ORCCA Rational Functions and Equations Chapter Review

Section 8.7 Rational Functions and Equations Chapter Review

Subsection 8.7.1 Introduction to Functions

In Section 8.1 we learned about relations. In particular, we learned about a particular type of relation called a function. We learned an informal definition of a function, how to use function notation, how to evaluate a function, and how to find the domain and range of a function that is defined by a table or list of ordered pairs.

Example 8.7.1. Evaluating Functions.

Find the given function values for a function \(f\) where \(f(x)=3x^2-4x-7\text{.}\)

\(\displaystyle f(-4)\)
\(\displaystyle f(1)\)
\(\displaystyle f(3)\)

Explanation

We find \(f(-4)\) by replacing all the \(x\)'s in the formula for \(f\) with \(-4\) and then, using the order of operations, simplifying the right side as much as possible.

\begin{align*} f(\substitute{-4}) \amp= 3(\substitute{-4})^2-4(\substitute{-4}-7\\ \amp= 3(16)+16-7\\ \amp= 57 \end{align*}
We find \(f(1)\) by replacing all the \(x\)'s in the formula for \(f\) with \(1\) and then, using the order of operations, simplifying the right side as much as possible.

\begin{align*} f(\substitute{1}) \amp= 3(\substitute{1})^2-4(\substitute{1}-7\\ \amp= 3(1)-4-7\\ \amp= -8 \end{align*}
We find \(f(3)\) by replacing all the \(x\)'s in the formula for \(f\) with \(3\) and then, using the order of operations, simplifying the right side as much as possible.

\begin{align*} f(\substitute{3}) \amp= 3(\substitute{3})^2-4(\substitute{3}-7\\ \amp= 3(9)-12-7\\ \amp= 8 \end{align*}

Example 8.7.2. Domain and Range.

The function \(g\) is defined by the ordered pairs

\begin{equation*} \{(0,2), (-13,-1), (15,2), (-9,-8), (-15,-16)\}\text{.} \end{equation*}

Determine the domain and range of \(g\text{.}\)

Explanation

The ordered pairs tell us that \(g(0)=2\text{,}\) \(g(-13)=-1\text{,}\) etc. So the valid input values are \(0\text{,}\) \(-13\text{,}\) \(15\text{,}\) \(-9\text{,}\) and \(-15\text{.}\) This means the domain is the set \(\{0,-13,15,-9,-15\}\text{.}\)

Similarly, the ordered pairs tell us that \(2\text{,}\) \(-1\text{,}\) \(-8\text{,}\) and \(-16\) are possible output values. Notice that the output \(2\) happened twice, but it only needs to be listed in this collection once. The range of \(g\) is \(\{2,-1,-8,-16\}\text{.}\)

Subsection 8.7.2 Rational Functions

In Section 8.2 we learned about rational functions and how to find the domain of a rational function. We also learned how to simplify a rational function and include any necessary explicit domain restrictions.

Example 8.7.3. Exploring a Rational Function.

In an apocalypse, a zombie infestation begins with \(1\) zombie and spreads rapidly. The population of zombies can be modeled by \(Z(x)=\frac{200000x+100}{5x+100}\text{,}\) where \(x\) is the number of days after the apocalypse began. Use the function to answer these questions:

How many zombies are there \(2\) days after the apocalypse began?
What is the domain of this function?
Can this function be simplified?

Explanation

To find the number of zombies after \(2\) days, we substitute \(2\) in place of \(x\) in the function and simplify.

\begin{align*} Z(\substitute{2})=\frac{200000(\substitute{2})+100}{5(\substitute{2})+100}\\ \amp=\frac{400100}{110}\\ \amp\approx3637 \end{align*}

Thus, after \(2\) days, there will be about \(3637\) zombies.
To find the domain of this function, we need to find the values of \(x\) that will make the denominator equal to zero. Those value of \(x\) will make the rational function undefined and therefore, must be excluded from the domain.

