ORCCA Rational Functions

Section 8.2 Rational Functions

In this section, we will learn about rational functions, which are ratios of two polynomial functions. Creating this ratio inherently requires division and we'll explore the effect this has on the domain of the function.

Subsection 8.2.1 Introduction to Rational Functions

Definition 8.2.1. Rational Function.

A rational function \(f\) is a function in the form

\begin{equation*} f(x)=\frac{P(x)}{Q(x)} \end{equation*}

where \(P\) and \(Q\) are polynomial functions, but \(Q\) is not the constant zero function.

Example 8.2.2.

The following functions are examples of rational functions.

\begin{equation*} f(x)=\frac{3x-5}{4x^2+7x-1} \end{equation*}

\begin{equation*} g(x)=\frac{2}{x+5} \end{equation*}

\begin{equation*} h(x)=\frac{5x^2-8x+9}{3x^3+2x^2-4x+10} \end{equation*}

Notice that each of these functions is the result of dividing a polynomial expression, which can include a constant, by another polynomial expression (except for zero). This is what makes each of them a rational function.

Checkpoint 8.2.3.

Subsection 8.2.2 Finding the Domain of a Rational Function

Recall that the domain of a function is the collection of all valid input values for that function. Since a rational function is the quotient of two polynomials, there is a possibility that there will be division by zero, which will make the rational expression undefined. This will occur whenever the polynomial in the denominator is equal to zero.

Thus, the domain of a rational function will be restricted to all real numbers except for any values of the variable which will make the denominator equal to zero. Hence, to find the domain of a rational function we need only set the denominator equal to zero and then solve the resulting equation. That will tell us which values of the variable to eliminate from the set of real numbers, to give us the domain.

Example 8.2.4.

Let \(f\) be a function where \(f(x)=\frac{1}{x-2}\text{.}\) Find the function's domain.

By setting the denominator, \(x-2\text{,}\) equal to zero and solving that equation, we obtain a solution of \(x=2\text{.}\) This tells us that \(x=2\) makes the denominator \(0\text{,}\) so the function will be undefined for \(x=2\text{.}\) Therefore, the domain of function \(f\) is all real numbers except \(x=2\text{.}\) In set-builder notation, the domain is \(\left\{x \mid x\ne2\right\}\text{.}\)

Example 8.2.5.

Algebraically find the domain of \(g(x)=\frac{3x^2}{x^2-2x-24}\text{.}\)

To find a rational function's domain, we set the denominator equal to \(0\) and solve:

\begin{align*} x^2-2x-24\amp=0\\ (x-6)(x+4)\amp=0 \end{align*}

\begin{align*} x-6\amp=0\amp\text{or}\amp\amp x+4\amp=0\\ x\amp=6\amp\text{or}\amp\amp x\amp=-4 \end{align*}

Since \(x=6\) and \(x=-4\) will cause the denominator to be \(0\text{,}\) they are excluded from the domain. The function's domain is \(\left\{x \mid x\ne6, x\ne-4\right\}\text{.}\)

Subsection 8.2.3 Simplifying Rational Expressions

Consider the two rational functions below. At first glance, which function looks simpler?

\begin{equation*} f(x)=\frac{8x^3-12x^2+8x-12}{2x^3-3x^2+10x-15} \end{equation*}

\begin{equation*} g(x)=\frac{4(x^2+1)}{x^2+5}, \text{ for }x\neq \frac{3}{2} \end{equation*}

It can be argued that the function \(g\) is simpler, at least with regard to the ease with which we can determine its domain, quickly evaluate it, and also determine where its function value is zero. All of these things are considerably more difficult with the function \(f\text{.}\)

These two functions are actually the same function. Using factoring and the same process of canceling that's used with numerical ratios, we will learn how to simplify the function \(f\) into the function \(g\text{.}\) (The full process for simplifying \(f(x)=\frac{8x^3-12x^2+8x-12}{2x^3-3x^2+10x-15}\) will be shown in Example 8.2.13.)

To see a simple example of the process for simplifying a rational function or expression, let's look at simplifying \(\frac{14}{21}\) and \(\frac{(x+2)(x+7)}{(x+3)(x+7)}\) by canceling common factors:

\begin{align*} \frac{14}{21}\amp=\frac{2\cdot \cancelhighlight{7}}{3\cdot \cancelhighlight{7}}\\ \amp=\frac{2}{3} \end{align*}

\begin{align*} \frac{(x+2)(x+7)}{(x+3)(x+7)}\amp=\frac{(x+2)\cancelhighlight{(x+7)}}{(x+3)\cancelhighlight{(x+7)}}\\ \amp=\frac{x+2}{x+3}, \text{ for }x\neq -7 \end{align*}

Remark 8.2.6. Adding Domain Restrictions.

