE

To determine

To discuss: the given rational function.

Answer to Problem 26RE

  

Explanation of Solution

Given:

  $H\left(x\right)=\frac{x}{x^2-1}$

Explanations:

Let us consider the following rational function,

  $H\left(x\right)=\frac{x}{x^2-1}$

analyze the graph of the given function.

For this follow the following steps:

Step $1$ :

Determination the domain of the function. Domain of a function is defined as the set of values of $x$ for which the function is defined.

So, let us first find the values of $x$ for which $R$ is not defined.

  $R$ is not defined when the denominator is zero. So, find the value of $x$ for which the denominator is zero.

  $\begin{array}{l}{x}^2-1=0\\ x^2=1\\ x=\sqrt{1}\\ x=\pm 1\end{array}$

So, at $x=1$ and $x=-1$ , $H$ is not defined.

Therefore, the domain of the function is $\left\{x|x\ne 1,x\ne -1\right\}$

Step $2$ :

Write $H$ in the lowest terms.

Here, since there is no common factor between the numerator and the denominator, $H$ is in the lowest terms.

Step $3$ :

Next, approximate $x$ and $y$ intercepts of the graph.

  $x-\mathrm{int}ercept$ of the graph is found by determining the zeros of the numerator.

So,

  $x=0$

Since there is only one zero of the numerator, there is only one $x-\mathrm{int}ercept$ at $x=0$ .

Also, since $x=0$ is in the domain of $R$ , $y-\mathrm{int}ercept$ is given by,

  $\begin{array}{l}H\left(0\right)=\frac{\left(0\right)}{{\left(0\right)}^2-1}\\ =0\end{array}$

So, the $y-\mathrm{int}ercept$ is at $y=0$

Step $4$ :

Determine whether the function is odd, even or neither.

A function is even if,

  $f\left(x\right)=f\left(-x\right)$

Also, a function is odd if,

  $f\left(-x\right)=-f\left(x\right)$

So, let us first find the values of $f\left(-x\right)$

  $\begin{array}{l}f\left(-x\right)=\frac{\left(-x\right)}{{\left(-x\right)}^2-1}\\ =\frac{-x}{x^2-1}\\ =-\left(\frac{x}{x^2-1}\right)\\ -f\left(x\right)=-\left(\frac{x}{x^2-1}\right)\\ So,f\left(-x\right)=-f\left(x\right)\end{array}$

Therefore, the function is not odd.

So, the graph is symmetric about origin.

Step $5$ :

Find the vertical asymptotes.

The vertical asymptotes of a rational function are located at the real zeros of the denominator.

As found above, the real zero of the denominator are $1$ and $-1$ .

So, the vertical asymptote of the function lies at $x=1$ and $x=-1$ .

Step $6$ :

Find the horizontal or oblique asymptotes. Also, locate the points where the asymptote intersects the graph.

  $H\left(x\right)=\frac{x}{x^2-1}$

Here, the degree of the numerator is smaller than the denominator. Therefore, this is a proper rational function. So, $y=0$ is the horizontal asymptote of $H$ .

Therefore, the horizontal asymptote of the function lies at $y=0$ .

To determine where the graph intersects the horizontal asymptote, solve $H\left(x\right)=0$

So,

  $\begin{array}{l}\frac{x}{x^2-1}=0\\ x=0\end{array}$

So, the graph intersects the horizontal asymptote at $\left(0,0\right)$ .

Step $7$ ;

Graph $R$ using graphing utility.

  

Step $8$ :

Complete the graph using all the above information.

  

Conclusion:

Therefore,the given rational function is analyzed.