Chapter 2, Problem 53RE

(a)

To determine

To find: The domain of the function $f\left(x\right)=\frac{x}{\left|x^2-9\right|}$ .

Answer to Problem 53RE

The domain of the given function is $ℝ-\left\{-3,3\right\}$ .

Explanation of Solution

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

Calculation:

Consider the function.

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

The domain is the set of input values for which the function is defined. The function is not defined for:

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

So, the domain of the function is the set of all real numbers except 3 and $-3$ .

Therefore, thedomain of the given function is $ℝ-\left\{-3,3\right\}$ .

(b)

To determine

To find:The equation of each vertical asymptote for the graph of $f\left(x\right)=\frac{x}{\left|x^2-9\right|}$ .

Answer to Problem 53RE

The vertical asymptote of a function are $x=-3$ and $x=3$ .

Explanation of Solution

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

Calculation:

The value of x for which the denominator of a function is equal to zero is known as the vertical asymptote of the function.

The denominator of the function $f\left(x\right)=\frac{x}{\left|x^2-9\right|}$ is 0 for $x=\pm 3$ .

Therefore, the vertical asymptote of a function are $x=-3$ and $x=3$ .

(c)

To determine

To find: The equation of each horizontal asymptote of the given function.

Answer to Problem 53RE

The horizontal asymptotefor the graph of given function is $y=0$

Explanation of Solution

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

Calculation:

The graph of a function has horizontal asymptote at $y=0$ , if the degree of denominator is greater than the degree of the numerator.

For the function $f\left(x\right)=\frac{x}{\left|x^2-9\right|}$ , the degree of denominator is 2 that is greater than the degree of the numerator.

Therefore, the horizontal asymptote for the graph of given function is $y=0$

(d)

To determine

To check:Whether the givenfunction isodd, even or neither.

Answer to Problem 53RE

The given function is an odd function.

Explanation of Solution

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

Calculation:

Substitute $-x$ for $x$ in the given function.

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

Therefore, the given function is an odd function

(e)

To determine

To find: The value of $x$ such that the given function is discontinuous. Also, write the type of discontinuity.

Answer to Problem 53RE

The given function is discontinuous for $x=3,-3$ and has non removable discontinuity.

Explanation of Solution

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

The function is not defined for $x=3$ and $x=-3$ as the denominator of the function becomes zero. The discontinuity will be infinite discontinuity that is not removable.

Therefore, the given function is discontinuous for $x=3,-3$ and has non removable discontinuity.