Chapter 1.2, Problem 69E

Solution

(a)

To explain: this function's vertical asymptote is not intersected by the graph of the function.

This is shown as an empty circle in the graph.

Given information:

Graph the function $f\left(x\right)=\frac{x-\left|x\right|}{x^2}+1.$

Calculation:

The graph $f\left(x\right)=\frac{x-\left|x\right|}{x^2}+1.$ .

  

The function has a vertical asymptote at $x=0$ , but it does not intersect that asymptote, because $x=0$ is not part of the function's domain. This is shown an empty circle in the graph.

(b)

To show: how you can add a single point to the graph of $f$ and obtain a graph with its vertical asymptote intersected.

Given information:

The function $g\left(x\right)=\frac{|x-1|}{x^2-4}$ .

Calculation:

The graph to intersect the vertical asymptote by defining a point of the graph where $x=0$ .

Here, let $f\left(0\right)=1$ , shown by the filled circle in the graph, but it could have any other $y-$ value.

  

(c)

To find: is the graph in $\left(b\right)$ the graph of a function?

Yes.

Given information:

The function $h\left(x\right)=\frac{x}{\sqrt{x^2-4}}$ .

Calculation:

Start from the assumption that $f$ is a function, since it was given a function in part $\left(a\right)$ .

If it is a function, then each $x$ in its domain is mapped to a single $y$ in its range.

  $x=0$ had not been originally defined in $f$ and with us added a single definition for $x=0$ , then there is no ambiguous definition of $y$ for any $x$ in the domain. The new graph is still a function.