# 7.3: The Inverse of a Rational Function

Finding the inverse of a rational function
Finding the inverse of a rational function

### Learning Objectives

• Graph the inverse of a rational function
• Find the equation of the inverse function of a one-to-one rational function

In chapter 3, we discussed that every function has an inverse, but only a one-to-one function has an inverse function. Some rational functions are one-to-one functions such as $f(x)=\dfrac{1}{x}$ or $f(x)=\dfrac{x-1}{x+4}$. Therefore, their inverse is a function. Some rational functions are many-to-one functions such as $f(x)=\dfrac{1}{x^2-1}$ or $f(x)=\dfrac{x^3-4x^2+2}{x^2-1}$. Therefore, their inverse is not a function (Figure 1).

 One-to-one rational functions that have an inverse function Many-to-one rational functions that do not have an inverse function Figure 1. Graphs of rational functions

Notice that three of the graphs in figure 1 have horizontal and vertical asymptotes but the 4th graph has two vertical asymptotes and a slant asymptote. A slant asymptote is a line of the form $y=mx+b$ that is neither vertical nor horizontal but that the graph gets closer and closer to as $x$ approaches positive and negative infinity. Slant asymptotes occur in the graph of a rational function $f(x)=\dfrac{P(x)}{Q(x)}$ when the degree of $P(x)$ is one more than the degree of $Q(x)$. For example, in the function $f(x)=\dfrac{x^3-4x^2+2}{x^2-1}$ (Figure 1), $P(x)=x^3-4x^2+2$ and has degree 3, while $Q(x)=x^2-1$ which has degree 2. Since 3 is one more than 2, there is a slant asymptote.

## Graphing the Inverse Function of a Rational Function

We may graph the inverse of a rational function by creating and using its inverse table. For example, given the function $f(x)=\dfrac{1}{x}$, we may graph the function by creating a table of values (Table 1).

 $x$ $y=\dfrac{1}{x}$ $–2$ $-\dfrac{1}{2}$ $–1$ $–1$ $-\dfrac{1}{2}$ $–2$ $\dfrac{1}{2}$ $2$ $1$ $1$ $2$ $\dfrac{1}{2}$ $3$ $\dfrac{1}{3}$ Table 1. Table of values for $f(x)=\dfrac{1}{x}$

The inverse of the function is found by switching the values of the $x$ and $y$ columns so that the inputs become the values of $y$ and the outputs become the values of $x$. The table after switching the values of the $x$ and $y$ columns is the inverse table (Table 2).

 $x$ $y=\dfrac{1}{x}$ $-\dfrac{1}{2}$ $-2$ $-1$ $-1$ $-2$ $-\dfrac{1}{2}$ $2$ $\dfrac{1}{2}$ $1$ $1$ $\dfrac{1}{2}$ $2$ $\dfrac{1}{3}$ $3$ Table 2. The inverse table for $f^{-1}(x)=\dfrac{1}{x}$

Figure 2 shows the graph of the inverse of the function $f(x)=\dfrac{1}{x}$ drawn from its inverse table. Notice that the graph of the inverse is exactly the same as the graph of the original function $f(x)=\dfrac{1}{x}$. In other words, the function $f(x)=\dfrac{1}{x}$ is the inverse of the function itself. The inverse function is a reflection of the original function with respect to the line of symmetry $y=x$.

Figure 2. The inverse of the function $f(x)=\dfrac{1}{x}$ is $f^{-1}(x)=\dfrac{1}{x}$.

Since the graph of the function $f(x)=\dfrac{1}{x}$ is symmetric across the line $y=x$, the inverse function is identical to the original function.

## Determining the Inverse Function of an One-to-One Rational Function

To determine the equation of the inverse function of a one-to-one rational function, we use the same idea of switching the input and output. We start by writing $y=f(x)$, switch $x$ and $y$, and then solve for $y$.

