Lesson 16
Finding and Interpreting Inverse Functions
16.1: Shopping for Cookbooks (5 minutes)
Warmup
Earlier in the course, students have spent some time writing equations in one variable, two variables, and multiple variables to represent relationships and constraints. They have also rearranged equations to isolate a particular variable. This warmup activates that prior knowledge, preparing students to write equations for linear functions that involve two or more operations, and then to write their inverses.
To ensure access, the equations students are expected to write are scaffolded—starting with one operation and moving to two operations, from using numbers and letters to using only letters. The scaffolding also allows students to see structure in the equations, which would in turn be helpful when they try to reverse the process.
Student Facing
Lin is comparing the cost of buying cookbooks at different online stores.
 Store A sells them at \$9 each and offers free shipping.
 Store B sells them at \$9 each and charges \$5 for shipping.
 Store C sells them at \(p\) dollars and charges \$5 for shipping.
 Store D sells them at \(p\) dollars and charges \(f\) dollars for shipping.
 Write an equation to represent the total cost, \(T\), in dollars as a function of \(n\) cookbooks bought at each store.
 Write an equation to find the number of books, \(n\), that Lin could buy if she spent \(T\) dollars at each store.
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Activity Synthesis
Invite students to share their equations. Record and display them for all to see. Ask students who wrote the second set of equations how they went about doing so. Highlight two main points:
 Writing the second set of equations essentially involves undoing each operation in the first equation and doing so in reversed order.
 Each equation in the second set is the inverse of the corresponding equation in the first question. Each isolated variable is now an output, while previously it was an input. The variable that was previously isolated is now an input.
16.2: From Celsius to Fahrenheit (10 minutes)
Activity
In this activity, students investigate inverse functions in the context of temperature scales.
Students are given an equation that defines the temperature in degrees Fahrenheit as a function of the temperature in degrees Celsius (\(F = \frac59 C+32\)) and are asked to find some temperatures in both units. Using this equation to find a temperature in Celsius involves substituting the temperature in Fahrenheit for \(F\) and solving the equation. Students do so several times over. The regularity in repeated reasoning with numerical values (MP8) paves the way for students to write the equation for the inverse function, which entails solving for \(C\) before substituting values of \(F\).
Students are also prompted to verify that two unfamiliar equations represent inverse functions. To do so, students need to analyze the structure of one equation, use it to reverse the process that defines the function, and see if the reversal leads to the other equation (MP7).
Launch
Explain to students that the Rankine scale is a temperature scale that is sometimes used in engineering systems, typically alongside measurements in Fahrenheit.
Provide access to calculators, in case requested.
Student Facing
If we know the temperature in degrees Celsius, \(C\), we can find the temperature in degrees Fahrenheit, \(F\), using the equation:
\(F = \frac95 C + 32\)
 Complete the table with temperatures in degrees Fahrenheit or degrees Celsius.
\(C\) 0 100 25 \(F\) 104 50 62.6  The equation \(F =\frac95 C + 32\) represents a function. Write an equation to represent the inverse function. Be prepared to explain your reasoning.

The equation \(R = \frac{9}{5} (C + 273.15)\) defines the temperature in degrees Rankine as a function of the temperature in degrees Celsius.
Show that the equation \(C = (R − 491.67) \boldcdot \frac{5}{9}\) defines the inverse of that function.
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Student Facing
Are you ready for more?
It was cold enough in Alaska one day so that the temperature was the same in degrees Fahrenheit and degrees Celsius. How cold was it? Explain or show how you know.
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Anticipated Misconceptions
Some students may struggle to solve for \(C\) because of the rational coefficient of \(C\) in \(F=\frac95 C + 32\). They may be able to rearrange the equation to \(F32=\frac95 C\) but then get stuck at that point. Some strategies for supporting their reasoning:
 Think of \(\frac95\) simply as a number being multiplied to \(C\), so to find \(C\), we can divide both sides of the equation by \(\frac 95\). (Students can still reason this way even if they do not recall or know that dividing by \(\frac95\) gives the same result as multiplying by \(\frac59\). Their resulting equation may look like: \((F32) \div \frac95 = C\).)
 Think of \(\frac95 C\) as \(9 \boldcdot \frac15 \boldcdot C\) and then undo one operation at a time.
 Write \(\frac95\) as a decimal, 1.8, and divide both sides of the equation by 1.8. (Clarify that, depending on the fraction, rewriting as a decimal may not always be the most helpful way to go.)