The statement “for \(x\neq -7\)” was added when the factor of \(x+7\) was canceled from both the numerator and denominator. This is because \(\frac{(x+2)(x+7)}{(x+3)(x+7)}\) was undefined where \(x=-7\text{,}\) so the simplified version must also be undefined for \(x=-7\text{.}\)

Remark 8.2.7.

It may be tempting to want to try to simplify \(\frac{x+2}{x+3}\) into \(\frac{2}{3}\) by canceling each \(x\) that appears. These \(x\)'s are terms that are added though, and canceling (a process of dividing) is only possible with factors.

The process of canceling factors will be key to simplifying rational expressions and functions. If the expression or function is not given in factored form, then this will be our first step. We'll now look at a few more examples.

Example 8.2.8.

Simplify the rational function \(Q(x)=\frac{3x-12}{x^2+x-20}\) and state the domain of this function.

Explanation

To start, we'll factor the numerator and denominator. We'll then cancel any factors common to both the numerator and denominator.

\begin{align*} Q(x)\amp=\frac{3x-12}{x^2+x-20}\\ Q(x)\amp=\frac{3\cancelhighlight{(x-4)}}{(x+5)\cancelhighlight{(x-4)}}\\ Q(x)\amp=\frac{3}{x+5}, \text{ for }x\neq 4 \end{align*}

The domain of this function will incorporate the explicit domain restriction \(x\neq 4\) that was stated when the factor of \(x-4\) was canceled from both the numerator and denominator. We will also exclude \(-5\) from the domain as this value would make the denominator zero. Thus, the domain of \(Q\) is \(\left\{x\mid x\neq -5,4\right\}\text{.}\)

Warning 8.2.9.

When simplifying the function \(Q\) in Example 8.2.8, we cannot simply write \(Q(x)=\frac{3}{x+5}\text{.}\) The reason is that this would result in our simplified version of the function \(Q\) having a different domain than the original \(Q\text{.}\) More specifically, for our original function \(Q\text{,}\) it held that \(Q(4)\) was undefined, and this still needs to be true for the simplified form of \(Q\text{.}\)

Example 8.2.10.

Simplify the rational function \(R(y)=\frac{-y-2y^2}{2y^3-y^2-y}\) and state the domain of this function.

Explanation

\begin{align*} R(y)\amp=\frac{-y-2y^2}{2y^3-y^2-y}\\ R(y)\amp=\frac{-2y^2-y}{y(2y^2-y-1)}\\ R(y)\amp=\frac{-\cancelhighlight{y}\cancelhighlight{(2y+1)}}{\cancelhighlight{y}\cancelhighlight{(2y+1)}(y-1)}\\ \amp=-\frac{1}{y-1}, \text{ for }y\neq 0, y\neq -\frac{1}{2} \end{align*}

The domain of this function will incorporate the explicit restrictions \(y\neq 0, y\neq -\frac{1}{2}\) that were stated when the factors of \(y\) and \(2y+1\) were canceled from both the numerator and denominator. Since the factor \(y-1\) is still in the denominator, we also need the restriction that \(y\neq 1\text{,}\) but only when stating the domain of \(R\text{.}\) Therefore, the domain of \(R\) is \(\left\{y\mid y\neq -\frac{1}{2},0,1\right\}\text{.}\)

Example 8.2.11.

Simplify the expression \(\frac{9y+2y^2-5}{y^2-25}\text{.}\)

Explanation

To start, we need to recognize that \(9y+2y^2-5\) is not written in standard form (where terms are written from highest degree to lowest degree). Before attempting to factor this expression, we'll re-write it as \(2y^2+9y-5\text{.}\)

\begin{align*} \frac{9y+2y^2-5}{y^2-25}\amp=\frac{2y^2+9y-5}{y^2-25}\\ \amp=\frac{(2y-1)\cancelhighlight{(y+5)}}{\cancelhighlight{(y+5)}(y-5)}\\ \amp=\frac{2y-1}{y-5}, \text{ for } y\neq -5 \end{align*}

Example 8.2.12.

