4.5: Rational Functions - Multiplication and Division

Learning Objectives

• Identify restrictions to the domain of a rational function.
• Simplify rational functions.
• Multiply and divide rational functions.

Identifying Restrictions and Simplifying Rational Functions

Rational functions²⁵ have the form

$r ( x ) = \dfrac { p ( x ) } { q ( x ) }$,

where $p(x)$ and $q(x)$ are polynomials and $q(x)≠0$. The domain of a rational function²⁶ consists of all real numbers $x$ except those where the denominator  $q(x)=0$. Restrictions²⁷ are the real numbers for which the expression is not defined. We often express the domain of a rational function in terms of its restrictions. For example, consider the function

$f ( x ) = \dfrac { x ^ { 2 } - 4 x + 3 } { x ^ { 2 } - 5 x + 6 }$

which can be written in factored form

$f ( x ) = \dfrac { ( x - 1 ) ( x - 3 ) } { ( x - 2 ) ( x - 3 ) }$

Because rational expressions are undefined when the denominator is $0$, we wish to find the values for $x$ that make it $0$. To do this, apply the zero-product property. Set each factor in the denominator equal to $0$ and solve.

 $\begin{array} { c } { ( x - 2 )\: ( x - 3 ) = 0 } \\ { x - 2 = 0 \quad \text { or } \quad x - 3 = 0 } \\ { x = 3 }\quad\quad\quad\quad {x=3} \end{array}$

Therefore, the original function is defined for any real number except $2$ and $3$. We can express its domain using notation as follows:

$\begin{array} { l l } { \color{Cerulean} { Set-builder\: notation } } & { \color{Cerulean} { Interval \:notation } } \\ \{ x | x \neq 2,3 \} & {( - \infty , 2 ) \cup ( 2,3 ) \cup ( 3 , \infty ) } \end{array}$

The restrictions to the domain of a rational function are determined by the denominator. Once the restrictions are determined we can cancel factors and obtain an equivalent function as follows:

Figure $\PageIndex{1}$

It is important to note that $1$ is not a restriction to the domain because the expression is defined as $0$ when the numerator is $0$. In fact, $x=1$ is a root. This function is graphed below:

Figure $\PageIndex{2}$

Notice that there is a vertical asymptote at the restriction $x=2$ and the graph is left undefined at the restriction $x=3$ as indicated by the open dot, or hole, in the graph. Graphing rational functions in general is beyond the scope of this textbook. However, it is useful at this point to know that the restrictions are an important part of the graph of rational functions.

Example $\PageIndex{1}$:

State the restrictions and simplify: $g ( x ) = \dfrac { 24 x ^ { 7 } } { 6 x ^ { 5 } }$

Solution

In this example, the function is undefined where $x$ is $0$.

$g ( 0 ) = \dfrac { 24 ( 0 ) ^ { 7 } } { 6 ( 0 ) ^ { 5 } } = \dfrac { 0 } { 0 } \quad \color{Cerulean}{undefined}$

Therefore, the domain consists of all real numbers $x$, where $x≠0$. With this understanding, we can simplify by reducing the rational expression to lowest terms. Cancel common factors.

$g ( x ) = \dfrac { \stackrel{4\:\:\:\: x ^ { 2 }}{\cancel{24}\:\:\: \cancel{x^{7}}} } { \cancel{6}\:\:\:\:\: \cancel{x ^ { 8 }} } = 4 x ^ { 2 }$

Answer

$g ( x ) = 4 x ^ { 2 }$, where $x≠0$

Example $\PageIndex{2}$:

State the restrictions and simplify: $f ( x ) = \dfrac { 2 x ^ { 2 } + 5 x - 3 } { 4 x ^ { 2 } - 1 }$.

Solution

First, factor the numerator and denominator.

$f ( x ) = \dfrac { 2 x ^ { 2 } + 5 x - 3 } { 4 x ^ { 2 } - 1 } = \dfrac { ( 2 x - 1 ) ( x + 3 ) } { ( 2 x + 1 ) ( 2 x - 1 ) }$

Any $x$-value that makes the denominator zero is a restriction. To find the restrictions, first set the denominator equal to zero and then solve

$( 2 x + 1 ) ( 2 x - 1 ) = 0$

$\begin{array} { r l } { 2 x + 1 = 0 } & { \text { or } \quad 2 x - 1 = 0 } \\ { 2 x = - 1 } & \quad\quad\quad\quad{ 2 x = 1 } \\ { x = - \dfrac { 1 } { 2 } } &\quad\quad\quad\quad\: { x = \dfrac { 1 } { 2 } } \end{array}$

Therefore, $x≠±\dfrac{1}{2}$. With this understanding, we can cancel any common factors.

$\begin{aligned} f ( x ) & = \dfrac {\cancel{ ( 2 x - 1 )} ( x + 3 ) } { ( 2 x + 1 ) \cancel{( 2 x -1)} } \\ & = \dfrac { x + 3 } { 2 x + 1 } \end{aligned}$

Answer:

$f ( x ) = \dfrac { x + 3 } { 2 x + 1 }$, where $x \neq \pm \dfrac { 1 } { 2 }$

We define the opposite of a polynomial $P$ to be $−P$. Finding the opposite of a polynomial requires the application of the distributive property. For example, the opposite of the polynomial $(x−3)$ is written as

$\begin{aligned} - ( x - 3 ) & = - 1 \cdot ( x - 3 ) \\ & = - x + 3 \\ & = 3 - x \end{aligned}$

This leads us to the opposite binomial property²⁸, $−(a−b)=(b−a)$. Care should be taken not to confuse this with the fact that $(a+b)=(b+a)$. This is the case because addition is commutative. In general,

$\begin{array} { c |} -(a-b)=(b-a)\\\text{or}\\\dfrac{b-a}{a-b}=-1 \end{array} \begin{array}{c} (a-b)=(a+b)\\\text{or}\\ \dfrac{b+a}{a+b}=1\end{array}$

Also, it is important to recall that

$\dfrac { - a } { b } = - \dfrac { a } { b } = \dfrac { a } { - b }$

In other words, a negative fraction is shown by placing the negative sign in either the numerator, in front of the fraction bar, or in the denominator. Generally, negative denominators are avoided.