To find those values, we set the denominator equal to \(0\) and then solve that resulting equation.

\begin{gather*} 5x+100=0\\ 5x=-100\\ x=-20 \end{gather*}

Thus, \(x=-20\) needs to be excluded from the domain. Therefore, the domain is all real numbers except \(x=-20\text{.}\) In set-builder notation, the domain is \(\{x \mid x\ne-20\}\text{.}\)
To determine whether or not this function can be simplified, we must factor the numerator and denominator to see if there are any common factors that can be canceled.

\begin{align*} \frac{200000x+100}{5x+100}=\frac{100(2000x+1)}{5(x+20)}\\ \amp=\frac{\cancelhighlight{5}\cdot20(2000x+1)}{\cancelhighlight{5}(x+20)}\\ \amp=\frac{20(2000x+1)}{x+20} \end{align*}

Thus, the function \(Z\) simplifies to \(Z(x)=\frac{20(2000x+1)}{x+20}\text{.}\) Note that there are no explicit domain restrictions to include since the only factor that was canceled was a constant.

Example 8.7.4. Simplifying Rational Expressions.

Simplify the expression \(\frac{8t+4t^2-12t^3}{1-t}\text{.}\)

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{8t+4t^2-12t^3}{1-t}\amp=\frac{-12t^3+4t^2+8t}{-t+1}\\ \amp=\frac{-4t(3t^2-t-2)}{-(t-1)}\\ \amp=\frac{-4t(3t+2)(t-1)}{-(t-1)}\\ \amp=\frac{-4t(3t+2)\cancelhighlight{(t-1)}}{-\cancelhighlight{(t-1)}}\\ \amp=\frac{-4t(3t+2)}{-1}, \text{ for } t\neq 1\\ \amp=4t(3t+2), \text{ for } t\neq 1 \end{align*}

Subsection 8.7.3 Multiplication and Division of Rational Expressions

In Section 8.3 we learned how to multiply and divide rational expressions, and include any explicit domain restrictions necessary.

Example 8.7.5. Multiplication of Rational Expressions in One Variable.

Multiply the rational expressions: \(\frac{x^2-7x}{x^2-9}\cdot\frac{9-6x+x^2}{42+x-x^2}\text{.}\)

Explanation

Note that to factor the second rational expression, we'll want to re-write the terms in descending order for both the numerator and denominator. In the denominator, we'll first factor out \(-1\) as the leading term is \(-x^2\text{.}\)

\begin{align*} \frac{x^2-7x}{x^2-9}\cdot\frac{9-6x+x^2}{42+x-x^2}\amp=\frac{x^2-7x}{x^2-9}\cdot\frac{x^2-6x+9}{-x^2+x+42}\\ \amp=\frac{x^2-7x}{x^2-9}\cdot\frac{x^2-6x+9}{-(x^2-x-42)}\\ \amp=\frac{x\cancelhighlight{(x-7)}}{(x+3)\cancelhighlight{(x-3)}} \cdot\frac{(x-3)\cancelhighlight{(x-3)}}{-(x+6)\cancelhighlight{(x-7)}}\\ \amp=-\frac{x(x-3)}{(x+3)(x+6)}, \text{ for } x\neq3,x\neq 7 \end{align*}

Example 8.7.6. Multiplication of Rational Expressions in Two Variables.

Multiply the rational expressions: \(\frac{r^3s}{4t}\cdot\frac{2t^2}{r^2s^3}\text{.}\)

Explanation

Note that we won't need to factor anything in this problem, and can simply multiply across and then simplify, using the rules of exponents. With multivariable expressions, this textbook ignores domain restrictions.

\begin{align*} \frac{r^3s}{4t}\cdot\frac{2t^2}{r^2s^3}\amp=\frac{r^3s\cdot2t^2}{4t\cdot r^2s^3}\\ \amp=\frac{2r^3st^2}{4r^2s^3t}\\ \amp=\frac{rt}{2s^2} \end{align*}

Example 8.7.7. Division of Rational Expressions in One Variable.

Divide the rational expressions: \(\frac{y^2-2y-15}{y^2+4y+3}\div\frac{y-5}{y+8}\text{.}\)

Explanation

\begin{align*} \frac{y^2-2y-15}{y^2+4y+3}\div\frac{y-5}{y+8}\amp=\frac{y^2-2y-15}{y^2+4y+3}\cdot\frac{y+8}{y-5}, \text{ for }y\neq -8\\ \amp=\frac{\cancelhighlight{(y-5)}\cancelhighlight{(y+3)}}{(y+1)\cancelhighlight{(y+3)}}\cdot\frac{y+8}{\cancelhighlight{y-5}}, \text{ for }y\neq -8\\ \amp=\frac{y+8}{y+1}, \text{ for }y\neq -8, y\neq -3, y\neq 5 \end{align*}

Example 8.7.8. Division of Rational Expressions in Two Variables.

