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SectionAlgebraic Fractions

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SubsectionIdentifying Restrictions and Simplifying

Algebraic fractions, or rational functions, have the form

\begin{equation*}
r(x)=\frac{p(x)}{q(x)}\text{,}
\end{equation*}

where \(p(x)\) and \(q(x)\) are polynomials and \(q(x)\neq 0\text{.}\) The domain of an algebraic fraction will be identified with the domain or the rational function it defines and consists of all real numbers \(x\) except those where the denominator \(q(x)=0\text{.}\) Restrictions on the domain are real numbers for which the expression is undefined. For example, consider the function

\begin{equation*}
f(x)=\frac{(x-1)(x-3)}{(x-2)(x-3)}\text{.}
\end{equation*}

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

\begin{equation*}
\begin{aligned}
x-2\amp =0 \: \: \: \: \: \amp \text{ or }\: \: \: \: \: x-3\amp =0\\
x\amp =2\amp x\amp =3
\end{aligned}
\end{equation*}

Therefore, the original function is defined for any real number except \(2\) and \(3\text{.}\) We can express its domain using notation.

\begin{equation*}
x\neq 2,3 \: \: \: \: \: \text{ or }\: \: \: \: \: (-\infty, 2)\cup(2,3)\cup(3,\infty)
\end{equation*}

The restrictions to the domain of an algebraic fraction are determined by the denominator. Once the restrictions are determined we can cancel factors and obtain an equivalent function.

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 \(0\text{.}\) In fact, \(x=1\) is a root. This function is graphed below.

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 algebraic fractions in general is beyond the scope of this course. However, it is useful at this point to know that the restrictions are an important part of the graph of algebraic fractions.

######
Example167

State the restrictions and simplify the algebraic fraction \(g(x)=\frac{24x^7}{6x^5}\text{.}\)

SolutionSince \(g(x)\) is an algebraic fraction, it is undefined, or restricted, where its denominator, \(6x^5\) is zero. This means the function is undefined where \(x\) is \(0\text{.}\)

\begin{equation*}
g(0)=\frac{24(0)^7}{6(0)^5}=\frac{0}{0} \: \: \: \alert{\text{Undefined}}
\end{equation*}

Therefore, the domain consists of all real numbers \(x\text{,}\) where \(x\neq 0\text{.}\) With this understanding, we can simplify by reducing the rational expression to lowest terms. Cancel common factors.

\begin{equation*}
g(x)=\frac{24x^7}{6x^5}=\frac{4\cdot 6 x^2\cdot x^5}{6x^5}=4x^2
\end{equation*}

Our answer is \(g(x)=4x^2\text{,}\) where \(x\neq 0\text{.}\)

######
Example168

State the restrictions and simplify the algebraic fraction \(g(x)=\frac{2x^2+5x-3}{4x^2-1}\text{.}\)

SolutionFirst, we factor the numerator and denominator.

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

Since \(g(x)\) is an algebraic fraction, it is undefined, or restricted, where its denominator is zero.

\begin{equation*}
\begin{aligned}
2x+1\amp =0 \: \: \: \: \: \amp \text{ or }\: \: \: \: \: 2x-1\amp =0\\
2x\amp =-1\amp 2x\amp =1\\
x\amp =-\frac{1}{2}\amp x\amp =\frac{1}{2}
\end{aligned}
\end{equation*}

Therefore, the domain consists of all real numbers \(x\text{,}\) where \(x\neq -\frac{1}{2},\frac{1}{2}\text{.}\) With this understanding, we can simplify by reducing the rational expression to lowest terms. Cancel common factors.

\begin{equation*}
g(x)=\frac{(2x-1)(x+3)}{(2x+1)(2x-1)}=\frac{x+3}{2x+1}
\end{equation*}

Our answer is \(g(x)=\frac{x+3}{2x+1}\text{,}\) where \(x\neq -\frac{1}{2},\frac{1}{2}\text{.}\)

Occasionally we will have to factor out a negative one to simplify an algebraic fraction as much as possible.

######
Example169

State the restrictions and simplify the algebraic fraction \(g(x)=\frac{25-x^2}{x^2-10x+25}\text{.}\)

SolutionFirst, we factor the numerator and denominator.

\begin{equation*}
\begin{aligned}
g(x)\amp =\frac{25-x^2}{x^2-10x+25}\\
\amp =\frac{(5-x)(5+x)}{(x-5)(x-5)}\\
\amp = \frac{\alert{-1\cdot (x-5)}(5+x)}{(x-5)(x-5)}\\
\amp=\frac{-1(5+x)}{(x-5)}\\
\amp = -\frac{x+5}{x-5}\\
\end{aligned}
\end{equation*}

Our answer is \(g(x)=-\frac{x+5}{x-5}\text{,}\) where \(x\neq 5\text{.}\)

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

######
Exercise170

State the restrictions and simplify \(f(x)=\frac{x-2x^2}{4x^4-x^2}\text{.}\)

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.