Example $\PageIndex{3}$:

State the restrictions and simplify: $\dfrac { 25 - x ^ { 2 } } { x ^ { 2 } - 10 x + 25 }$.

Solution

Begin by factoring the numerator and denominator.

$\begin{aligned} \dfrac { 25 - x ^ { 2 } } { x ^ { 2 } - 10 x + 25 } & = \dfrac { ( 5 - x ) ( 5 + x ) } { ( x - 5 ) ( x - 5 ) } \\ & = \dfrac { \color{Cerulean}{- 1 \cdot ( x - 5 ) }\color{black}{(} 5 + x ) } { ( x - 5 ) ( x - 5 ) }\quad\quad\quad\quad\color{Cerulean}{Opposite\:binomial\:property.} \\ & = \dfrac { - 1 \cdot \cancel{( x - 5 )} ( 5 + x ) } { \cancel{( x - 5 )} ( x - 5 ) } \quad\color{Cerulean}{Cancel.} \\ & = - \dfrac { x + 5 } { x - 5 } \end{aligned}$

Answer:

$- \dfrac { x + 5 } { x - 5 }$, where $x≠5$

It is important to remember that we can only cancel factors of a product. A common mistake is to cancel terms. For example,

$\dfrac {\cancel{ x ^ { 2} } + 7 x - 30 } { \cancel{x ^ { 2} } - 7 x + 12 } \\ \color{red}{incorrect!}$ $\begin{array} { c } { \dfrac { \cancel{x} + 10 } { \cancel{x} - 4 } } \\ { \color{red} { incorrect! } } \end{array}$ $\dfrac { 2 \cancel{x-1}} {\cancel{x-1} } \\ \color{red}{incorrect!}$

Exercise $\PageIndex{1}$

State the restrictions and simplify: $\dfrac { x - 2 x ^ { 2 } } { 4 x ^ { 4 } - x ^ { 2 } }$.

Answer

$- \dfrac { 1 } { x ( 2 x + 1 ) }$, where $x \neq 0 , \pm \dfrac { 1 } { 2 }$

www.youtube.com/v/Ek2qw7OgxbY

In some examples, we will make a broad assumption that the denominator is nonzero. When we make that assumption, we do not need to determine the restrictions.

Example $\PageIndex{4}$:

Simplify: $\dfrac { x ^ { 3 } - 2 x ^ { 2 } y + 4 x y ^ { 2 } - 8 y ^ { 3 } } { x ^ { 4 } - 16 y ^ { 4 } }$. (Assume all denominators are nonzero.)

Solution

Factor the numerator by grouping. Factor the denominator using the formula for a difference of squares.

$\begin{aligned} \dfrac { x ^ { 3 } + 4 x y ^ { 2 } - 2 x ^ { 2 } y - 8 y ^ { 3 } } { x ^ { 4 } - 16 y ^ { 4 } } & = \dfrac { x \left( x ^ { 2 } + 4 y ^ { 2 } \right) - 2 y \left( x ^ { 2 } + 4 y ^ { 2 } \right) } { \left( x ^ { 2 } + 4 y ^ { 2 } \right) \left( x ^ { 2 } - 4 y ^ { 2 } \right) } \\ & = \dfrac { \left( x ^ { 2 } + 4 y ^ { 2 } \right) ( x - 2 y ) } { \left( x ^ { 2 } + 4 y ^ { 2 } \right) ( x + 2 y ) ( x - 2 y ) } \end{aligned}$

Next, cancel common factors.

$= \dfrac { \stackrel{1}{\cancel{\left( x ^ { 2 } + 4 y ^ { 2 } \right)}}\stackrel{1}{ \cancel{\left( x - 2y \right)}} } { { \cancel{\left( x ^ { 2 } + 4 y ^ { 2 } \right) }} ( x + 2 y )\cancel{ ( x - 2 y )} } \\ = \dfrac{1}{x+2y}$

Note

When the entire numerator or denominator cancels out a factor of $1$ always remains.

Answer:

$\dfrac { 1 } { x + 2 y }$

Example $\PageIndex{5}$:

Given $f ( x ) = x ^ { 2 } - 2 x + 5$, simplify $\dfrac { f ( x ) - f ( 3 ) } { x - 3 }$.

Solution

Begin by calculating $f(3)$.

$\begin{aligned} f ( 3 ) & = ( 3 ) ^ { 2 } - 2 ( 3 ) + 5 \\ & = 9 - 6 + 5 \\ & = 3 + 5 \\ & = 8 \end{aligned}$

Next, substitute into the quotient that is to be simplified.

$\begin{aligned} \dfrac { f ( x ) - f ( 3 ) } { x - 3 } & = \dfrac { x ^ { 2 } - 2 x + 5 - 8 } { x - 3 } \\ & = \dfrac { x ^ { 2 } - 2 x - 3 } { x - 3 } \\ & = \dfrac { ( x + 1 ) ( x - 3 ) } { ( x - 3 ) } \\ & = x + 1 \end{aligned}$

Answer:

$x+1$, where $x≠3$

An important quantity in higher level mathematics is the difference quotient²⁹:

$\dfrac { f ( x + h ) - f ( x ) } { h }$, where $h \neq 0$

This quantity represents the slope of the line connecting two points on the graph of a function. The line passing through the two points is called a secant line³⁰.

Figure $\PageIndex{3}$

Calculating the difference quotient for many different functions is an important skill to learn in intermediate algebra. We will encounter this quantity often as we proceed in this textbook. When calculating the difference quotient we assume the denominator is nonzero.

Example $\PageIndex{6}$:

Given $g ( x ) = - 2 x ^ { 2 } + 1$, simplify $\dfrac { g ( x + h ) - g ( x ) } { h }$.

Solution

$\begin{aligned} \dfrac { g ( x + h ) - g ( x ) } { h } & = \dfrac { \left( - 2 ( x + h ) ^ { 2 } + 1 \right) - \left( - 2 x ^ { 2 } + 1 \right) } { h } \\ & = \dfrac { - 2 \left( x ^ { 2 } + 2 x h + h ^ { 2 } \right) + 1 + 2 x ^ { 2 } - 1 } { h } \\ &= { \dfrac { - 2 x ^ { 2 } - 4 x h - 2 h ^ { 2 } + 1 + 2 x ^ { 2 } - 1 } { h } } \\ & = \dfrac { - 4 x h - 2 h ^ { 2 } } { h } \\ & = - 4 x - 2 h \end{aligned}$

Answer:

$-4x-2h$

Exercise $\PageIndex{2}$

Given $f ( x ) = x ^ { 2 } - x - 1$, simplify $\dfrac { f ( x + h ) - f ( x ) } { h }$.