In the course of solving for a variable, students may neglect to apply the distributive property correctly when multiplying a number to an expression. For example, when performing the last step to solve \(F32=\frac95 C\) for \(C\), they may write \(\frac59F  32=C\), instead of \(\frac59 (F32)\). Remind students that any operation performed on both sides of an equation needs to be applied to all the parts on each side.
Activity Synthesis
Discuss with students:
 "How did you find the inverse of the first function?" (By solving the equation for \(F\). By noting what is done to the temperature in Celsius to get the temperature in Fahrenheit and then reversing the steps.)
 "Suppose we want to find the temperature in Celsius when it is 10 degrees Fahrenheit and when it is 90 degrees Fahrenheit. Which equation would you use? Why?" (The second equation, because it doesn't involve solving an equation, which tends to be quicker.)
 "How did you verify that the two equations that relate the temperature in Rankine and the temperature in Celsius are inverse functions?" (By solving the first equation for \(C\) and see if it matches the second equation, or by solving the second equation for \(R\) and see if it matches the first equation.)
 "What does each inverse function tell us?" (The first one tells us how to find the temperature in degrees Celsius if we know the temperature in degrees Fahrenheit. The second tells us how to find the temperature in degrees Celsius if we know the temperature in degrees Rankine.)
If time permits, display the graphs of the two functions relating temperature in Celsius and in Fahrenheit. Emphasize how the input and output variables are switched.
The two points on each graph mark the boiling temperature and freezing temperature. In both graphs, the boiling temperature, \(100 ^\circ C\), is paired with \(212 ^\circ F\), and the freezing temperature, \(0 ^\circ C\), is paired with \(32^\circ F\), but the \(C\) and \(F\)values show up in a different order in the coordinate pairs.)
Design Principle(s): Optimize output; Cultivate conversation
Supports accessibility for: Attention; Socialemotional skill
16.3: Info Gap: Custom Mugs (20 minutes)
Activity
This info gap activity gives students an opportunity to determine and request the information needed to evaluate and find the inverse of a function.
The info gap structure requires students to make sense of problems by determining what information is necessary, and then to ask for information they need to solve it. This may take several rounds of discussion if their first requests do not yield the information they need (MP1). It also allows them to refine the language they use and ask increasingly more precise questions until they get the information they need (MP6).
Here is the text of the cards for reference and planning:
Launch
Design Principle(s): Cultivate Conversation
Supports accessibility for: Memory; Organization
Student Facing
Your teacher will give you either a problem card or a data card. Do not show or read your card to your partner.
If your teacher gives you the problem card:
 Silently read your card and think about what information you need to answer the question.
 Ask your partner for the specific information that you need.
 Explain to your partner how you are using the information to solve the problem.
 Solve the problem and explain your reasoning to your partner.
If your teacher gives you the data card:
 Silently read the information on your card.
 Ask your partner “What specific information do you need?” and wait for your partner to ask for information. Only give information that is on your card. (Do not figure out anything for your partner!)
 Before telling your partner the information, ask “Why do you need that information?”
 After your partner solves the problem, ask them to explain their reasoning and listen to their explanation.
Pause here so your teacher can review your work. Ask your teacher for a new set of cards and repeat the activity, trading roles with your partner.
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Activity Synthesis
After students have completed their work, share the correct answers and ask students to discuss the process of solving the problems. Display both equations for all to see and ask students to compare them. Discuss questions such as:
 "Do both equations define functions? How can we tell?" (Yes. For each price per mug, there is only one possible number of mugs that can be bought. For each number of mugs that the club buys, there is only one possible original price.)
 "In which situation is the first equation more useful?" (When we know the price and want to know how many mugs we can buy.)
 "In which situation is the second equation more useful?" (When we know the number of mugs bought and want to know what the original price was.)
 "If so, are they inverse functions? How do you know?" (Yes, the two equations “undo” each other. The input of the first function is the output in the second function. The output of the first function becomes the input in the second function.)
Highlight for students how solving the first equation for \(p\) gives us the second equation, and solving the second equation for \(n\) gives us the first equation.
16.4: Tables and Seats (15 minutes)
Optional activity
This optional activity allows students to practice writing and interpreting inverse functions. It also prompts students to consider the domain and range of a function and of its inverse, and to do so in terms of a situation. Students think about the set of possible outputs of an original function and then consider, based on what the quantities represent, whether the same set of values makes sense as the domain of the inverse function.
Launch
Design Principle(s): Support sensemaking
Student Facing
At a party, hexagonal tables are placed side by side along one side, as shown here.
