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SectionLong-Run Behavior of Rational Functions

A rational function is the quotient of two polynomials. (As with rational numbers, the word rational refers to a ratio.)

Rational Functions

A rational function is one of the form

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

where \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x)\neq 0\text{.}\)

The graphs of rational functions can be quite different from the graphs of polynomials.

Example329

Queueing theory is used to predict your waiting time in a line, or queue. For example, suppose the attendant at a toll booth can process \(6\) vehicles per minute. The average total time spent by a motorist negotiating the toll booth depends on the rate, \(r\text{,}\) at which vehicles arrive, according to the formula

\begin{equation*} T=g(r)=\frac{12-r}{12(6-r)}. \end{equation*}

  1. What is the average time spent at the toll booth if vehicles arrive at a rate of \(3\) vehicles per minute?

  2. Graph the function on the domain \([0, 6)\text{.}\)

Solution

  1. If the vehicles arrice at a rate of \(3\) vehicles per minute, then \(r=3\) so

    \begin{equation*} g(3)=\dfrac{12-3}{12(6-3)}=\dfrac{9}{36}=0.25 \end{equation*}

    The average total time is \(0.25\) minutes, or \(15\) seconds.

  2. rational function
Example330

EarthCare decides to sell T-shirts to raise money. The company makes an initial investment of $100 to pay for the design of the T-shirt and to set up the printing process. After that, the T-shirts cost $5 each for labor and materials.

  1. Express the average cost per T-shirt as a function of the number of T-shirts EarthCare produces.

  2. Make a table of values for the function.

  3. Graph the function and explain what it tells you about the cost of the T-shirts.

Solution

  1. If EarthCare produces \(x\) T-shirts, the total costs will be \(100 + 5x\) dollars. To find the average cost per T-shirt, we divide the total cost by the number of T-shirts produced, to get

    \begin{equation*} C = g(x) = \frac{100 + 5x}{x}. \end{equation*}

  2. We evaluate the function for several values of \(x\text{,}\) as shown in the table

    \(x\) \(1\) \(2\) \(4\) \(5\) \(10\) \(20\)
    \(C\) \(105\) \(55\) \(40\) \(25\) \(15\) \(10\)

    If EarthCare makes only one T-shirt, its cost is $105. But if more than one T-shirt is made, the cost of the original $100 investment is distributed among them. For example, the average cost per T-shirt for \(\alert{2}\) T-shirts is

    \begin{equation*} \frac{100 + 5(\alert{2})}{\alert{2}}=55 \end{equation*}

    and the average cost for \(\alert{5}\) T-shirts is

    \begin{equation*} \frac{100 + 5(\alert{5})}{\alert{5}}=25. \end{equation*}

  3. The graph is shown in Figure331. You can use a graphing utility with the window

    \begin{align*} \text{Xmin} \amp = 0 \amp\amp \text{Xmax} = 470\\ \text{Ymin} \amp = 0 \amp\amp \text{Ymax} = 30. \end{align*}

    to verify the graph.

    graph of average cost of production
    Figure331

    The graph shows that as the number of T-shirts increases, the average cost per shirt continues to decrease, but not as rapidly as at first. Eventually the average cost levels off and approaches $5 per T-shirt. For example, if EarthCare produces \(\alert{400}\) T-shirts, the average cost per shirt is

    \begin{equation*} \frac{100+5(\alert{400})}{\alert{400}}=5.25. \end{equation*}

The horizontal line \(C = 5\) on the graph of \(C = \displaystyle{\frac{100 + 5x}{x}}\) is a horizontal asymptote. As \(x\) increases, the graph approaches the line \(C = 5\text{.}\) The average price per T-shirt will always be slightly more than $5. Horizontal asymptotes are also important in sketching the graphs of rational functions.

Example332

Delbert prepares a \(20\%\) saturated glucose solution of by mixing \(2\) mL of saturated glucose with \(8\) mL of water. If he adds \(x\) ml of saturated glucose to the solution, its concentration is given by

\begin{equation*} C(x) = \frac{2 + x}{8+x}. \end{equation*}

  1. How many milliliters of saturated glucose should Delbert add to increase the concentration to \(50\%\text{?}\)

  2. Graph the function on the domain \([0, 100]\text{.}\)

  3. What is the horizontal asymptote of the graph? What does it tell you about the solution?

Solution

  1. To obtain a concentration of \(50/%\text{,}\) we solve \(C(x)=0.50\text{:}\)

    \begin{equation*} \begin{aligned} C(x) \amp= 0.50 \\ \dfrac{2+x}{8+x} \amp= 0.50 \\ 2+x \amp= 0.50(8+x) \\ 2+x \amp= 4 + 0.50x \\ 0.5x \amp= 2 \\ x \amp= 4 \end{aligned} \end{equation*}

    He needs to add \(4\) ml.

  2. rational function

  3. \(C=1\text{.}\) As Delbert adds more glucose to the mixture, its concentration increases toward \(100\%\text{.}\)

SubsectionLong Run Behavior for Rational Functions

Recall in the start of this chapter we discussed the long run behavior of polynomial functions. Recall too that given a polynomial

\begin{equation*} f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots + a_1x+a_0 \end{equation*}

the leading term of \(f(x) \) is the term \(a_nx^n \text{.}\)

Long Run Behavior

Suppose that \(f(x)=\frac{P(x)}{Q(x)}\) is a rational function. Let \(ax^n\) be the leading term of \(P(x)\) and \(bx^m \) be the leading term of \(Q(x)\text{.}\) Then the long run behavior of \(f(x)\) is the same as the long run behavior of the power function

\begin{equation*} \frac{ax^n}{bx^m}=\frac{a}{b}x^{n-m}. \end{equation*}

If the resulting power function happens to be a constant function (so \(n=m\) above) then the rational function has a horizontal asymptote at \(y=\frac{a}{b}\text{.}\)

Example333

Identify the long run behavior of the rational function

\begin{equation*} f(x)=\frac{-2x^7+4x^2-1}{x^3+8x}. \end{equation*}
Solution

First, we note that the leading term of the numerator is \(-2x^7\) and the leading term of the denominator is \(x^3\text{.}\) So, the long run behavior of \(f \) is given by

\begin{equation*} \frac{-2x^7}{x^3}=-2x^4. \end{equation*}

We know that the long run behavior of \(x^4\) is

\begin{equation*} x^4\to \infty\ \text{as}\ x\to\pm\infty. \end{equation*}

Since the graph of \(-2x^4\) is the graph of \(x^4 \) stretched vertically by a factor of \(2\) and then reflected about the \(x\)-axis, get that the long run behavior of \(-2x^4\) is

\begin{equation*} -2x^4\to -\infty\ \text{as}\ x\to\pm\infty. \end{equation*}

Thus, the long run behavior of \(f(x) \) is

\begin{equation*} f(x)\to -\infty\ \text{as}\ x\to\pm\infty. \end{equation*}

SubsectionSupplemental Videos

SubsectionExercises