Answer

$2 x - 1 + h$

www.youtube.com/v/xYofIrdNwWs

Multiplying and Dividing Rational Functions

When multiplying fractions, we can multiply the numerators and denominators together and then reduce. Multiplying rational expressions is performed in a similar manner. In general, given polynomials $P, Q, R$, and $S$, where $Q≠0$ and $S≠0$, we have

$\dfrac { P } { Q } \cdot \dfrac { R } { S } = \dfrac { P R } { Q S }$

The restrictions to the domain of a product consist of the restrictions of each function.

Example $\PageIndex{7}$:

Given $f ( x ) = \dfrac { 9 x ^ { 2 } - 25 } { x - 5 }$ and $g ( x ) = \dfrac { x ^ { 2 } - 2 x - 15 } { 3 x + 5 }$, find $( f \cdot g ) ( x )$ and determine the restrictions to the domain.

Solution

In this case, the domain of $f$ consists of all real numbers except $5$, and the domain of $g$ consists of all real numbers except $−\dfrac{5}{3}$. Therefore, the domain of the product consists of all real numbers except $5$ and $−\dfrac{5}{3}$. Multiply the functions and then simplify the result.

$\begin{aligned} ( f \cdot g ) ( x ) & = f ( x ) \cdot g ( x ) \\ & = \dfrac { 9 x ^ { 2 } - 25 } { x - 5 } \cdot \dfrac { x ^ { 2 } - 2 x - 15 } { 3 x + 5 } \\ & = \dfrac { ( 3 x + 5 ) ( 3 x - 5 ) } { x - 5 } \cdot \dfrac { ( x - 5 ) ( x + 3 ) } { 3 x + 5 } \quad\quad\color{Cerulean}{Factor.} \\ & = \dfrac {\cancel{ ( 3 x + 5 )} ( 3 x - 5 ) \cancel{( x - 5 )} ( x + 3 ) } { \cancel{( x - 5 )}\cancel{ ( 3 x + 5 )} }\quad\quad\quad\color{Cerulean}{Cancel.} \\ & = ( 3 x - 5 ) ( x + 3 ) \end{aligned}$

Answer:

$( f \cdot g ) ( x ) = ( 3 x - 5 ) ( x + 3 )$, where $x \neq 5 , - \dfrac { 5 } { 3 }$

To divide two fractions, we multiply by the reciprocal of the divisor. Dividing rational expressions is performed in a similar manner. In general, given polynomials $P, Q, R$, and $S$, where $Q≠0, R≠0$, and $S≠0$, we have

$\dfrac { P } { Q } \div \dfrac { R } { S } = \dfrac { P } { Q } \cdot \dfrac { S } { R } = \dfrac { P S } { Q R }$

The restrictions to the domain of a quotient will consist of the restrictions of each function as well as the restrictions on the reciprocal of the divisor.

Example $\PageIndex{8}$:

Given $f ( x ) = \dfrac { 2 x ^ { 2 } + 13 x - 7 } { x ^ { 2 } - 4 x - 21 }$ and $g ( x ) = \dfrac { 2 x ^ { 2 } + 5 x - 3 } { 49 - x ^ { 2 } }$, find $( f / g ) ( x )$ and determine the restrictions.

Solution

$\begin{aligned} ( f / g ) ( x ) & = f ( x ) \quad \div g ( x ) \\ & = \dfrac { 2 x ^ { 2 } + 13 x - 7 } { x ^ { 2 } - 4 x - 21 } \div \dfrac { 2 x ^ { 2 } + 5 x - 3 } { 49 - x ^ { 2 } } \\ & = \dfrac { 2 x ^ { 2 } + 13 x - 7 } { x ^ { 2 } - 4 x - 21 } \cdot \dfrac { 49 - x ^ { 2 } } { 2 x ^ { 2 } + 5 x - 3 } \quad\quad\quad\color{Cerulean}{Multiply\:by\:the\:reciprocal} \\ & = \dfrac { ( 2 x - 1 ) ( x + 7 ) } { ( x + 3 ) ( x - 7 ) } \cdot \dfrac { ( 7 + x ) ( 7 - x ) } { ( 2 x - 1 ) ( x + 3 ) } \quad\color{Cerulean}{Factor.} \\ & = \dfrac { \cancel{(2x-1)}(x+7)(7+x)(-1)\cancel{(x-7)} } {(x+3)\cancel{(x-7)}\cancel{(2x-1)}(x+3) } \:\:\:\color{Cerulean}{Cancel.}\\ & = - \dfrac { ( x + 7 ) ^ { 2 } } { ( x + 3 ) ^ { 2 } } \end{aligned}$

In this case, the domain of $f$ consists of all real numbers except $−3$ and $7$, and the domain of $g$ consists of all real numbers except $7$ and $−7$. In addition, the reciprocal of $g(x)$ has a restriction of $−3$ and $\dfrac{1}{2}$. Therefore, the domain of this quotient consists of all real numbers except $−3, \dfrac{1}{2}$, and $±7$.

Answer:

$( f / g ) ( x ) = - \dfrac { ( x + 7 ) ^ { 2 } } { ( x + 3 ) ^ { 2 } }$, where $x \neq - 3 , \dfrac { 1 } { 2 } , \pm 7$

Recall that multiplication and division operations are to be performed from left to right.

Example $\PageIndex{9}$:

Simplify: $\dfrac { 4 x ^ { 2 } - 1 } { 6 x ^ { 2 } + 3 x } \div \dfrac { 2 x + 1 } { x ^ { 2 } + 2 x + 1 } \cdot \dfrac { 27 x ^ { 4 } } { 2 x ^ { 2 } + x - 1 }$ (Assume all denominators are nonzero.)

Solution

Begin by replacing the factor that is to be divided by multiplication of its reciprocal.

