Chapter 14, Problem 55CCSR

Solution

a.

To draw: The graph of the given function.

  

Given:

  $f\left(x\right)=\frac{1}{x}$

Calculation:

The given function passes through the points $\left(1,1\right),\left(2,0.5\right),\left(-2,-0.5\right),\left(-1,-1\right)$ .

Since it is an odd function, therefore the graph is symmetric with respect to the origin.

Since the value of the denominator is 0 when $x=0$ , therefore the vertical asymptote is $x=0$ .

Since the degree of numerator (0) is less than the degree of denominator (1), therefore the horizontal asymptote is $y=0$ .

Now the graph is,

  

b.

To compare: The graph of the given function and $f\left(x\right)$ .

The graph of $g\left(x\right)=\frac{4}{x+2}$ is obtained by shifting the graph of $f\left(x\right)=\frac{1}{x}$ to the left side by 2 units and then vertically stretching by a factor of 4 units.

Given:

  $g\left(x\right)=\frac{4}{x+2}$

Calculation:

The graph of $y=\frac{1}{x+2}$ is obtained by shifting the graph of $y=\frac{1}{x}$ to the left side by 2 units, and then the graph of $y=\frac{4}{x+2}$ is obtained by vertically stretching the graph of $y=\frac{1}{x+2}$ by 4 units.

Hence, the graph of $g\left(x\right)=\frac{4}{x+2}$ is obtained by shifting the graph of $f\left(x\right)=\frac{1}{x}$ to the left side by 2 units and then vertically stretching by a factor of 4 units.

c.

To find: The horizontal asymptotes of the graph of $g\left(x\right)$ .

The horizontal asymptote is $y=0$ .

Given:

  $g\left(x\right)=\frac{4}{x+2}$

Calculation:

Here, the degree of numerator is 0 and the degree of denominator is 1.

Since the degree of numerator is less than the degree of denominator, therefore the horizontal asymptote is $y=0$ .

d.

To find: The vertical asymptotes of the graph of $g\left(x\right)$ .

The vertical asymptote is $x=-2$ .

The vertical asymptotes must be excluded from the domain of $g\left(x\right)$ .

Given:

  $g\left(x\right)=\frac{4}{x+2}$

Calculation:

The value of the denominator is 0 when $x=-2$ .

Therefore, the vertical asymptote is $x=-2$ .

Since the value of the denominator is zero at the vertical asymptotes, therefore the value of the function is undefined at the vertical asymptotes.

Hence, the vertical asymptotes must be excluded from the domain of $g\left(x\right)$ .