Lesson 4
Representing Functions at Rational Inputs
4.1: Math Talk: Unknown Exponents (5 minutes)
Warmup
This Math Talk encourages students to think about rules of exponents in order to solve exponential equations. These exponential rules elicited here will be helpful later in the lesson and in future lessons when students study the values of exponential functions at rational inputs and use these functions to model different phenomena.
Launch
Display one problem at a time. Give students quiet think time for each problem and ask them to give a signal when they have an answer and a strategy. Keep all problems displayed throughout the talk. Follow with a wholeclass discussion.
Supports accessibility for: Memory; Organization
Student Facing
Solve each equation mentally.
 \(5^q=125\)
 \(\frac{1}{5^r} = \frac{1}{125}\)
 \(5^t = \frac{1}{125}\)
 \(125^u = 5\)
Student Response
Teachers with a valid work email address can click here to register or sign in for free access to Student Response.
Activity Synthesis
Ask students to share their strategies for each problem. Record and display their responses for all to see. To involve more students in the conversation, consider asking:
 “Who can restate \(\underline{\hspace{.5in}}\)’s reasoning in a different way?”
 “Did anyone have the same strategy but would explain it differently?”
 “Did anyone solve the problem in a different way?”
 “Does anyone want to add on to \(\underline{\hspace{.5in}}\)’s strategy?”
 “Do you agree or disagree? Why?”
If not mentioned in students’ explanations, emphasize that these equations can all be solved by writing each expression as a power of the same base, and then examining the exponents of those expressions. For example, \(\frac{1}{125}\) is the same as \(\frac{1}{5^3}\).
Design Principle(s): Optimize output (for explanation)
4.2: Population of Nigeria (15 minutes)
Activity
In this activity, students practice constructing an exponential function and evaluating it at a fractional input value as they translate between decades and years. They also think about how to express a percent increase over a fractional interval of input—specifically, how to express the annual growth rate of a population given the growth rate for one decade. They analyze and articulate why the growth rate for one year is not one tenth of that of one decade (MP3).
Launch
Tell students that, as of 2017, Nigeria is the seventhmost populous country in the world. It has grown at a fairly steady (and relatively high) rate since 1980.
Design Principle(s): Maximize metaawareness; Support sensemaking
Student Facing
In 1990, Nigeria had a population of about 95.3 million. By 2000, there were about 122.4 million people, an increase of about 28.4%. During that decade, the population can be reasonably modeled by an exponential function.
 Express the population of Nigeria \(f(d)\), in millions of people, \(d\) decades since 1990.
 Write an expression to represent the population of Nigeria in 1996.
 A student said, “The population of Nigeria grew at a rate of 2.84% every year.”
 Explain or show why the student’s statement is incorrect.
 Find the correct annual growth rate. Explain or show your reasoning.
Student Response
Teachers with a valid work email address can click here to register or sign in for free access to Student Response.
Anticipated Misconceptions
If students struggle to identify the input to their function where \(d\) is not an integer, remind them that there are 10 years in a decade. Then ask them how many decades is 6 years? What about one year?
Activity Synthesis
Focus the discussion on why the annual growth rate is not \(\frac1{10}\) of 28.4%. Consider asking students:
 “Why might someone think the annual population growth rate is 2.84%?” (Because 28.4 divided by 10 is 2.84, and they assume that population was growing linearly, or by the same amount each year, rather than by the same factor each year.)
 “How can we determine the growth factor for one year?” (We know that when the annual growth factor, say \(b\), is raised to an exponent of 10, it must equal 1.284, or \(b^{10} = 1.284\), so \(b\) must be \(1.284^\frac{1}{10}\) or \(\sqrt[10]{1.284}\) because raising that number to an exponent of 10 gives 1.284.)
 “How do we express the factor as a growth rate? ( \(1.284^\frac{1}{10} \approx 1.0258\), so the growth rate is 2.58%.)
Supports accessibility for: Visualspatial processing; Conceptual processing
4.3: Got Caffeine? (15 minutes)
Activity
This task prompts students to carefully interpret a description and some expressions representing exponential decay. It further reinforces the distinction between how exponential and linear functions change over intervals of time, which was explored in the previous activity.
The decay factor is not given for a familiar unit of input (that is, the caffeine decays by half every 6 hours, instead of every hour). To correctly represent the decay with functions, students need to notice that the \(\frac12\) does not correspond to the change in 1 hour, and that the decay factor in 1 hour is not \(\frac16\) of \(\frac12\).
Launch
Ask students to close their books or devices. Tell the class they are going to do some calculations around how the amount of caffeine in the body changes over time after consuming a certain amount. Before students read the task statement or do any calculations, ask “How much caffeine do you think is left in a healthy adult 24 hours after they consume 100 mg (if no additional caffeine is consumed)?” Poll the class for their estimates and display the result for all to see. Students will consider this question again during the wholeclass discussion.
Tell students to open their books or devices and read the opening statement of the activity. Explain that the term halflife refers to the time it takes for a certain quantity (such as the concentration of a substance in a body) that is decaying exponentially to decrease by half. In this context, a halflife of 6 hours means that the amount of caffeine in the body is cut in half 6 hours after it is consumed.
Pause students after they complete the first set of questions. Discuss what the 100, the \(\frac12\), and the exponent in each expression represent, and review students’ responses to the questions before they continue with the task.
Supports accessibility for: Language; Organization
Student Facing
In healthy adults, caffeine has an average halflife of about 6 hours. Let's suppose a healthy man consumes a cup of coffee that contains 100 mg of caffeine at noon.
 Each of the following expressions describes the amount of caffeine in the man's body some number of hours after consumption. How many hours after consumption?
 \(100 \boldcdot \left(\frac12\right)^1\)
 \(100 \boldcdot \left(\frac12\right)^3\)
 \(100 \boldcdot \left(\frac12\right)^{\frac16}\)
 \(100 \boldcdot \left(\frac12\right)^t\)

