In this chapter, we will learn rational functions, which looks like \(f(x)=\frac{P(x)}{Q(x)}\text{.}\) We can see rational functions are needed when division is involved. In this lesson, we will look at a rational function's domain, range and graph.
Figure13.1.1Alternative Video Lesson
Subsection13.1.1Graph of Rational Functions
Scenario: When a drug is injected into a patient, the drug's concentration in the patient's bloodstream can be modeled by the function
where \(C\) stands for the drug's concentration in milligrams per liter, and \(t\) stands for the number of hours since the injection. A new injection is needed when the concentration falls to \(\frac{1}{3}\) milligrams per liter. After how many hours since the first inject should the next injection be given?
The function \(C(t)=\frac{3t}{t^2+8}\) is a rational function. Let's look at its formal definition:
where \(P(x)\) and \(Q(x)\) are polynomials, but \(Q(x)\) is not the constant zero function.
Let's look at the graph of \(C(t)=\frac{3t}{t^2+8}\text{:}\)
Figure13.1.3Graph of \(C(t)=\frac{3t}{t^2+8}\)
We can see this rational function's graph is a smooth curve. In this context, only positive values make sense, so we focus on the first quadrant. Since the drug is injected, its concentration in the bloodstream increases, and starts to decrease at about \(3\) hours. We need to find out when the concentration will become \(\frac{1}{3}\) milligrams per liter, so we substitute \(C(t)\) in the function with \(\frac{1}{3}\) and solve for \(t\text{:}\)
\begin{align*}
C(t)\amp=\frac{3t}{t^2+8}\\
\frac{1}{3}\amp=\frac{3t}{t^2+8}\\
1(t^2+8)\amp=3(3t)\amp\text{cross multiply}\\
t^2+8\amp=9t\\
t^2-9t+8\amp=0\\
(t-1)(t-8)\amp=0\\
t-1\amp=0\text{ or }t-8=0\\
t\amp=1\text{ or }t=8
\end{align*}
The concentration will be \(\frac{1}{3}\) milligrams per liter at \(1\) hour since the injection, and then at \(8\) hours again. We can verify those two solutions by \((1,\frac{1}{3})\) and \((8,\frac{1}{3})\) on the curve.
Solution: The patient should receive a second injection \(8\) hours since the first injection.
Next, let's study the long-term behavior of \(C(t)\) with a calculator.
A rational function's graph is not always smooth. It could have breaks. Let's look at another example.
Example13.1.4
Two towns are connected by a \(12\)-mile-long river, which flows from Town A to Town B at the velocity of \(2\) miles per hour. A boat will travel at a constant velocity, \(v\) miles per hour, from Town A to Town B, and then back to Town A. Due to the river's flow, the boat's actual speed is \(v+2\) miles per hour from Town A to Town B, and \(v-2\) miles per hour from Town B back to Town A. If the boat driver plans to spend \(8\) hours for the whole trip, how fast should he/she drive the boat?
We need to review 3 formulas for constant-rate movement:
where \(d\) stands for distance, \(v\) stands for velocity, and \(t\) stands for time. According to the third formula, the time it takes the boat to travel from Town A to Town B is \(\frac{12}{v+2}\text{,}\) and \(\frac{12}{v-2}\) from Town B to Town A. The function to model the whole trip's time is
where \(t\) stands for time in hours. Let's look at the graph of this function:
Figure13.1.5Graph of \(t(v)=\frac{12}{v-2}+\frac{12}{v+2}\)
The function passes \((4,8)\text{.}\) This implies if the boat drives at \(4\) miles per hour, it would take \(8\) hours to complete the trip. From Town A to Town B, it would take \(\frac{12}{v+2}=\frac{12}{4+2}=2\) hours; and from Town B to Town A, it would take \(\frac{12}{v-2}=\frac{12}{4-2}=6\) hours. In a later section, we will learn how to solve this equation algebraically.
In the example, the curve is still smooth, but there are two breaks at \(x=-2\) and \(x=2\text{.}\) Those two vertical lines are called asymptotes. A function's graph approaches an asymptote infinitely close, but never touches it.
In math, we use \(\infty\) to represent infinity. In the graph, the function's value approaches \(\infty\) as its \(x\) value approaches \(2\) from the right side, and approaches \(-\infty\) as its \(x\) value approaches \(2\) from the left side. Let's investigate why this is happening.
We will build a table of values for \(t(v)\text{:}\)
\(v\)
\(t(v)\)
interpretation
\(2.1\)
\(122.92\)
If the boat drives at \(2.1\) miles per hour, the trip would take approximately \(122.93\) hours.