######
Example171

Given \(f(x)=x^2-2x+5\text{,}\) simplify \(\frac{f(x)-f(3)}{ax-3a}\text{.}\) Assume all denominators are nonzero.

SolutionFirst, we calculate \(f(3)\text{.}\)

\begin{equation*}
\begin{aligned}
f(3)\amp = (3)^2-2(3)+5\\
\amp = 9-6+5\\
\amp =3+5\\
\amp =8.
\end{aligned}
\end{equation*}

Now we substitute into the quotient.

\begin{equation*}
\begin{aligned}
\frac{f(x)-f(3)}{ax-3a}\amp = \frac{x^2-2x+5-8}{ax-3a}\\
\amp = \frac{x^2-2x-3}{ax-3a}\\
\amp =\frac{(x+1)(x-3)}{a(x-3)}\\
\amp =\frac{x+1}{a}
\end{aligned}
\end{equation*}

Our answer is \(\frac{f(x)-f(3)}{ax-3a}=\frac{x+1}{a}\text{.}\)

###
SubsectionMultiplying and Dividing Algebraic Fractions

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 \(S\text{,}\) where \(Q\neq 0\) and \(S\neq 0\text{,}\) we have

\begin{equation*}
\frac{P}{Q}\cdot\frac{R}{S}=\frac{PR}{QS}\text{.}
\end{equation*}

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

######
Example172

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

SolutionIn this case, the domain of \(f\) consists of all real numbers except \(5\text{,}\) and the domain of \(g\) consists of all real numbers except \(-\frac{5}{3}\text{.}\) Therefore, the domain of the product consists of all real numbers except \(5\) and \(-\frac{5}{3}\text{.}\) Multiply the functions and then simplify the result.

\begin{equation*}
\begin{aligned}
(f\cdot g)(x)\amp = f(x)\cdot g(x)\\
\amp =\frac{9x^2-25}{x-5}\cdot \frac{x^2-2x-15}{3x+5}\\
\amp =\frac{(3x+5)(3x-5)}{x-5}\cdot\frac{(x-5)(x+3)}{3x+5}\\
\amp=\frac{(3x-5)(x+3)}{1}\\
\amp =(3x-5)(x+3)
\end{aligned}
\end{equation*}

Our answer is \((f\cdot g)(x)=(3x-5)(x+3)\) where \(x\neq 5,-\frac{5}{3}\text{.}\)

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 \(P, Q, R\text{,}\) and \(S\text{,}\) where \(Q\neq 0\text{,}\) \(R\neq 0\text{,}\) and \(S\neq 0\text{,}\) we have

\begin{equation*}
\frac{P}{Q}\div\frac{R}{S}=\frac{\frac{P}{Q}}{\frac{R}{S}}=\frac{P}{Q}\cdot \frac{S}{R}=\frac{PS}{QR}\text{.}
\end{equation*}

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.

######
Example173

Given \(f(x)=\frac{2x^2+13x-7}{x^2-4x-21}\) and \(g(x)=\frac{2x^2+5x-3}{49-x^2}\text{,}\) find \(\left(\frac{f}{g}\right)(x)\) and determine the restrictions on the domain.

Solution
\begin{equation*}
\begin{aligned}
\left(\frac{f}{g}\right)(x)\amp = f(x)\div g(x)\\
\amp =\frac{2x^2+13x-7}{x^2-4x-21}\div \frac{2x^2+5x-3}{49-x^2}\\
\amp =\frac{2x^2+13x-7}{x^2-4x-21}\cdot\frac{49-x^2}{2x^2+5x-3}\\
\amp=\frac{(2x-1)(x+7)(7+x)(-1)(x-7)}{(x+3)(x-7)(2x-1)(x+3)}\\
\amp =-\frac{(x+7)^2}{(x+3)^2}
\end{aligned}
\end{equation*}