 Write a function \(g\) to represent the amount of caffeine left in the body, \(h\) hours after it enters the bloodstream.
 The function \(f\) represents the amount of caffeine left in the body after \(t\) 6hour periods. Explain why \(g(6) = f(1)\).
Student Response
Teachers with a valid work email address can click here to register or sign in for free access to Student Response.
Student Facing
Are you ready for more?
What percentage of the initial amount of caffeine do you expect to break down in the first 3 hours: less than 25%, exactly 25%, or more than 25%? Explain or show your reasoning.
Student Response
Teachers with a valid work email address can click here to register or sign in for free access to Extension Student Response.
Anticipated Misconceptions
Students may struggle viewing groups of 6 hours as a unit for the halflife of caffeine. Identifying that \(100 \boldcdot \left(\frac{1}{2}\right)^{\frac{1}{6}}\) is the number of mg of caffeine left after 1 hour is the critical step. If students do not see what this expression means in the situation, ask them how any times the scale factor \(\left(\frac{1}{2}\right)^{\frac{1}{6}}\) has to be applied for the caffeine to be cut in half. Then ask them how many hours it takes for the caffeine to be cut in half.
Activity Synthesis
Begin the discussion by repeating the earlier question, “How much caffeine do you think is left in a healthy adult 24 hours after they consume 100 mg (if no additional caffeine is consumed)?” Invite students to share their reasoning, making sure they see that “24 hours later” is 4 periods of 6 hours, so the amount of caffeine will have been halved 4 times: \(100 \boldcdot \left(\frac12\right)^4 = 100 \boldcdot \frac{1}{16} = 6.25\).
Next, focus the discussion on the distinction between expressing the amount of caffeine as a function of time in groups of 6 hours (function \(f\)) and time in groups of 1 hour (function \(g\)). Discuss with students:
 “What decay factor did you use to write the expression for function \(g\)? What is the approximate decay rate per hour?” (Raise \(\frac12\) to an exponent of \(\frac 16\), which is approximately 0.89, so the decay rate is about 11%.)
 “Why not divide \(\frac12\) by 6 (or multiply it by \(\frac16\)) to get the decay factor for an hour?” (Because when that factor is raised to an exponent of 6, it needs to equal \(\frac12\). \(\left(\frac{1}{12}\right)^6\) is not \(\frac12\).)

