Chapter 2.2, Problem 27E

(a)

To determine

To find: The vertical asymptotes of the graph of $f\left(x\right)=\frac{1}{x^2-4}$ .

Answer to Problem 27E

The vertical asymptotes of the graph of $f\left(x\right)=\frac{1}{x^2-4}$ are $x=2$ and $x=-2$ .

Explanation of Solution

Given information:The function is $f\left(x\right)=\frac{1}{x^2-4}$ .

Calculation:

To graph a function $y=\frac{1}{x^2-4}$ , follow the steps using graphing calculator.

First press “ON” button on graphical calculator, press $\boxed{\text{Y}=}$ key and enter right hand side of the equation $y=\frac{1}{x^2-4}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{1}\boxed{÷}\boxed{(}\boxed{\text{X}}\boxed{^}\boxed{2}\boxed{-}\boxed{4}\boxed{)}$ .

The display will show the equation,

  ${\text{Y}}_{\text{1}}=1/\left(\text{X}^2-4\right)$

Press the window key and adjust the window to $\left[-4,4\right]$ by $\left[-3,3\right]$ .Now, press the $\boxed{\text{GRAPH}}$ key and $\boxed{\text{TRACE}}$ key to produce the graph of given function in standard window as shown below.

  

Figure (1)

As observed from graph of function $f\left(x\right)=\frac{1}{x^2-4}$ , the vertical asymptotes to the graph are $x=-2$ and $x=2$ .

Therefore, the vertical asymptotes of the graph of $f\left(x\right)=\frac{1}{x^2-4}$ are $x=2$ and $x=-2$ .

(b)

To determine

To find:The behavior of the function $f\left(x\right)=\frac{1}{x^2-4}$ to the right and left of the vertical asymptotes.

Answer to Problem 27E

The value of $\underset{x\to -2^{-}}{\lim }f\left(x\right)$ , $\underset{x\to -2^{+}}{\lim }f\left(x\right)$ are $\infty$ and $-\infty$ respectively and the value of $\underset{x\to 2^{-}}{\lim }f\left(x\right)$ , $\underset{x\to 2^{+}}{\lim }f\left(x\right)$ are $-\infty$ and $\infty$ respectively.

Explanation of Solution

Given information:The function is $f\left(x\right)=\frac{1}{x^2-4}$ .

Calculation:

First take the vertical asymptote $x=-2$ of function $f\left(x\right)=\frac{1}{x^2-4}$ .

As observed from the graph of function, the function $f\left(x\right)=\frac{1}{x^2-4}$ approaches to $\infty$ as $x$ approaches to $-2$ from the left side.So, the value of $\underset{x\to -2^{-}}{\lim }f\left(x\right)$ is $\infty$ .

As observed from the graph of function, the function $f\left(x\right)=\frac{1}{x^2-4}$ approaches to $-\infty$ as $x$ approaches to $-2$ from the right side.So, the value of $\underset{x\to -2^{+}}{\lim }f\left(x\right)$ is $-\infty$ .

Now take the vertical asymptote $x=2$ of the function $f\left(x\right)=\frac{1}{x^2-4}$ .

As observed from the graph of function, the function $f\left(x\right)=\frac{1}{x^2-4}$ approaches to $-\infty$ as $x$ approaches to $2$ from the left side. So, the value of $\underset{x\to 2^{-}}{\lim }f\left(x\right)$ is $-\infty$ .

As observed from the graph of function, the function $f\left(x\right)=\frac{1}{x^2-4}$ approaches to $\infty$ as $x$ approaches to $2$ from the right side.So, the value of $\underset{x\to 2^{+}}{\lim }f\left(x\right)$ is $\infty$ .

Therefore, the value of $\underset{x\to -2^{-}}{\lim }f\left(x\right)$ , $\underset{x\to -2^{+}}{\lim }f\left(x\right)$ are $\infty$ and $-\infty$ respectively and the value of $\underset{x\to 2^{-}}{\lim }f\left(x\right)$ , $\underset{x\to 2^{+}}{\lim }f\left(x\right)$ are $-\infty$ and $\infty$ respectively.