Simplify the expression \(\frac{-48z+24z^2-3z^3}{4-z}\text{.}\)

Explanation

To begin simplifying this expression, we will rewrite each polynomial in descending order. Then we'll factor out the GCF, including the constant \(-1\) from both the numerator and denominator because their leading terms are negative.

\begin{align*} \frac{-48z+24z^2-3z^3}{4-z}\amp=\frac{-3z^3+24z^2-48z}{-z+4}\\ \amp=\frac{-3z(z^2-8z+16)}{-(z-4)}\\ \amp=\frac{-3z(z-4)^2}{-(z-4)}\\ \amp=\frac{-3z(z-4)\cancelhighlight{(z-4)}}{-\cancelhighlight{(z-4)}}\\ \amp=\frac{3z(z-4)}{1}\\ \amp=3z(z-4), \text{ for } z\neq 4 \end{align*}

Example 8.2.13.

Simplify the rational function \(f(x)=\frac{8x^3-12x^2+8x-12}{2x^3-3x^2+10x-15}\) and state the domain of this function.

Explanation

To simplify this rational function, we'll first note that both the numerator and denominator have four terms. To factor them we'll need to use factoring by grouping. (Note that if this technique didn't work, very few other approaches would be possible.) Once we've used factoring by grouping, we'll cancel any factors common to both the numerator and denominator and state the associated restrictions.

\begin{align*} f(x)\amp=\frac{8x^3-12x^2+8x-12}{2x^3-3x^2+10x-15}\\ \amp=\frac{4(2x^3-3x^2+2x-3)}{2x^3-3x^2+10x-15}\\ \amp=\frac{4(x^2(2x-3)+(2x-3))}{x^2(2x-3)+5(2x-3)}\\ \amp=\frac{4\cancelhighlight{(2x-3)}(x^2+1)}{\cancelhighlight{(2x-3)}(x^2+5)}\\ \amp=\frac{4(x^2+1)}{x^2+5}, \text{ for }x\neq \frac{3}{2} \end{align*}

In determining the domain of this function, we'll need to account for any implicit and explicit restrictions. When the factor \(2x-3\) was canceled, the explicit statement of \(x\neq \frac{3}{2}\) was given. The denominator in the final simplified form of this function has \(x^2+5\text{.}\) There is no value of \(x\) for which \(x^2+5=0\text{,}\) so the only restriction is that \(x\neq \frac{3}{2}\text{.}\) Therefore, the domain is \(\left\{ x\mid x\neq \frac{3}{2}\right\}\text{.}\)

Example 8.2.14.

Simplify the expression \(\frac{3y-x}{x^2-xy-6y^2}\text{.}\) (In this example, there are two variables. It is still possible that in examples like this, there can be domain restrictions when simplifying rational expressions. However, since we are not studying functions of more than one variable, this textbook ignores domain restrictions with examples containing more than one variable like this one.)

Explanation

\begin{align*} \frac{3y-x}{x^2-xy-6y^2}\amp=\frac{-1\cancelhighlight{(x-3y)}}{\cancelhighlight{(x-3y)}(x+2y)}\\ \amp=\frac{-1}{x+2y}\\ \amp=-\frac{1}{x+2y} \end{align*}

Note that the expressions \(3y-x\) and \(x-3y\) are opposites, since they differ by a factor of \(-1\text{.}\) Binomials that are opposites can be easily identified as being of the form \(a-b\) and \(b-a\text{.}\) By factoring out a factor of \(-1\text{,}\) this enables us to identify a common factor that can be cancelled, which is what we did in the first step.

Exercises 8.2.4 Exercises

Rational Functions Exercises

1.

Select all rational functions. There are several correct answers.

\(\displaystyle s(x)=\frac{\sqrt{2} x^2 + 6 x - 5}{7 - 8 x^{2}}\)
\(\displaystyle h(x)=\frac{7}{2 x^2 + 6 x - 5}\)
\(\displaystyle r(x)=\frac{2 x^2 + 6 x - 5}{7 - 8 x^{-2}}\)
\(\displaystyle a(x)=\frac{2 x^2 + 6 x - 5}{7 - 8 x^{2}}\)
\(\displaystyle c(x)=\frac{2 x^2 + 6 x - 5}{7 + \lvert x \rvert }\)
\(\displaystyle b(x)=\frac{2 x^2 + 6 x - 5}{7}\)
\(\displaystyle t(x)=\frac{7 - 8 x^3}{2 x^{0.7} + 6 x - 5}\)
\(\displaystyle n(x)=\frac{2 x^2 + 6 \sqrt{x} - 5}{7 - 8 x^{2}}\)
\(\displaystyle m(x)=\frac{2 x + 6}{2 x +6}\)

2.