“How can we show that the expression for \(g\), which is \(100 \boldcdot \left(\left(\frac12\right)^{\frac16}\right)^h\), is equivalent to the expression for \(f\), which is \(100 \boldcdot \left(\frac12\right)^t\)?” (We know that \(h=6t\). If we substitute \(6t\) in the expression for \(g\), we have \(100 \boldcdot \left(\frac12\right)^{(\frac16)^{6t}}\), which equals \(100 \boldcdot \left(\frac12\right)^t\).)
Time permitting, consider showing the two graphs representing the two functions.
Here are some possible questions for discussion about the graphs:

“How are these two graphs alike and different?” (They seem to be the same curve. They have the same vertical intercept. The scales for the horizontal axis are different. One curve goes through \((1,50)\) and the other goes through \((6,50)\).)

“On the graph representing \(f\), what does the point at \((1,50)\) mean in the context? What is the corresponding point on the graph of \(g\)?” (The amount of caffeine in the body after 1 period of 6 hours is 50 milligrams. This is the same as the point \((6,50)\) on the graph of \(g\).)

“How could these two graphs represent the same decay, if the coordinates they go through are not the same?” (They use different units for the input. 1 unit of the input in \(f\) is 6 units of the input in \(g\).)

“To find the amount of caffeine 3 hours after consumption, what input value should we use for each function?” (\(\frac12\) for \(f\) and 3 for \(g\)) “What about for 30 minutes after consumption?” (\(\frac {1}{12}\) for \(f\) and \(\frac12\) for \(g\))
Design Principle(s): Optimize output (for explanation); Maximize metaawareness
Lesson Synthesis
Lesson Synthesis
Tell students that sometimes we want to know how a quantity is changing for different intervals of input (or time, in this case), and those intervals might not be whole numbers. For example, a population of a town \(P\), in thousands, is modeled by \(P=20 \boldcdot (1.5)^t\) where \(t\) is the number of centuries since 1800.
Here are some questions for discussion:
 “What is the growth rate or percent increase per century for this population?” (50%)
 “Why is the percent increase for each decade not 5%?” (The function is exponential, so the percent increase for one decade is not one tenth of the increase for a century. In other words, a 5% growth rate per decade would mean the growth factor would be 1.05 per decade, and that would make the growth factor per century about 1.63, not 1.5.)
 “How do we find the growth factor per decade? How do you know?” (When the growth factor per decade is raised to an exponent of 10, it must equal 1.5. This means the growth factor per decade is \((1.5)^{\frac{1}{10}}\) since \(\left(1.5^{\frac{1}{10}}\right)^{10} = 1.5\).)
 “How do we find the growth rate or percent increase per decade?” (\((1.5)^{\frac{1}{10}} \approx 1.04\), so the growth rate is about 4% per decade.)
4.4: Cooldown  Blog Subscribers (5 minutes)
CoolDown
Teachers with a valid work email address can click here to register or sign in for free access to CoolDowns.
Student Lesson Summary
Student Facing
Imagine a medicine has a halflife of 3 hours. If a patient takes 200 mg of the medicine, then the amount of medicine in their body, in mg, can be modeled by the function \(\displaystyle f(t) = 200 \boldcdot \left(\frac{1}{2}\right)^t\). In this model, \(t\) represents a unit of time. Notice that the 200 represents the initial dose the patient took. The number \(\frac{1}{2}\) indicates that for every 1 unit of time, the amount of medicine is cut in half. Because the halflife is 3 hours, this means that \(t\) must measure time in groups of 3 hours.
But what if we wanted to find the amount of medicine in the patient’s body each hour after taking it? We know there are 3 equal groups of 1 hour in a 3hour period. We also know that because the medicine decays exponentially, it decays by the same factor in each of those intervals. In other words, if \(b\) is the decay factor for each hour, then \(\displaystyle b \boldcdot b \boldcdot b = \frac12\) or \(\displaystyle b^3=\frac12\) . This means that over each hour, the medicine must decay by a factor of \(\sqrt[3]\frac12\), which can also be written as \(\left(\frac12\right)^\frac13\). So if \(h\) is time in hours since the patient took the medicine, we can express the amount of medicine in mg, \(g\), in the person’s body as \(\displaystyle g(h) = 200 \boldcdot \left(\sqrt[3]{\frac{1}{2}}\right)^h\) or \(g(h)=200 \boldcdot \left(\frac12\right)^\frac{h}{3}\).