Chapter 4, Problem 34RE

To determine

To discuss: the given rational function

Answer to Problem 34RE

The graph of this function is as follows:

  

Explanation of Solution

Given:

  $F(x)=\frac{{(x-1)}^2}{x^2-1}$

Calculation:

Let us consider the following polynomial function.

  $F(x)=\frac{{(x-1)}^2}{x^2-1}$

We apply 8 steps to analyze the rational function.

Step 1

We factor the numerator and denominator of the function and find the domain of the function,

  $F(x)=\frac{{(x-1)}^2}{x^2-1}=\frac{(x-1)(x-1)}{(x+1)(x-1)}=\frac{(x-1)}{(x+1)}$

Thus, the domain of the function is $\{x\mid x\ne -1\}$

Step 2

Now, we write $F$ in lowest terms.

Since, there are no more common factors between numerator and denominator therefore $F$ isin lowest terms.

Step 3

The $x-$ intercepts are found by determining real zeros of the numerator of $F$ written in lowestterms, equating numerator to zero we get,

  $\begin{array}{r}\hfill x-1=0\\ \hfill x=1\end{array}$

So, $x-$ intercept of $F$ is 1 .

Since, zero is the domain of $F$ , the $y-$ intercept is $R(0)=-1$

Step 4

Now, we test for symmetry if $F(-x)=F(x)$ , the function is even and its graph will be symmetric with respect to $y-$ axis. If $F(-x)=-F(x)$ , the function is odd and its graph will be symmetricwith respect to origin.

So

  $F(-x)=\frac{(-x-1)}{(-x+1)}$

We conclude that, neither $F(-x)=F(x)$ nor $F(-x)=-F(x)$ so, no symmetry with $y-$ axisor origin.

Step 5

We locate the vertical asymptotes by finding the zeros of the denominator with the rationalfunction in lowest terms. With $F$ written in lowest terms, we find that the graph of $F$ has a vertical asymptote $x=-1$

Step 6

Since the degree of numerator equals degree of numerator, therefore the graph is horizontal. The leading coefficient of numerator is 1 and the leading coefficient of denominator is 1. To determine the graph of $F$ intersects the horizontal asymptote, we solve the equation $R(x)=0$

  $\frac{x-1}{x+1}=0$

  $x-1=0$

  $x=1$

The only solution $x=1,$ so the graph intersects at (1,0)

Step 7

From the graph we can see that graph of $F(x)$ is in connected mode. The graph has vertical line at $x=1.x-$ intercept is 1 and $y-$ intercept is -1

  

Step 8:

The graph of this function is as follows:

  

Conclusion:

Thus, the given relation is discussed