Select all rational functions. There are several correct answers.

\(\displaystyle h(x)=\frac{5}{3 x^2 + 3 x - 8}\)
\(\displaystyle c(x)=\frac{3 x^2 + 3 x - 8}{5 + \lvert x \rvert }\)
\(\displaystyle a(x)=\frac{3 x^2 + 3 x - 8}{5 - 8 x^{3}}\)
\(\displaystyle t(x)=\frac{5 - 8 x^3}{3 x^{0.7} + 3 x - 8}\)
\(\displaystyle m(x)=\frac{3 x + 3}{3 x +3}\)
\(\displaystyle s(x)=\frac{\sqrt{3} x^2 + 3 x - 8}{5 - 8 x^{3}}\)
\(\displaystyle r(x)=\frac{3 x^2 + 3 x - 8}{5 - 8 x^{-3}}\)
\(\displaystyle n(x)=\frac{3 x^2 + 3 \sqrt{x} - 8}{5 - 8 x^{3}}\)
\(\displaystyle b(x)=\frac{3 x^2 + 3 x - 8}{5}\)

3.

Find the domain of the function \(a\) defined by \(\displaystyle{a(x)=\frac{x - 4}{x^{5}}}\)

4.

Find the domain of the function \(m\) defined by \(\displaystyle{m(x)=\frac{x - 1}{x^{4}}}\)

5.

Find the domain of the function \(y\) defined by \(\displaystyle{y(x)=\frac{x+1}{x^2+36}}\)

6.

Find the domain of the function \(A\) defined by \(\displaystyle{A(x)=\frac{x+3}{x^2+64}}\)

7.

Find the domain of the function \(A\) defined by \(\displaystyle{A(x)=\frac{x+6}{x + 6}}\)

8.

Find the domain of the function \(r\) defined by \(\displaystyle{r(x)=\frac{x+8}{x + 8}}\)

Simplifying Rational Expressions with One Variable

9.

Select all correct simplifications, ignoring possible domain restrictions.

\(\displaystyle \frac{9 x+10}{9}=x+10\)
\(\displaystyle \frac{x+10}{x}=10\)
\(\displaystyle \frac{10}{x+10}=\frac{1}{x}\)
\(\displaystyle \frac{10}{x+10}=\frac{1}{x+1}\)
\(\displaystyle \frac{9 x+10}{x + 10}=9\)
\(\displaystyle \frac{x+ 10}{x+10}=1\)
\(\displaystyle \frac{x}{9 x}=\frac{1}{9}\)
\(\displaystyle \frac{10 x}{x}=10\)
\(\displaystyle \frac{9(x-10)}{x -10}=9\)
\(\displaystyle \frac{x+10}{10}=x\)
\(\displaystyle \frac{x+10}{x+9}=\frac{10}{9}\)

10.

Select all correct simplifications, ignoring possible domain restrictions.

\(\displaystyle \frac{3 x+2}{3}=x+2\)
\(\displaystyle \frac{x+2}{2}=x\)
\(\displaystyle \frac{x+2}{x}=2\)
\(\displaystyle \frac{3(x-2)}{x -2}=3\)
\(\displaystyle \frac{x}{3 x}=\frac{1}{3}\)
\(\displaystyle \frac{2 x}{x}=2\)
\(\displaystyle \frac{x+2}{x+3}=\frac{2}{3}\)
\(\displaystyle \frac{2}{x+2}=\frac{1}{x}\)
\(\displaystyle \frac{2}{x+2}=\frac{1}{x+1}\)
\(\displaystyle \frac{3 x+2}{x + 2}=3\)
\(\displaystyle \frac{x+ 2}{x+2}=1\)

11.

Simplify the following expressions, and if applicable, write the restricted domain on the simplified expression.

\(\displaystyle{{\frac{x+3}{x+3}}=}\)
\(\displaystyle{{\frac{x+3}{3+x}}=}\)
\(\displaystyle{{\frac{x-3}{x-3}}=}\)
\(\displaystyle{{\frac{x-3}{3-x}}=}\)

12.

Simplify the following expressions, and if applicable, write the restricted domain on the simplified expression.

\(\displaystyle{{\frac{y+8}{y+8}}=}\)
\(\displaystyle{{\frac{y+8}{8+y}}=}\)
\(\displaystyle{{\frac{y-8}{y-8}}=}\)
\(\displaystyle{{\frac{y-8}{8-y}}=}\)