$\begin{array} { l } { \dfrac { 4 x ^ { 2 } - 1 } { 6 x ^ { 2 } + 3 x } \color{Cerulean}{\div \dfrac { 2 x + 1 } { x ^ { 2 } + 2 x + 1 }}\color{black}{ \cdot} \dfrac { 27 x ^ { 4 } } { 2 x ^ { 2 } + x - 1 } } \\ { = \dfrac { 4 x ^ { 2 } - 1 } { 6 x ^ { 2 } + 3 x } \cdot \dfrac { x ^ { 2 } + 2 x + 1 } { 2 x + 1 } \cdot \dfrac { 27 x ^ { 4 } } { 2 x ^ { 2 } + x - 1 } } \\ { = \dfrac { ( 2 x + 1 ) ( 2 x - 1 ) } { 3 x ( 2 x + 1 ) } \cdot \dfrac { ( x + 1 ) ( x + 1 ) } { ( 2 x + 1 ) } \cdot \dfrac { 27 x ^ { 3 } } { ( 2 x - 1 ) ( x + 1 ) } } \\ = \dfrac{\cancel{(2x+1)}\cancel{(2x-1)}\cancel{(x+1)}(x+1)\cdot \stackrel{9}{\cancel{27}} \stackrel{x^{3}}{\cancel{x^{4}}}}{\cancel{3}\cancel{ x}\cancel{ (2x+1)}(2x+1)\cancel{(2x-1)}\cancel{(x+1)}} \\ { = \dfrac { 9 x ^ { 3 } ( x + 1 ) } { ( 2 x + 1 ) } } \end{array}$

Answer:

$\dfrac { 9 x ^ { 3 } ( x + 1 ) } { ( 2 x + 1 ) }$

Exercise $\PageIndex{3}$

Given $f ( x ) = \dfrac { 2 x + 5 } { 3 x ^ { 2 } + 14 x - 5 }$ and $g ( x ) = \dfrac { 6 x ^ { 2 } + 13 x - 5 } { x + 5 }$, calculate $( f / g ) ( x )$ and determine the restrictions.

Answer

$( f / g ) ( x ) = \dfrac { 1 } { ( 3 x - 1 ) ^ { 2 } }$, where $x \neq - 5 , - \dfrac { 5 } { 2 } , \dfrac { 1 } { 3 }$

www.youtube.com/v/UCPL1ZQIvWE

If a cost function $C$ represents the cost of producing $x$ units, then the average cost³¹ $\overline{C}$ is the cost divided by the number of units produced.

$\overline { C } ( x ) = \dfrac { C ( x ) } { x }$

Example $\PageIndex{10}$:

A manufacturer has determined that the cost in dollars of producing sweaters is given by \9C ( x ) = 0.01 x ^ { 2 } - 3 x + 1200\), where $x$ represents the number of sweaters produced daily. Determine the average cost of producing $100, 200$, and $300$ sweaters per day.

Solution

Set up a function representing the average cost.

$\overline { C } ( x ) = \dfrac { C ( x ) } { x } = \dfrac { 0.01 x ^ { 2 } - 3 x + 1200 } { x }$

Next, calculate $\overline { C } (100)$, $\overline { C } (200)$, and $\overline { C } (300)$.

$\begin{array} { l } { \overline { C } ( 100 ) = \dfrac { 0.01 ( \color{OliveGreen}{100}\color{black}{ )} ^ { 2 } - 3 (\color{OliveGreen}{ 100}\color{black}{ )} + 1200 } { (\color{OliveGreen}{100}\color{black}{ )} } = \dfrac { 100 - 300 + 1200 } { 100 } = \dfrac { 1000 } { 100 } = 10.00 } \\ { \overline { C } ( 200 ) = \dfrac { 0.01 ( \color{OliveGreen}{200}\color{black}{ )} ^ { 2 } - 3 ( \color{OliveGreen}{200}\color{black}{ )} + 1200 } { ( \color{OliveGreen}{200} ) } = \dfrac { 400 - 600 + 1200 } { 200 } = \dfrac { 1000 } { 200 } = 5.00 } \\ { \overline { C } ( 300 ) = \dfrac { 0.01 ( \color{OliveGreen}{300}\color{black}{ )} ^ { 2 } - 3 ( \color{OliveGreen}{300}\color{black}{ )} + 1200 } { ( \color{OliveGreen}{300}\color{black}{ )} } = \dfrac { 900 - 900 + 1200 } { 300 } = \dfrac { 1200 } { 300 } = 4.00 } \end{array}$

Answer:

The average cost of producing $100$ sweaters per day is $$10.00$ per sweater. If 200 sweaters are produced, the average cost per sweater is $$5.00$. If $300$ are produced, the average cost per sweater is $$4.00$.

Key Takeaways

• Simplifying rational expressions is similar to simplifying fractions. First, factor the numerator and denominator and then cancel the common factors. Rational expressions are simplified if there are no common factors other than 1 in the numerator and the denominator.
• Simplified rational functions are equivalent for values in the domain of the original function. Be sure to state the restrictions unless the problem states that the denominators are assumed to be nonzero.
• After multiplying rational expressions, factor both the numerator and denominator and then cancel common factors. Make note of the restrictions to the domain. The values that give a value of $0$ in the denominator for all expressions are the restrictions.
• To divide rational expressions, multiply the numerator by the reciprocal of the divisor.
• The restrictions to the domain of a product consist of the restrictions to the domain of each factor.

Exercise $\PageIndex{4}$

Simplify the function and state its domain using interval notation.