 Explain why the equation \(S=4n+2\) represents the number of seats, \(S\), as a function of the number of tables, \(n\).
 What domain and range make sense for this function?
 Write an equation to represent the inverse of the given function. Explain what this inverse function tells us.

How many tables are needed if the following number of people are attending the party? Be prepared to explain your reasoning.
 94 people
 95 people
 What domain makes sense for the inverse function? Is it the same set of values as the range of the original function? Explain your reasoning.
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Activity Synthesis
Invite students to share their response and reasoning. Make sure students see that because the number of seats, \(S\), is found by multiplying the number of table, \(n\), by 4 and then adding 2, the inverse function can be found by subtracting 2 from \(S\) and then dividing the result by 4.
Discuss with students:
 "Why are 23 tables needed for 94 guests but 24 tables needed for just 1 more guest?" (Entering 95 for \(S\) in the equation \(n=\frac{S2}{4}\) and gives \(n=23\frac14\), but a fractional value of \(n\) does not make sense in this situation, so one more table is needed, though there will be open seats.)
 "The range of the first function, the set of \(S\)values, are numbers that are 2 more than multiples of 4 (6, 10, 14, . . .). In the inverse function, is the domain limited to the same set of \(S\)values? Or are numbers like 17 and 85 also possible inputs?"
Highlight that, just as solutions to equations need to be interpreted in context, so do the domain and range of a function and its inverse.
 The equation for the first function, \(S=4n+2\), tells us the number of seats at \(n\) tables, assuming all the tables are filled completely and regardless of whether the seats are occupied by guests. The range values of 6, 10, 14, . . . reflect these assumptions.
 The equation for the inverse function, \(n=\frac{S2}{4}\), gives the number of tables if we know the number of seats, \(S\), but we don't assume that the seats completely fill each table. The domain of the inverse function (the set of \(S\)values) could include any whole number (not just 6, 10, 14, . . . ). When an input gives a fractional number of table for the output (such as \(n=23\frac14\)), we need to interpret it in terms of the situation.
Lesson Synthesis
Lesson Synthesis
Display for all to see the equations students saw in this lesson: the given equation for the temperature function and that for its inverse, as well as the two equations from the mug problem.
\(F = \frac95 C + 32\)
\(C = (F32) \boldcdot \frac59\)
Ask students questions such as:
 "How can we use the first equation to find out the temperature in degrees Celsius when the temperature is 50 degrees Fahrenheit? What about when it is 78 degrees Fahrenheit?" (Substitute \(F\) for 50 and 78, respectively, and then solve the equation.)
 "How can we use the second equation to answer the same questions?" (Evaluate that equation at \(F=50\) and \(F=78\).)
 "When is the first equation more useful?" (When we know the temperature in degrees Celsius and want to find the equivalent in degrees Fahrenheit.)
 "When is the second equation more useful?" (When we know the temperature in degrees Fahrenheit and want to find it in degrees Celsius.)
Highlight that finding the inverse of a function is particularly useful when we have multiple outputs for which we wish to find the inputs. Evaluating a function at a value (using the equation for the inverse) is much more efficient than solving an equation (using the original equation).
16.5: Cooldown  Carnival Functions (5 minutes)
CoolDown
Cooldowns for this lesson are available at one of our IM Certified Partners
Student Lesson Summary
Student Facing
It is helpful to interpret the inverse of a function in terms of a situation and the quantities it represents.
Suppose a linear function gives the dollar cost, \(C\), of renting some equipment for \(n\) hours. The function is defined by this equation:
\(C = 8.25n+30\)
If we know the number of hours of rental, \(n\), we can substitute it into the expression \(8.25n+30\) and evaluate it to find the cost, \(C\).
What is the inverse of this function, and what does it tell us about the length and cost of rental?
To find the inverse, let's solve for \(n\):
\(\begin {align} 8.25n + 30 &= C\\ 8.25n &=C30\\ n&=\dfrac {C30}{8.25} \end{align}\)
If we know the cost of rental, \(C\), we can substitute it into the expression \(\dfrac {C30}{8.25}\) and evaluate it to find the hours of rental, \(n\).
Notice that the equation defining the inverse function is found by reversing the process that defines the original linear function.
 The original rule, \(C=8.25n + 30\), tells us to multiply the input, \(n\), by 8.25 and add 30 to the result to get the output, \(C\).
 The rule of the inverse function, \(\dfrac {C30}{8.25}\), suggests that we subtract 30 from the input and then divide the result by 8.25 to get the output \(n\).