In this case, the domain of \(f\) consists of all real numbers except \(-3\) and \(7\text{,}\) and the domain of \(g\) consists of all real numbers except \(7\) and \(-7\text{.}\) In addition, the reciprocal of \(g(x)\) has restrictions on the domain of \(-3\) and \(\frac{1}{2}\text{.}\) Therefore, the domain of this quotient consists of all real numbers except \(-3, \frac{1}{2}, 7\) and \(-7\text{.}\)

Our answer is \(\left(\frac{f}{g}\right)(x)=-\frac{(x+7)^2}{(x+3)^2}\) where \(x\neq -3,\frac{1}{2},7,-7\text{.}\)

######
Exercise174

Given \(f(x)=\frac{2x+5}{3x^2+14x-5}\) and \(g(x)=\frac{6x^2+13x-5}{x+5}\text{,}\) find \(\left(\frac{f}{g}\right)(x)\) and determine the restrictions on the domain.

If a cost function \(C\) represents the cost of producing \(x\) units, then the average cost \(A\) is the cost divided by the number of units produced.

\begin{equation*}
A(x)=\frac{C(x)}{x}
\end{equation*}

######
Example175

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

SolutionFirst we set up a function representing the average cost.

\begin{equation*}
A(x)=\frac{C(x)}{x}=\frac{0.01x^2-3x+1200}{x}
\end{equation*}

Next, we calculate \(A(100), A(200)\text{,}\) and \(A(300)\text{.}\)

\begin{equation*}
\begin{aligned}
A(100)\amp =\frac{0.01(\alert{100})^2-3(\alert{100})+1200}{(\alert{100})}=\frac{100-300+1200}{100}=\frac{1000}{100}=10\\
A(200)\amp =\frac{0.01(\alert{200})^2-3(\alert{200})+1200}{(\alert{200})}=\frac{400-600+1200}{200}=\frac{1000}{200}=5\\
A(300)\amp =\frac{0.01(\alert{300})^2-3(\alert{300})+1200}{(\alert{300})}=\frac{900-900+1200}{300}=\frac{1200}{300}=4\\
\end{aligned}
\end{equation*}

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 \($5.00\text{.}\) If \(300\) are produced, the average cost per sweater is \($4.00\text{.}\)

###
SubsectionAdding and Subtracting Algebraic Fractions

Adding and subtracting rational expressions is similar to adding and subtracting fractions. Recall that if the denominators are the same, we can add or subtract the numerators and write the result over the common denominator. When working with rational expressions, the common denominator will be a polynomial. In general, given polynomials \(P, Q\text{,}\) and \(R\text{,}\) where \(Q\neq 0\text{,}\) we have the following:

\begin{equation*}
\frac{P}{Q}\pm\frac{R}{Q}=\frac{P\pm R}{Q}\text{.}
\end{equation*}

The set of restrictions to the domain of a sum or difference of rational expressions consists of the restrictions to the domains of each expression.

######
Example176

Perform the subtraction \(\frac{4x}{x^2-64}-\frac{3x+8}{x^2-64}\) and keep track of restrictions on the domain.

SolutionSince the denominators are the same, we can subtract the numerators and write the result over the common denominator. Take care to distribute the negative \(1\text{.}\)

\begin{equation*}
\begin{aligned}
\frac{4x}{x^2-64}-\frac{3x+8}{x^2-64}\amp = \frac{4x-(3x+8)}{x^2-64}\\
\amp = \frac{4x-3x-8}{x^2-64}\\
\amp = \frac{x-8}{(x+8)(x-8)}\\
\amp = \frac{1}{x+8}
\end{aligned}
\end{equation*}

Our solution is \(\frac{1}{x+8}\) for all real numbers \(x\neq \pm 8\text{.}\)

To add or subtract rational expressions with unlike denominators, we first find equivalent expressions with common denominators. We do this just as we have with fractions. If the denominators of fractions are relatively prime, then the least common denominator (LCD) is their product. For example,

\begin{equation*}
\frac{1}{x}+\frac{1}{y} \implies \text{LCD}=x\cdot y=xy.
\end{equation*}

Multiply each fraction by the appropriate form of \(1\) to obtain equivalent fractions with a common denominator.

\begin{equation*}
\begin{aligned}
\frac{1}{x}+\frac{1}{y}\amp =\frac{1}{x}\cdot\alert{\frac{y}{y}}+\frac{1}{y}\cdot\alert{\frac{x}{x}}\\
\amp = \frac{y}{xy}+\frac{x}{xy}\\
\amp = \frac{y+x}{xy}
\end{aligned}
\end{equation*}

In general, given polynomials \(P, Q, R\text{,}\) and \(S\text{,}\) where \(Q\neq 0\) and \(S\neq 0\text{,}\) we have the following:

\begin{equation*}
\frac{P}{Q}\pm\frac{R}{S}=\frac{PS\pm QR}{QS}
\end{equation*}

######
Example177

Given \(f(x)=\frac{5x}{3x+1}\) and \(g(x)=\frac{2}{x+1}\text{,}\) find \(f+g\) and determine the restrictions.