1. $f ( x ) = \dfrac { 25 x ^ { 9 } } { 5 x ^ { 5 } }$
2. $f ( x ) = \dfrac { 64 x ^ { 8 } } { 16 x ^ { 3 } }$
3. $f ( x ) = \dfrac { x ^ { 2 } - 64 } { x ^ { 2 } + 16 x + 64 }$
4. $f ( x ) = \dfrac { x ^ { 2 } + x - 20 } { x ^ { 2 } - 25 }$
5. $g ( x ) = \dfrac { 9 - 4 x ^ { 2 } } { 2 x ^ { 2 } - 5 x + 3 }$
6. $g ( x ) = \dfrac { x - 3 x ^ { 2 } } { 9 x ^ { 2 } - 6 x + 1 }$
7. $g ( x ) = \dfrac { 2 x ^ { 2 } - 8 x - 42 } { 2 x ^ { 2 } + 5 x - 3 }$
8. $g ( x ) = \dfrac { 6 x ^ { 2 } + 5 x - 4 } { 3 x ^ { 2 } + x - 4 }$
9. $h ( x ) = \dfrac { x ^ { 3 } + x ^ { 2 } - x - 1 } { x ^ { 2 } + 2 x + 1 }$
10. $h ( x ) = \dfrac { 2 x ^ { 3 } - 5 x ^ { 2 } - 8 x + 20 } { 2 x ^ { 2 } - 9 x + 10 }$

Answer

1. $f ( x ) = 5 x ^ { 4 }$; Domain: $( - \infty , 0 ) \cup ( 0 , \infty )$

3. $f ( x ) = \dfrac { x - 8 } { x + 8 }$; Domain: $( - \infty , - 8 ) \cup ( - 8 , \infty )$

5. $g ( x ) = - \dfrac { 2 x + 3 } { x - 1 }$; Domain: $( - \infty , 1 ) \cup \left( 1 , \dfrac { 3 } { 2 } \right) \cup \left( \dfrac { 3 } { 2 } , \infty \right)$

7. $g ( x ) = \dfrac { 2 ( x - 7 ) } { 2 x - 1 }$; Domain: $( - \infty , - 3 ) \cup \left( - 3 , \dfrac { 1 } { 2 } \right) \cup \left( \dfrac { 1 } { 2 } , \infty \right)$

9. $h ( x ) = x - 1$; Domain: $( - \infty , - 1 ) \cup ( - 1 , \infty )$

Exercise $\PageIndex{5}$

State the restrictions and simplify the given rational expressions.

1. $\dfrac { 66 x ( 2 x - 5 ) } { 18 x ^ { 3 } ( 2 x - 5 ) ^ { 2 } }$
2. $\dfrac { 26 x ^ { 4 } ( 5 x + 2 ) ^ { 3 } } { 20 x ^ { 5 } ( 5 x + 2 ) }$
3. $\dfrac { x ^ { 2 } + 5 x + 6 } { x ^ { 2 } - 5 x - 14 }$
4. $\dfrac { x ^ { 2 } - 8 x + 12 } { x ^ { 2 } - 2 x - 24 }$
5. $\dfrac { 1 - x ^ { 2 } } { 5 x ^ { 2 } + x - 6 }$
6. $\dfrac { 4 - 9 x ^ { 2 } } { 3 x ^ { 2 } - 8 x + 4 }$
7. $\dfrac { 4 x ^ { 2 } + 15 x + 9 } { 9 - x ^ { 2 } }$
8. $\dfrac { 6 x ^ { 2 } + 13 x - 5 } { 25 - 4 x ^ { 2 } }$
9. $\dfrac { x ^ { 2 } - 5 x + 4 } { x ^ { 3 } - x ^ { 2 } - 16 x + 16 }$
10. $\dfrac { x ^ { 4 } + 4 x ^ { 2 } } { x ^ { 3 } + 3 x ^ { 2 } + 4 x + 12 }$

Answer

1. $\dfrac { 11 } { 3 x ^ { 2 } ( 2 x - 5 ) } ; x \neq 0 , \dfrac { 5 } { 2 }$

3. $\dfrac { x + 3 } { x - 7 } ; x \neq - 2,7$

5. $- \dfrac { x + 1 } { 5 x + 6 } ; x \neq - \dfrac { 6 } { 5 } , 1$

7. $- \dfrac { 4 x + 3 } { 3 - x } ; x \neq \pm 3$

9. $\dfrac { 1 } { x + 4 } ; x \neq 1 , \pm 4$

Exercise $\PageIndex{6}$

Simplify the given rational expressions. Assume all variable expressions in the denominator are nonzero.

1. $\dfrac { 50 a b ^ { 3 } ( a + b ) ^ { 2 } } { 200 a ^ { 2 } b ^ { 3 } ( a + b ) ^ { 3 } }$
2. $\dfrac { 36 a ^ { 5 } b ^ { 7 } ( a - b ) ^ { 2 } } { 9 a ^ { 3 } b ( a - b ) }$
3. $\dfrac { a ^ { 2 } - b ^ { 2 } } { a ^ { 2 } + 2 a b + b ^ { 2 } }$
4. $\dfrac { a ^ { 2 } - 2 a b + b ^ { 2 } } { a ^ { 2 } - b ^ { 2 } }$
5. $\dfrac { 6 x ^ { 2 } - x y } { 6 x ^ { 2 } - 7 x y + y ^ { 2 } }$
6. $\dfrac { y - x } { 2 x ^ { 3 } - 4 x ^ { 2 } y + 2 x y ^ { 2 } }$
7. $\dfrac { x ^ { 2 } y ^ { 2 } - 2 x y ^ { 3 } } { x ^ { 2 } y ^ { 2 } - x y ^ { 3 } - 2 y ^ { 4 } }$
8. $\dfrac { x ^ { 4 } y - x ^ { 2 } y ^ { 3 } } { x ^ { 3 } y + 2 x ^ { 2 } y ^ { 2 } + x y ^ { 3 } }$
9. $\dfrac { x ^ { 3 } - x ^ { 2 } y + x y ^ { 2 } - y ^ { 3 } } { x ^ { 4 } - y ^ { 4 } }$
10. $\dfrac { y ^ { 4 } - x ^ { 4 } } { x ^ { 3 } + x ^ { 2 } y + x y ^ { 2 } + y ^ { 3 } }$
11. $\dfrac { a ^ { 2 } - ( b + c ) ^ { 2 } } { ( a + b ) ^ { 2 } - c ^ { 2 } }$
12. $\dfrac { ( a + b ) ^ { 2 } - c ^ { 2 } } { ( a + c ) ^ { 2 } - b ^ { 2 } }$
13. $\dfrac { x ^ { 3 } + y ^ { 3 } } { x ^ { 2 } + 2 x y + y ^ { 2 } }$
14. $\dfrac { x ^ { 3 } y + x ^ { 2 } y ^ { 2 } + x y ^ { 3 } } { x ^ { 3 } - y ^ { 3 } }$

Answer

1. $\dfrac { 1 } { 4 a ( a + b ) }$

3. $\dfrac { a - b } { a + b }$

5. $\dfrac { x } { x - y }$

7. $\dfrac { x } { x + y }$

9. $\dfrac { 1 } { x + y }$

11. $\dfrac { a - b - c } { a + b - c }$

13. $\dfrac { x ^ { 2 } - x y + y ^ { 2 } } { x + y }$

Exercise $\PageIndex{7}$

Given the function, simplify the rational expression.