\(2.01\)
\(1202.99\)
If the boat drives at \(2.01\) miles per hour, the trip would take approximately \(1202.99\) hours.
Table13.1.6Values of \(t(v)=\frac{12}{v-2}+\frac{12}{v+2}\)
To understand the interpretations, we need to find a pattern of division:
When the boat drives from Town B back to Town A at \(2.1\) miles per hour, the river's current would decrease the actual velocity to \(0.1\) mile per hour. It would take the boat \(\frac{12}{0.1}=120\) hours to complete the travel from Town B back to Town A.
Similarly, as the boat drives from Town B to Town A at a velocity closer and closer to \(2\text{,}\) the division \(\frac{12}{v-2}\) would produce a larger and larger value, approaching \(\infty\text{.}\) When we graph the function, we would plot \((2.1,122.93), (2.01,1202.99), (2.001,12003), \ldots\text{.}\) This is why we see a vertical asymptote at \(x=2\text{.}\)
Remark13.1.7
The rational function \(f(x)=\frac{1}{x-a}\) has a vertical asymptote at \(x=a\text{,}\) as this value makes the denominator \(0\text{.}\)
Now we understand why \(t(v)=\frac{12}{v-2}+\frac{12}{v+2}\) has two vertical asymptotes at \(x=-2\) and \(x=2\text{.}\) The domain of \(t(v)\) is \((-\infty,-2)\cup(-2,2)\cup(2,\infty)\) in interval notation, and {\(x \mid x\ne-2, x\ne2\)} in set notation.
Let's look at some examples of rational functions.
Example13.1.8
Build a table and sketch the graph of the function \(f(x)=\frac{1}{x}\text{.}\) Find the function's domain and range.
Since \(x=0\) makes the denominator \(0\text{,}\) the function has a vertical asymptote at \(x=0\text{.}\) This implies, as \(x\)-values approaches to \(0\text{,}\) the function's values approaches \(\infty\) or \(-\infty\text{.}\) We will choose \(x\) values properly and build a table:
\(x\)
\(f(x)\)
Points on the Curve
\(-10\)
\(\frac{1}{-10}=-0.1\)
\((-10,-0.1)\)
\(-2\)
\(\frac{1}{-2}=-0.5\)
\((-2,-0.5)\)
\(-1\)
\(\frac{1}{-1}=-1\)
\((-1,-1)\)
\(-0.5\)
\(\frac{1}{-0.5}=-2\)
\((-0.5,-2)\)
\(-0.1\)
\(\frac{1}{-0.1}=-10\)
\((-0.1,-10)\)
\(0.1\)
\(\frac{1}{0.1}=10\)
\((0.1,10)\)
\(0.5\)
\(\frac{1}{0.5}=2\)
\((0.5,2)\)
\(1\)
\(\frac{1}{1}=1\)
\((1,1)\)
\(2\)
\(\frac{1}{2}=0.5\)
\((2,0.5)\)
\(10\)
\(\frac{1}{10}=0.1\)
\((10,0.1)\)
Table13.1.9Values of \(f(x)=\frac{1}{x}\)
Next, we will plot those points and then connect them:
Figure13.1.10Graph of \(f(x)=\frac{1}{x}\)
The function \(f(x)=\frac{1}{x}\) has a vertical asymptote at \(x=0\text{.}\) As a result, its domain is \((-\infty,0)\cup(0,\infty)\) in interval notation, or {\(x \mid x\ne0\)} in set notation.
By the graph, the function's value cannot be \(0\text{,}\) so its range is also \((-\infty,0)\cup(0,\infty)\text{.}\) In other words, the function \(f(x)\) has a horizontal asymptote at \(y=0\text{.}\)
Example13.1.11
Use technology to build a table and graph the function \(f(x)=\frac{1}{x^2}\text{.}\) Find the function's domain and range.
With technology, we will choose the starting value of \(-2\) and an increment of \(0.5\text{,}\) and look at values in a table:
\(x\)
\(f(x)\)
\(-2\)
\(0.25\)
\(-1.5\)
\(0.444\ldots\)
\(-1\)
\(1\)
\(-0.5\)
\(4\)
\(0\)
\(\)
\(0.5\)
\(4\)
\(1\)
\(1\)
\(1.5\)
\(0.444\ldots\)
\(2\)
\(0.25\)
Table13.1.12Values of \(f(x)=\frac{1}{x^2}\) in a table
Next, we will use technology to graph the function:
Figure13.1.13Graph of \(f(x)=\frac{1}{x^2}\)
The function \(f(x)=\frac{1}{x^2}\) has a vertical asymptote at \(x=0\text{.}\) As a result, its domain is \((-\infty,0)\cup(0,\infty)\) in interval notation, or {\(x \mid x\ne0\)} in set notation.
