Chapter 1, Problem 55EP

(a)

To determine

To find: The domain of $f$ .

Answer to Problem 55EP

All, real numbers except $x=-3$ and

  $x=3$

Explanation of Solution

Given information: The function is $f(x)=\frac{x}{\left|x^2-9\right|}$

Calculation:

The given function is:

  $f(x)=\frac{x}{\left|x^2-9\right|}$

When the denominator is 0, the function is undefined.

  $\begin{array}{c}{x}^2-9=0\\ \Rightarrow x^2=9\\ \Rightarrow x=\pm 3\end{array}$

Therefore the required all, real numbers except $x=-3$ and

  $x=3$ .

(b)

To determine

To find: The equation of vertical asymptote for the graph of $f$ .

Answer to Problem 55EP

  $x=-3$ and

  $x=3$ .

Explanation of Solution

Given information: The function is $f(x)=\frac{x}{\left|x^2-9\right|}$

Calculation:

The given function is:

  $f(x)=\frac{x}{\left|x^2-9\right|}$

Where the function is undefined, vertical asymptotes occur, which was determined in section (a) to be:

  $x=-3$ and

  $x=3$ .

Therefore, the required equation of vertical asymptote for the graph of $f$ are $x=-3$ and

  $x=3$ .

(c)

To determine

To find: The equation of horizontal asymptote for the graph of $f$ .

Answer to Problem 55EP

  $y=0$

Explanation of Solution

Given information: The function is $f(x)=\frac{x}{\left|x^2-9\right|}$

Calculation:

The given function is:

  $f(x)=\frac{x}{\left|x^2-9\right|}$

By assessing the limit as $x$ gets closer to $\infty$ , determine the appropriate final behavior:

  $\begin{array}{c}\underset{x\to \infty }{\lim }\frac{x}{x^2-9}\\ =\underset{x\to \infty }{\lim }\frac{\frac{1}{x}}{1-\frac{9}{x^2}}\\ =\frac{0}{1-0}\\ =0\end{array}$

As $x$ gets closer to $-\infty$ , evaluate the limit to determine the left end behavior:

  $\begin{array}{c}\underset{x\to -\infty }{\lim }\frac{x}{x^2-9}\\ =\underset{x\to -\infty }{\lim }\frac{\frac{1}{x}}{1-\frac{9}{x^2}}\\ =\frac{0}{1-0}\\ =0\end{array}$

Therefore the required equation of horizontal asymptote for the graph of $f$ is $y=0$ .

(d)

To determine

To find: The $f$ is odd, even or neither.

Answer to Problem 55EP

The function is odd.

Explanation of Solution

Given information: The function is $f(x)=\frac{x}{\left|x^2-9\right|}$

Calculation:

The given function is:

  $f(x)=\frac{x}{\left|x^2-9\right|}$

To determine whether the function is odd or even, use $-x$ rather than $x$ substitute.

  $\begin{array}{c}f(-x)=\frac{-x}{\left|{(-x)}^2-9\right|}\\ =-\frac{x}{\left|x^2-9\right|}\\ =-f(x)\end{array}$

Therefore the required function is odd.

(e)

To determine

To find: The all values of $x$ for which $f$ is discontinuous and classify each discontinuity as removable or non removable.

Answer to Problem 55EP

  $x=3$ and

  $x=-3$

Both are infinite discontinuities; non removable.

Explanation of Solution

Given information: The function is $f(x)=\frac{x}{\left|x^2-9\right|}$

Calculation:

The given function is:

  $f(x)=\frac{x}{\left|x^2-9\right|}$

At $x=-3$ and $x=3$ , the function exhibits two places of discontinuity. Due to the fact that both of these are vertical asymptotes, there is no finite number $L$ such that:

  $\underset{x\to -3}{\lim }f(x)=L$

Or,

  $\underset{x\to 3}{\lim }f(x)=L$

As there are no finite limits at the points of discontinuity, then there is no value we can assign to $f(3)$ or $f(-3)$ to make the function continuous at those points. Both discontinuities are non removable.

Therefore, the required $x=3$ and

  $x=-3$ Both are infinite discontinuities; non removable.