1. Given $f ( x ) = x ^ { 2 } - 8$, simplify $\dfrac { f ( x ) - f ( 5 ) } { x - 5 }$.
2. Given $f ( x ) = x ^ { 2 } + 4 x - 1$, simplify $\dfrac { f ( x ) - f ( 2 ) } { x - 2 }$.
3. Given $g ( x ) = x ^ { 2 } - 3 x + 1$, simplify $\dfrac { g ( x ) - g ( - 1 ) } { x + 1 }$.
4. Given $g ( x ) = x ^ { 2 } - 2 x$, simplify $\dfrac { g ( x ) - g ( - 4 ) } { x + 4 }$.
5. Given $f ( x ) = 4 x ^ { 2 } + 6 x + 1$, simplify $\dfrac { f ( x ) - f \left( \dfrac { 1 } { 2 } \right) } { 2 x - 1 }$.
6. Given $f ( x ) = 9 x ^ { 2 } + 1$, simplify $\dfrac { f ( x ) - f \left( - \dfrac { 1 } { 3 } \right) } { 3 x + 1 }$.

Answer

1. $x+5$, where $x \neq 5$

3. $x-4$, where $x \neg -1$

5. $2(x+2)$, where $x \neq \dfrac{1}{2}$

Exercise $\PageIndex{8}$

For the given function, simplify the difference quotient $\dfrac { f ( x + h ) - f ( x ) } { h }$, where $h\neq 0$.

1. $f ( x ) = 5 x - 3$
2. $f ( x ) = 3 - 2 x$
3. $f ( x ) = x ^ { 2 } - 3$
4. $f ( x ) = x ^ { 2 } + 8 x$
5. $f ( x ) = x ^ { 2 } - x + 5$
6. $f ( x ) = 4 x ^ { 2 } + 3 x - 2$
7. $f ( x ) = a x ^ { 2 } + b x + c$
8. $f ( x ) = a x ^ { 2 } + b x$
9. $f ( x ) = x ^ { 3 } + 1$
10. $f ( x ) = x ^ { 3 } - x + 2$

Answer

1. $5$

3. $2x+h$

5. $2x-1+h$

7. $2ax+b+ah$

9. $3 x ^ { 2 } + 3 x h + h ^ { 2 }$

Exercise $\PageIndex{9}$

Simplify the product $f \cdot g$ and state its domain using interval notation.

1. $f ( x ) = \dfrac { 52 x ^ { 4 } } { ( x - 2 ) ^ { 2 } } , g ( x ) = \dfrac { ( x - 2 ) ^ { 3 } } { 12 x ^ { 5 } }$
2. $f ( x ) = \dfrac { 46 ( 2 x - 1 ) ^ { 3 } } { 15 x ^ { 6 } } , g ( x ) = \dfrac { 25 x ^ { 3 } } { 23 ( 2 x - 1 ) }$
3. $f ( x ) = \dfrac { 10 x ^ { 3 } } { x ^ { 2 } + 4 x + 4 } , g ( x ) = \dfrac { x ^ { 2 } - 4 } { 50 x ^ { 4 } }$
4. $f ( x ) = \dfrac { 25 - x ^ { 2 } } { 46 x ^ { 5 } } , g ( x ) = \dfrac { 12 x ^ { 3 } } { x ^ { 2 } + 10 x + 25 }$
5. $f ( x ) = \dfrac { 5 - 3 x } { x ^ { 2 } - 10 x + 25 } , g ( x ) = \dfrac { x ^ { 2 } - 6 x + 5 } { 3 x ^ { 2 } - 8 x + 5 }$
6. $f ( x ) = \dfrac { 1 - 4 x ^ { 2 } } { 6 x ^ { 2 } + 3 x } , g ( x ) = \dfrac { 12 x ^ { 2 } } { 4 x ^ { 2 } - 4 x + 1 }$

Answer

1. $( f \cdot g ) ( x ) = \dfrac { 13 ( x - 2 ) } { 3 x }$; Domain: $( - \infty , 0 ) \cup ( 0,2 ) \cup ( 2 , \infty )$

3. $( f \cdot g ) ( x ) = \dfrac { x - 2 } { 5 x ( x + 2 ) }$; Domain: $( - \infty , - 2 ) \cup ( - 2,0 ) \cup ( 0 , \infty )$

5. $( f \cdot g ) ( x ) = - \dfrac { 1 } { x - 5 }$; Domain: $( - \infty , 1 ) \cup \left( 1 , \dfrac { 5 } { 3 } \right) \cup \left( \dfrac { 5 } { 3 } , 5 \right) \cup ( 5 , \infty )$

Exercise $\PageIndex{10}$

Simplify the quotient $f/g$ and state its domain using interval notation.