By the graph, the function's range is also \((0,\infty)\text{.}\) The function \(f(x)\) has a horizontal asymptote at \(y=0\text{.}\)
Remark13.1.14
It is important to be able to quickly sketch the graphs of the following types of basic functions:
To find a rational function's domain, we set the denominator to be \(0\text{,}\) and solve for the variable:
\begin{align*}
x^2-2x-24\amp=0\\
(x-6)(x+4)\amp=0\\
x-6\amp=0\text{ or }x+4=0\\
x\amp=6\text{ or }x=-4
\end{align*}
Since \(x=6\) and \(x=-4\) will cause the denominator to be \(0\text{,}\) we eliminate them from the domain. The function's domain is {\(x \mid x\ne6, x\ne-4\)}. The function has two vertical asymptotes at \(x=6\) and \(x=-4\text{.}\)
With a graping device, we can verity the vertical asymptotes:
Figure13.1.16Graph of \(g(x)=\frac{4x^2}{x^2-2x-24}\)
It's easy to verify the vertical asymptotes, \(x=-4\) and \(x=6\text{,}\) by the graph. However, it's difficult to see any horizontal asymptote. An educated guess is \(y=4\text{,}\) but we should verify this by looking at a table of values. Set both the start value and increment to be some big numbers, and we will see something like the following data:
\(x\)
\(g(x)\)
\(1000\)
\(4.008\ldots\)
\(2000\)
\(4.004\ldots\)
\(3000\)
\(4.002\ldots\)
\(4000\)
\(4.002\ldots\)
\(x\)
\(g(x)\)
\(-1000\)
\(3.992\ldots\)
\(-2000\)
\(3.996\ldots\)
\(-3000\)
\(3.997\ldots\)
\(-4000\)
\(3.998\ldots\)
Values in the tables have verified that the horizontal asymptote is indeed \(y=4\text{.}\)
Let's look at another example where a rational functions is used to model real life data.
Example13.1.19
The monthly operation cost of a shoe company is approximately \(\$300,000.00\text{.}\) The cost of producing each pair of shoes is \(\$30.00\text{.}\) As a result, the cost of producing \(x\) pairs of shoes is \(30x+300000\) dollars, and the average cost of producing each pair of shoes can be modeled by the function
We will graph the function with technology. After adjusting window settings, we have:
Figure13.1.20Graph of \(\bar{C}(x)=\frac{30x+300000}{x}\)
1. What's the average cost of producing \(100\) pairs of shoes? \(1,000\) pairs? \(10,000\) pairs? What's the pattern?
To answer this question, we locate the points where \(x\) values are \(100\text{,}\) \(1,000\) and \(10,000\text{.}\) They are \((100,3030), (1000,330)\) and \((10000,60)\text{.}\) They imply:
If the company produces \(100\) pairs of shoes, the average cost of producing one pair is \(\$3,030.00\text{.}\)
If the company produces \(1,000\) pairs of shoes, the average cost of producing one pair is \(\$330.00\text{.}\)
If the company produces \(10,000\) pairs of shoes, the average cost of producing one pair is \(\$60.00\text{.}\)
We can see the more shoes the company produces, the lower the average cost.
2. To make the average cost of producing each pair of shoes cheaper than \(\$50.00\text{,}\) at least how many pairs of shoes must the company produce?
To answer this question, we lcoate the point where its \(y\)-value is \(50\text{.}\) With technology, we graph both the function \(\bar{C}(x)\) and \(y=50\text{,}\) and locate their intersection.
Figure13.1.21Intersection of \(\bar{C}(x)=\frac{30x+300000}{x}\) and \(y=50\)
The intersection \((15000,50)\) implies the average cost of producing one pair is \(\$50.00\) if the company produces \(15,000\) pairs of shoes.
3. Assume the company's shoes are very popular. What's the average cost of producing infinitely many pairs of shoes?
To answer this question, we substitute \(x\) with some big numbers, and we have:
\((100000, 33),\)
\((1000000, 31),\)
\((10000000,30.03),\)
\((100000000,30.003)\)
We can guess that the average cost of producing one pair is getting closer and closer to \(\$30.00\) as the company produces more and more pairs.
Note that the cost of producing each pair is \(\$30.00\text{.}\) This implies, for big companies whose products are very popular, the cost of operations can be ignored when calculating the average cost of producing each unit of product.