For example, to determine the inverse function of the one-to-one rational function $g(x)=\dfrac{1}{x}$, we write $y=g(x)$ then switch $x$ and $y$:

\begin{aligned}g(x)&=\dfrac{1}{x}\\\\y&=\dfrac{1}{x}\\\\x&=\dfrac{1}{y}\end{aligned}

At this point, we have the inverse. We now need to solve for $y$ so we can write the inverse using function notation by replacing $y$ with $g^{-1}(x)$:

\begin{aligned}x&=\dfrac{1}{y}\\\\x\color{blue}{\cdot y}&=\dfrac{1}{y}\color{blue}{\cdot y}&&\text{Multiply both sides by }y\text{ to clear the fractions}\\\\xy&=1\\\\y&=\dfrac{1}{x}\\\\g^{-1}(x)&=\dfrac{1}{x}\end{aligned}

Therefore, the equation of the inverse function is $g^{-1}(x)=\dfrac{1}{x}$.

### Example 1

Determine the inverse function of the one-to-one rational function $h(x)=\dfrac{x-1}{x+4}$.

#### Solution

We start by writing $y=h(x)$ then switch$x$ and $y$ to get the inverse:

\begin{aligned}y&=\dfrac{x-1}{x+4}&&\text{Write }y\text{ for }h(x)\\\\x&=\dfrac{y-1}{y+4}&&\text{Switch }x\text{ and }y\\\\x\color{blue}{(y+4)}&=\dfrac{y-1}{y+4}\color{blue}{\cdot (y+4)}&&\text{Multiply both sides by }y+4\text{ to clear the fractions}\\\\x(y+4)&=y-1\\\\xy+4x &=y-1&&\text{Multiply the left side using the distributive property}\\\\xy-y &=-4x-1&&\text{Collect }y\text{ terms on the left side}\\\\y(x-1)&=-4x-1&&\text{Pull }y\text{ out as a common factor on the left side}\\\\y&=\dfrac{-4x-1}{x-1}&&\text{Divide both sides by }x-1\end{aligned}

Now write the inverse in function notation, $h^{-1}(x)=\dfrac{-4x-1}{x-1}$ or by pulling out $-1$ as a common factor on the numerator, $h^{-1}(x)=-\dfrac{4x+1}{x-1}$.

### Example 2

Determine the inverse function of the one-to-one rational function $h(x)=\dfrac{x+5}{x-1}$.

#### Solution

We start by writing $y=h(x)$ then switch $x$ and $y$ to get the inverse:

\begin{aligned}y&=\dfrac{x+5}{x-1}&&\text{Write }y\text{ for }h(x)\\\\x&=\dfrac{y+5}{y-1}&&\text{Switch }x\text{ and }y\\\\x\color{blue}{(y-1)}&=\dfrac{y+5}{y-1}\color{blue}{\cdot (y-1)}&&\text{Multiply both sides by }y-1\text{ to clear the fractions}\\\\x(y-1)&=y+5\\\\xy-x &=y+5&&\text{Multiply the left side using the distributive property}\\\\xy-y &=x+5&&\text{Collect }y\text{ terms on the left side}\\\\y(x-1)&=x+5&&\text{Pull }y\text{ out as a common factor on the left side}\\\\y&=\dfrac{x+5}{x-1}&&\text{Divide both sides by }x-1\end{aligned}

Now write the inverse in function notation, $h^{-1}(x)=\dfrac{x+5}{x-1}$.

### Try It 1

Determine the inverse function of the one-to-one rational function:

1. $h(x)=\dfrac{x+4}{x-6}$

2. $g(x)=\dfrac{x+7}{x+4}$

3. $f(x)=\dfrac{2x+3}{5x+4}$

1. $h^{-1}(x)=\dfrac{6x+4}{x-1}$
2. $h^{-1}(x)=\dfrac{-4x+7}{x-1}$
3. $h^{-1}(x)=\dfrac{-4x+3}{5x-2}$

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