SolutionHere the LCD is the product of the denominators \((3x+1)(x+1)\text{.}\) We multiply by the appropriate factors to obtain rational expressions with a common denominator before adding.

\begin{equation*}
\begin{aligned}
(f+g)(x)\amp = f(x)+g(x)\\
\amp =\frac{5x}{3x+1}+\frac{2}{x+1}\\
\amp =\frac{5x}{3x+1}\cdot\alert{\frac{(x+1)}{(x+1)}}+\frac{2}{x+1}\cdot\alert{\frac{(3x+1)}{(3x+1)}}\\
\amp = \frac{5x(x+1)}{(3x+1)(x+1)}+\frac{2(3x+1)}{(x+1)(3x+1)}\\
\amp=\frac{5x(x+1)+2(3x+1)}{(3x+1)(x+1)}\\
\amp = \frac{5x^2+5x+6x+2}{(3x+1)(x+1)}\\
\amp =\frac{5x^2+11x+2}{(3x+1)(x+1)}\\
\amp =\frac{(5x+1)(x+2)}{(3x+1)(x+1)}\\
\end{aligned}
\end{equation*}

The domain of \(f\) consists all real numbers except \(-\frac{1}{3}\text{,}\) and the domain of \(g\) consists of all real numbers except \(-1\text{.}\) Therefore, the domain of \(f + g\) consists of all real numbers except \(-1\) and \(-\frac{1}{3}\text{.}\)

Our solution is \(\frac{(5x+1)(x+2)}{(3x+1)(x+1)}\) for all real numbers \(x\neq -1,-\frac{1}{3}\text{.}\)

It is not always the case that the LCD is the product of the given denominators. Typically, the denominators are not relatively prime; thus determining the LCD requires some thought. Begin by factoring all denominators. The LCD is the product of all factors with the highest power.

######
Example178

Given \(f(x)=\frac{3x}{3x-1}\) and \(g(x)=\frac{4-14x}{3x^2-4x+1}\text{,}\) find \(f-g\) and determine the restrictions.

SolutionTo determine the LCD, factor the denominator of \(g\text{.}\)

\begin{equation*}
\begin{aligned}
(f-g)(x)\amp = f(x)-g(x)\\
\amp =\frac{3x}{3x-1} -\frac{4-14x}{3x^2-4x+1}\\
\amp=\frac{3x}{3x-1} -\frac{4-14x}{(3x-1)(x-1)}\\
\end{aligned}
\end{equation*}

In this case the LCD\(=(3x-1)(x-1)\text{.}\) Multiply \(f\) by \(1\) in the form of \(\frac{(x-1)}{(x-1)}\) to obtain equivalent algebraic fractions with a common denominator and then subtract.

\begin{equation*}
\begin{aligned}
(f-g)(x)\amp = f(x)-g(x)\\
\amp =\frac{3x}{3x-1}\cdot\alert{\frac{(x-1)}{(x-1)}} -\frac{4-14x}{(3x-1)(x-1)}\\
\amp=\frac{3x(x-1)-4+14x}{(3x-1)(x-1)}\\
\amp =\frac{(3x-1)(x+4)}{(3x-1)(x-1)}\\\\
\amp = \frac{(x+4)}{(x-1)}
\end{aligned}
\end{equation*}

The domain of \(f\) consists all real numbers except \(\frac{1}{3}\text{,}\) and the domain of \(g\) consists of all real numbers except \(1\) and \(\frac{1}{3}\text{.}\) Therefore, the domain of \(f - g\) consists of all real numbers except \(1\) and \(\frac{1}{3}\text{.}\)

Our solution is \(\frac{x+4}{x-1}\) for all real numbers \(x\neq 1, \frac{1}{3}\text{.}\)

######
Exercise179

Simplify \(\frac{-2x}{x+6}-\frac{3x}{6-x}-\frac{18(x-2)}{x^2-36}\) and determine the restrictions.

######
Exercise180

Simplify \(\frac{x+1}{(x-1^2)}-\frac{2}{x^2-1}-\frac{4}{(x+1)(x-1)^2}\) and determine the restrictions.