Chapter 3.7, Problem 7E

(a)

To determine

To complete the tables.

Answer to Problem 7E

Table 1

$x$	$r\left(x\right)$
1.5	$r\left(x\right)=\frac{1.5}{1.5-2}=-\frac{1.5}{0.5}=-3$
1.9	$-19$
1.99	$-199$
1.999	$-1999$

Table 2

$x$	$r\left(x\right)$
2.5	$r\left(x\right)=\frac{2.5}{2.5-2}=\frac{2.5}{0.5}=5$
2.1	$21$
2.01	$201$
2.001	$2001$

Table 3

$x$	$r\left(x\right)$
10	$r\left(x\right)=\frac{10}{10-2}=\frac{10}{8}=1.25$
50	$1.0417$
100	$1.0204$
1000	$1.0020$

Table 4

$x$	$r\left(x\right)$
$-10$	$r\left(x\right)=\frac{-10}{-10-2}=\frac{-10}{-12}=0.8333$
$-50$	$0.9615$
$-100$	$0.9804$
$-1000$	$0.9980$

Explanation of Solution

Given information:

The rational function is $r\left(x\right)=\frac{x}{x-2}$ .

Calculation:

To complete each table, substitute, each value of $x$ on $r\left(x\right)=\frac{x}{x-2}$ to determine the value of $r\left(x\right)$ ,

Therefore, the tables are obtained as:

Table 1

$x$	$r\left(x\right)$
1.5	$r\left(x\right)=\frac{1.5}{1.5-2}=-\frac{1.5}{0.5}=-3$
1.9	$-19$
1.99	$-199$
1.999	$-1999$

Table 2

$x$	$r\left(x\right)$
2.5	$r\left(x\right)=\frac{2.5}{2.5-2}=\frac{2.5}{0.5}=5$
2.1	$21$
2.01	$201$
2.001	$2001$

Table 3

$x$	$r\left(x\right)$
10	$r\left(x\right)=\frac{10}{10-2}=\frac{10}{8}=1.25$
50	$1.0417$
100	$1.0204$
1000	$1.0020$

Table 4

$x$	$r\left(x\right)$
$-10$	$r\left(x\right)=\frac{-10}{-10-2}=\frac{-10}{-12}=0.8333$
$-50$	$0.9615$
$-100$	$0.9804$
$-1000$	$0.9980$

(b)

To determine

To describe the behaviour of the function near its vertical asymptote.

Answer to Problem 7E

The behaviour of the function near its vertical asymptote is as $x\to 2^{+}$ , $r\to \infty$ and as $x\to 2^{-}$ , $r\to -\infty$ .

Explanation of Solution

Given information:

The rational function is $r\left(x\right)=\frac{x}{x-2}$ .

The $x=2$ is a vertical asymptote because the denominator of $r$ is zero when $x=2$ .

Use table 1 and table 2 to identify the behaviour of the function for $x$ -values to the left and right of 2.

Table 1

  $x\to 2^{-}$

$x$	$r\left(x\right)$
1.5	$r\left(x\right)=\frac{1.5}{1.5-2}=-\frac{1.5}{0.5}=-3$
1.9	$-19$
1.99	$-199$
1.999	$-1999$

Table 2

  $x\to 2^{+}$

$x$	$r\left(x\right)$
2.5	$r\left(x\right)=\frac{2.5}{2.5-2}=\frac{2.5}{0.5}=5$
2.1	$21$
2.01	$201$
2.001	$2001$

From the above table shown above, it can be concluded that as $x\to 2^{+}$ , $r\to \infty$ and as $x\to 2^{-}$ , $r\to -\infty$ .

Hence,

The behaviour of the function near its vertical asymptote is as $x\to 2^{+}$ , $r\to \infty$ and as $x\to 2^{-}$ , $r\to -\infty$ .

(c)

To determine

To find the horizontal asymptote.

Answer to Problem 7E

The horizontal asymptote is the line $y=1$ .

Explanation of Solution

Given information:

The rational function is $r\left(x\right)=\frac{x}{x-2}$ .

The horizontal asymptote is the value that $y$ approaches as $x\to \pm \infty$

Table 3

  $x\to +\infty$

$x$	$r\left(x\right)$
10	$r\left(x\right)=\frac{10}{10-2}=\frac{10}{8}=1.25$
50	$1.0417$
100	$1.0204$
1000	$1.0020$

Table 4

  $x\to -\infty$

$x$	$r\left(x\right)$
$-10$	$r\left(x\right)=\frac{-10}{-10-2}=\frac{-10}{-12}=0.8333$
$-50$	$0.9615$
$-100$	$0.9804$
$-1000$	$0.9980$

From the above table shown above, it can be concluded that as $x\to +\infty$ , $r\to 1^{+}$ and as $x\to -\infty$ , $r\to 1^{-}$ .

Hence,

The horizontal asymptote is the line $y=1$ .