Chapter 9.4, Problem 43HP

To determine

To write the rational function for the given graph

Answer to Problem 43HP

Both $f\left(x\right)$ and $g\left(x\right)$ are rational functions, and both of them have one and the same horizontal asymptote that is $y=0$ . They also have $x=0$ as vertical asymptote. The graph of $g\left(x\right)$ has four branches while $f\left(x\right)$ has only one.

The rational function of $g\left(x\right)$ is $g\left(x\right)=\frac{\left(x-1\right)\left(x+1\right)}{x\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}$

Explanation of Solution

Given information:

Consider the graph given in the question

The function $f\left(x\right)$ has the vertical asymptote $x=0$ and the horizontal asymptote of $y=0$ . The function is undefined when $x=-1\text{ and }x=1$ as shown by the holes in its graph.

Now, $g\left(x\right)$ can be written as,

  $g\left(x\right)=\frac{\left(x-1\right)\left(x+1\right)}{x\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}$

This function has the zeroes $\left(1,0\right)\text{ and }\left(-1,0\right)$ and has the vertical asymptotes $x=0,x=-\sqrt{2}\text{ and }x=\sqrt{2}$ and a horizontal asymptote of $y=0$ (refer the graph)

Therefore, both $f\left(x\right)$ and $g\left(x\right)$ are rational functions, and both of them have one and the same horizontal asymptote that is $y=0$ . They also have $x=0$ as vertical asymptote. The graph of $g\left(x\right)$ has four branches while $f\left(x\right)$ has only one.