Divide the rational expressions: \(\frac{2x^2+8xy}{x^2-4x+3}\div\frac{x^3+4x^2y}{x^2+4x-5}\text{.}\)

Explanation

To divide rational expressions, we multiply by the reciprocal of the second fraction. Then we will factor and cancel any common factors. With multivariable expressions, this textbook ignores domain restrictions.

\begin{align*} \frac{2x^2+8xy}{x^2-4x+3}\div\frac{x^3+4x^2y}{x^2+4x-5}\amp=\frac{2x^2+8xy}{x^2-4x+3}\cdot\frac{x^2+4x-5}{x^3+4x^2y}\\ \amp=\frac{2x\cancelhighlight{(x+4y)}}{\cancelhighlight{(x-1)}(x-3)}\cdot\frac{\cancelhighlight{(x-1)}(x+5)}{x^2\cancelhighlight{(x+4y)}}\\ \amp=\frac{2x}{x-3}\cdot\frac{x+5}{x^2}\\ \amp=\frac{2(x+5)}{x(x-3)} \end{align*}

Subsection 8.7.4 Addition and Subtraction of Rational Expressions

In Section 8.4 we covered how to add and subtract rational expressions.

Example 8.7.9. Addition and Subtraction of Rational Expressions with the Same Denominator.

Add the rational expressions: \(\dfrac{5x}{x+5}+\dfrac{25}{x+5}\text{.}\)

Explanation

These expressions already have a common denominator:

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

Note that we didn't stop at \(\frac{5x+25}{x+5}\text{.}\) If possible, we must simplify the expression by factoring the numerator and denominator and looking for common factors that can be canceled.

Example 8.7.10. Addition and Subtraction of Rational Expressions with Different Denominators.

Add and subtract the rational expressions: \(\dfrac{6y}{y+2}+\dfrac{y}{y-2}-7\)

Explanation

The denominators can't be factored, so we'll find the least common denominator and build each expression to that denominator. Then we will be able to combine the numerators and simplify the expression.

\begin{align*} \frac{6y}{y+2}+\frac{y}{y-2}-7\amp=\frac{6y}{y+2}\multiplyright{\frac{y-2}{y-2}}+\frac{y}{y-2}\multiplyright{\frac{y+2}{y+2}}-7\multiplyright{\frac{(y-2)(y+2)}{(y-2)(y+2)}}\\ \amp=\frac{6y(y-2)}{(y-2)(y+2)}+\frac{y(y+2)}{(y-2)(y+2)}-\frac{7(y-2)(y+2)}{(y-2)(y+2)}\\ \amp=\frac{6y^2-12y+y^2+2y-\highlight{\attention{(}}7(y^2-4)\highlight{\attention{)}}}{(y-2)(y+2)}\\ \amp=\frac{6y^2-12y+y^2+2y-7y^2+28}{(y-2)(y+2)}\\ \amp=\frac{-10y+28}{(y-2)(y+2)}\\ \amp=\frac{-2(5y-14)}{(y-2)(y+2)} \end{align*}

Subsection 8.7.5 Complex Fractions

In Section 8.5 we covered how to simplify a rational expression that has fractions in the numerator and/or denominator.

Example 8.7.11. Simplifying Complex Fractions.

Simplify the complex fraction \(\dfrac{\frac{2t}{t^2-9}+3}{\frac{6}{t+3}+\frac{1}{t-3}}\text{.}\)

Explanation

First, we check all quadratic polynomials to see if they can be factored and factor them:

\begin{equation*} \frac{\frac{2t}{t^2-9}+3}{\frac{6}{t+3}+\frac{1}{t-3}}=\frac{\frac{2t}{(t-3)(t+3)}+3}{\frac{6}{t+3}+\frac{1}{t-3}} \end{equation*}

Next, we identify the common denominator of the three fractions, which is \((t+3)(t-3)\text{.}\) We then multiply the main numerator and denominator by that expression:

\begin{align*} \frac{\frac{2t}{(t-3)(t+3)}+3}{\frac{6}{t+3}+\frac{1}{t-3}}\amp=\frac{\frac{2t}{(t-3)(t+3)}+3}{\frac{6}{t+3}+\frac{1}{t-3}}\multiplyright{\frac{(t-3)(t+3)}{(t-3)(t+3)}}\\ \amp=\frac{\frac{2t}{\highlight{\xcancel{(t-3)(t+3)}}}\highlight{\xcancel{(t-3)(t+3)}}+3(t-3)(t+3)}{\frac{6}{\highlight{\cancel{t+3}}}(t-3)\highlight{\cancel{(t+3)}}+\frac{1}{\lighthigh{\cancel{t-3}}}\lighthigh{\cancel{(t-3)}}(t+3)}\\ \amp=\frac{2t+3(t-3)(t+3)}{6(t-3)+1(t+3)} \text{ for }t\neq -3, t\neq 3\\ \amp=\frac{2t+3(t^2-9)}{6t-18+t+3} \text{ for }t\neq -3, t\neq 3\\ \amp=\frac{2t+3t^2-27}{7t-15} \text{ for }t\neq -3, t\neq 3\\ \amp=\frac{3t^2+2t-27}{7t-15} \text{ for }t\neq -3, t\neq 3 \end{align*}

Both the numerator and denominator are prime polynomials so this expression cannot simplify any further.

Subsection 8.7.6 Solving Rational Equations

In Section 8.6 we covered how to solve rational equations, by first clearing the fractions, much the same as we learned in Section 3.3. We also looked at application problems like uniform motion and shared work problems. Lastly, we solved rational equations for a specified variable.

Example 8.7.12. Solving Rational Equations.

Two pipes are being used to fill a large tank. Pipe B can fill the tank twice as fast as Pipe A can. When both pipes are turned on, it takes 12 hours to fill the tank. Write and solve a rational equation to answer the following questions:

If only Pipe A is turned on, how many hours would it take to fill the tank?
If only Pipe B is turned on, how many hours would it take to fill the tank?

Explanation

Since both pipes can fill the tank in \(12\) hours, they fill \(\frac{1}{12}\) of the tank together each hour. We will let \(a\) represent the number of hours it takes pipe A to fill the tank alone, so pipe A will fill \(\frac{1}{a}\) of the tank each hour. Pipe B can fill the tank twice as fast so it fills \(2\cdot \frac{1}{a}\) of the tank each hour or \(\frac{2}{a}\text{.}\) When they are both turned on, they fill \(\frac{1}{a}+\frac{2}{a}\) of the tank each hour.

Now we can write this equation:

\begin{equation*} \frac{1}{a}+\frac{2}{a}=\frac{1}{12} \end{equation*}

To clear away denominators, we multiply both sides of the equation by the common denominator of \(12\) and \(a\text{,}\) which is \(12a\text{:}\)

\begin{align*} \frac{1}{a}+\frac{2}{a}\amp=\frac{1}{12}\\ \multiplyleft{12a}\left(\frac{1}{a}+\frac{2}{a}\right)\amp=\multiplyleft{12a}\frac{1}{12}\\ 12a\cdot\frac{1}{a}+12a\cdot\frac{2}{a}\amp=12a\cdot\frac{1}{12}\\ 12+24\amp=a\\ 36\amp=a\\ a\amp=36 \end{align*}

The possible solution \(a=36\) should be checked

\begin{align*} \frac{1}{\substitute{36}}+\frac{2}{\substitute{36}}\amp\stackrel{?}{=}\frac{1}{12}\\ \frac{3}{36}\amp\stackrel{\checkmark}{=}\frac{1}{12} \end{align*}

So it is a solution.

If only Pipe A is turned on, it would take \(36\) hours to fill the tank.
Since Pipe B can fill the tank twice as fast, it would take half the time, or \(18\) hours to fill the tank.

Example 8.7.13. Solving Rational Equations for a Specific Variable.

Solve the rational equation \(y=\frac{2x+5}{3x-1}\) for \(x\text{.}\)

Explanation

To get the \(x\) out of the denominator, our first step will be to multiply each side by the LCD, which is \(3x-1\text{.}\) Then we'll isolate all terms containing \(x\text{,}\) factor out \(x\text{,}\) and then finish solving for that variable.

\begin{align*} y\amp=\frac{2x+5}{3x-1}\\ y\multiplyright{(3x-1)}\amp=\frac{2x+5}{\highlight{\cancel{3x-1}}}\multiplyright{\highlight{\cancel{(3x-1)}}}\\ 3xy-y\amp=2x+5\\ 3xy\amp=2x+5+y\\ 3xy-2x\amp=y+5\\ x(3y-2)\amp=y+5\\ \divideunder{x(3y-2)}{3y-2}\amp=\divideunder{y+5}{3y-2}\\ x\amp=\frac{y+5}{3y-2} \end{align*}