1. $f ( x ) = \dfrac { 12 x ^ { 3 } } { 5 ( 5 x - 1 ) ^ { 3 } } , g ( x ) = \dfrac { 6 x ^ { 2 } } { 25 ( 5 x - 1 ) ^ { 4 } }$
2. $f ( x ) = \dfrac { 7 x ^ { 2 } ( x + 9 ) } { ( x - 8 ) ^ { 2 } } , g ( x ) = \dfrac { 49 x ^ { 3 } ( x + 9 ) } { ( x - 8 ) ^ { 4 } }$
3. $f ( x ) = \dfrac { 25 x ^ { 2 } - 1 } { 3 x ^ { 2 } - 15 x } , g ( x ) = \dfrac { 25 x ^ { 2 } + 10 x + 1 } { x ^ { 3 } - 5 x ^ { 2 } }$
4. $f ( x ) = \dfrac { x ^ { 2 } - x - 6 } { 2 x ^ { 2 } + 13 x + 15 } , g ( x ) = \dfrac { x ^ { 2 } - 6 x + 9 } { 4 x ^ { 2 } + 12 x + 9 }$
5. $f ( x ) = \dfrac { x ^ { 2 } - 64 } { x ^ { 2 } } , g ( x ) = 2 x ^ { 2 } + 19 x + 24$
6. $f ( x ) = 2 x ^ { 2 } + 11 x - 6 , g ( x ) = 36 - x ^ { 2 }$

Answer

1. $( f / g ) ( x ) = 10 x ( 5 x - 1 )$; Domain: $( - \infty , 0 ) \cup \left( 0 , \dfrac { 1 } { 5 } \right) \cup \left( \dfrac { 1 } { 5 } , \infty \right)$

3. $( f / g ) ( x ) = \dfrac { x ( 5 x - 1 ) } { 2 ( 5 x + 1 ) }$; Domain: $( - \infty , 0 ) \cup \left( 0 , \dfrac { 1 } { 5 } \right) \cup \left( \dfrac { 1 } { 5 } , \infty \right)$

5. $( f / g ) ( x ) = \dfrac { x - 8 } { x ^ { 2 } ( 2 x + 3 ) }$; Domain: $( - \infty , - 8 ) \cup \left( - 8 , - \dfrac { 3 } { 2 } \right) \cup \left( - \dfrac { 3 } { 2 } , 0 \right) \cup ( 0 , \infty )$

Exercise $\PageIndex{11}$

Multiply or divide as indicated, state the restrictions, and simplify.

1. $\dfrac { 14 ( x + 12 ) ^ { 2 } } { 5 x ^ { 3 } } \cdot \dfrac { 45 x ^ { 4 } } { 2 ( x + 12 ) ^ { 3 } }$
2. $\dfrac { 27 x ^ { 6 } } { 20 ( x - 7 ) ^ { 3 } } \cdot \dfrac { ( x - 7 ) ^ { 5 } } { 54 x ^ { 7 } }$
3. $\dfrac { x ^ { 2 } - 64 } { 36 x ^ { 4 } } \cdot \dfrac { 12 x ^ { 3 } } { x ^ { 2 } + 4 x - 32 }$
4. $\dfrac { 50 x ^ { 5 } } { x ^ { 2 } + 6 x - 27 } \cdot \dfrac { x ^ { 2 } - 81 } { 125 x ^ { 3 } }$
5. $\dfrac { 2 x ^ { 2 } + 7 x + 5 } { 3 x ^ { 2 } } \cdot \dfrac { 15 x ^ { 3 } - 30 x ^ { 2 } } { 2 x ^ { 2 } + x - 10 }$
6. $\dfrac { 3 x ^ { 2 } + 14 x - 5 } { 2 x ^ { 2 } + 11 x + 5 } \cdot \dfrac { 4 x ^ { 2 } + 4 x + 1 } { 6 x ^ { 2 } + x - 1 }$
7. $\dfrac { x ^ { 2 } + 4 x - 21 } { 5 x ^ { 2 } + 10 x } \div \dfrac { x ^ { 2 } - 6 x + 9 } { x ^ { 2 } + 9 x + 14 }$
8. $\dfrac { x ^ { 2 } - 49 } { 9 x ^ { 2 } - 24 x + 16 } \div \dfrac { 2 x ^ { 2 } - 13 x - 7 } { 6 x ^ { 2 } - 5 x - 4 }$
9. $\dfrac { 5 x ^ { 2 } + x - 6 } { 4 x ^ { 2 } - 7 x - 15 } \div \dfrac { 1 - x ^ { 2 } } { 4 x ^ { 2 } + 9 x + 5 }$
10. $\dfrac { 6 x ^ { 2 } - 8 x - 8 } { 4 - 9 x ^ { 2 } } \div \dfrac { 3 x ^ { 2 } - 4 x - 4 } { 9 x ^ { 2 } + 12 x + 4 }$
11. $\dfrac { x ^ { 2 } + 4 x - 12 } { x ^ { 2 } - 2 x - 15 } \div \dfrac { 2 x ^ { 2 } - 13 x + 18 } { 6 x ^ { 2 } - 31 x + 5 }$
12. $\dfrac { 8 x ^ { 2 } + x - 9 } { 25 x ^ { 2 } - 1 } \div \dfrac { 2 x ^ { 2 } - x - 1 } { 10 x ^ { 2 } - 3 x - 1 }$

Answer

1. $\dfrac { 63 x } { x + 12 } ; x \neq - 12,0$

3. $\dfrac { x - 8 } { 3 x ( x - 4 ) } ; x \neq - 8,0,4$

5. $5 ( x + 1 ) ; x \neq - \dfrac { 5 } { 2 } , 0,2$

7. $\dfrac { ( x + 7 ) ^ { 2 } } { 5 x ( x - 3 ) } ; x \neq - 7 , - 2,0,3$

9. $- \dfrac { 5 x + 6 } { x - 3 } ; x \neq - \dfrac { 5 } { 4 } , - 1,1,3$

11. $\dfrac { ( x + 6 ) ( 6 x - 1 ) } { ( x + 3 ) ( 2 x - 9 ) } ; x \neq - 3 , \dfrac { 1 } { 6 } , 2 , \dfrac { 9 } { 2 } , 5$

Exercise $\PageIndex{12}$

Perform the operations and simplify. Assume all variable expressions in the denominator are nonzero.

1. $\dfrac { 1 } { 12 a b } \cdot \dfrac { 50 a ^ { 2 } ( a - b ) ^ { 2 } } { a ^ { 2 } - b ^ { 2 } } \cdot \dfrac { 6 b } { a ( a - b ) }$
2. $\dfrac { b ^ { 2 } - a ^ { 2 } } { ( a - b ) ^ { 2 } } \cdot \dfrac { 12 a ( a - b ) } { 36 a ^ { 2 } b } \cdot \dfrac { 9 a b ( a - b ) } { a + b }$
3. $\dfrac { x ^ { 3 } + y ^ { 3 } } { 5 x y } \cdot \dfrac { x ^ { 2 } - y ^ { 2 } } { x ^ { 2 } - 2 x y + y ^ { 2 } } \cdot \dfrac { 25 x ^ { 2 } y } { ( y + x ) ^ { 2 } }$
4. $\dfrac { 3 x y ^ { 2 } } { ( 2 y + x ) ^ { 2 } } \cdot \dfrac { 2 x ^ { 2 } + 5 x y + 2 y ^ { 2 } } { 9 x ^ { 2 } } \cdot \dfrac { x ^ { 3 } + 8 y ^ { 3 } } { 6 x y ^ { 2 } + 3 y ^ { 3 } }$
5. $\dfrac { 2 x + 5 } { x - 3 } \cdot \dfrac { x ^ { 2 } - 9 } { 5 x ^ { 4 } } \div \dfrac { 2 x ^ { 2 } + 15 x + 25 } { 25 x ^ { 5 } }$
6. $\dfrac { 5 x ^ { 2 } - 15 x } { 9 x ^ { 2 } - 4 } \cdot \dfrac { 3 x - 2 } { 20 x ^ { 3 } } \div \dfrac { x - 3 } { 3 x ^ { 2 } - x - 2 }$
7. $\dfrac { x ^ { 2 } + 5 x - 50 } { x ^ { 2 } + 5 x - 14 } \div \dfrac { x ^ { 2 } - 25 } { x ^ { 2 } - 49 } \cdot \dfrac { x - 2 } { x ^ { 2 } + 3 x - 70 }$
8. $\dfrac { x ^ { 2 } - x - 56 } { 4 x ^ { 2 } - 4 x - 3 } \div \dfrac { 2 x ^ { 2 } + 11 x - 21 } { 25 - 9 x ^ { 2 } } \cdot \dfrac { 4 x ^ { 2 } - 12 x + 9 } { 3 x ^ { 2 } - 19 x - 40 }$
9. $\dfrac { 20 x ^ { 2 } - 8 x - 1 } { 6 x ^ { 2 } + 13 x + 6 } \div \dfrac { 1 - 100 x ^ { 2 } } { 3 x ^ { 2 } - x - 2 } \cdot \dfrac { 10 x - 1 } { 2 x ^ { 2 } - 3 x + 1 }$
10. $\dfrac { 12 x ^ { 2 } - 13 x + 1 } { x ^ { 2 } + 18 x + 81 } \div \left( 144 x ^ { 2 } - 1 \right) \cdot \dfrac { x ^ { 2 } + 14 x + 45 } { 12 x ^ { 2 } - 11 x - 1 }$
11. A manufacturer has determined that the cost in dollars of producing bicycles is given by $C (x) = 0.5x^{ 2} − x + 6200$, where $x$ represents the number of bicycles produced weekly. Determine the average cost of producing $50, 100$, and $150$ bicycles per week.
12. The cost in dollars of producing custom lighting fixtures is given by the function $C (x) = x^{2} − 20x + 1200$, where $x$ represents the number of fixtures produced in a week. Determine the average cost per unit if $20, 40$, and $50$ units are produced in a week.
13. A manufacturer has determined that the cost in dollars of producing electric scooters is given by the function $C (x) = 3x (x − 100) + 32,000$, where $x$ represents the number of scooters produced in a month. Determine the average cost per scooter if $50$ are produced in a month.
14. The cost in dollars of producing a custom injected molded part is given by $C (n) = 1,900 + 0.01n$, where $n$ represents the number of parts produced. Calculate the average cost of each part if $2,500$ custom parts are ordered.
15. The cost in dollars of an environmental cleanup is given by the function $C ( p ) = \dfrac { 25,000 p } { 1 - p }$, where $p$ represents the percentage of the area to be cleaned up $(0 ≤ p < 1)$. Use the function to determine the cost of cleaning up $50$% of an affected area and the cost of cleaning up $80$% of the area.
16. The value of a new car is given by the function $V (t) = 16,500(t + 1)^{−1}$ where $t$ represents the age of the car in years. Determine the value of the car when it is $6$ years old.

Answer

1. $\dfrac { 25 } { a + b }$

3. $\dfrac { 5 x \left( x ^ { 2 } - x y + y ^ { 2 } \right) } { x - y }$

5. $\dfrac { 5 x ( x + 3 ) } { x + 5 }$

7. $\dfrac { 1 } { x + 5 }$

9. $- \dfrac { 1 } { 2 x + 3 }$

11. If $50$ bicycles are produced, the average cost per bicycle is $$148$. If $100$ are produced, the average cost is $$111$. If $150$ bicycles are produced, the average cost is $$115.33$.

13. If $50$ scooters are produced, the average cost of each is $$490$.

15. A $50$% cleanup will cost $$25,000$. An $80$% cleanup will cost $$100,000$.

Exercise $\PageIndex{13}$

1. Describe the restrictions to the rational expression $\dfrac { 1 } { x ^ { 2 } - y ^ { 2 } }$. Explain.
2. Describe the restrictions to the rational expression $\dfrac { 1 } { x ^ { 2 } + y ^ { 2 } }$. Explain.
3. Explain why $x = 5$ is a restriction to $\dfrac { 1 } { x + 5 } \div \dfrac { x - 5 } { x }$.
4. Explain to a beginning algebra student why we cannot cancel $x$ in the rational expression $\dfrac { x + 2 } { x }$.
5. Research and discuss the importance of the difference quotient. What does it represent and in what subject does it appear?

Answer

1. Answer may vary

3. Answer may vary

5. Answer may vary

Footnotes

²⁵Functions of the form $r ( x ) = \dfrac { p ( x ) } { q ( x ) }$, where $p(x)$ and $q(x)$ are polynomials and $q(x) ≠ 0$.

²⁶The set of real numbers for which the rational function is defined.

²⁷The set of real numbers for which a rational function is not defined.

²⁸If given a binomial $a-b$, then the opposite is $- ( a - b ) = b - a$.

²⁹The mathematical quantity $\dfrac { f ( x + h ) - f ( x ) } { h }$, where $h \neq 0$, which represents the slope of a secant line through a function $f$.

³⁰Line that intersects two points on the graph of a function.

³¹The total cost divided by the number of units produced, which can be represented by $\overline { C } ( x ) = \dfrac { C ( x ) } { x }$, where $C(x)$ is a cost function.