Mathematics II Unit 5
GPS Geometry Frameworks
Teacher’s Edition
Unit 6
Inverse, and Exponential Functions
1st Edition
March, 2011
Georgia Department of Education
Table of Contents
INTRODUCTION: 3
Notes on Please Tell Me in Dollar and Cents Learning Task 6
Notes on Growing by Leaps and Bounds Learning Task 33
GPS Geometry – Unit 6
Piecewise, Inverse, and Exponential Functions
Teacher’s Edition
INTRODUCTION:
In GPS Algebra, students expanded their knowledge of functions to include basic quadratic, cubic, absolute value, and rational functions. They learned to use the notation for functions and to describe many important characteristics of functions. In Unit 5 of GPS Algebra, students studied general quadratic functions in depth. In this unit, students apply their understanding of functions previously studied to explore the concept of inverse function. The exploration of inverse functions leads to investigation of: the operation of function composition, the concept of one-to-one function, and methods for finding inverses of previously studied functions. The unit ends with an examination of exponential functions, equations, and inequalities, with a focus on using basic exponential functions as models of real world phenomena.
ENDURING UNDERSTANDINGS:
• Functions with restricted domains can be combined to form a new function whose domain is the union of the functions to be combined as long as the function values agree for any input values at which the domains intersect.
• Step functions are specific piecewise functions; some well-known step functions can be defined using a single rule or correspondence.
• One-to-one functions have inverse functions.
• The inverse of a function is a function that reverses, or “undoes” the action of the original function.
• The graphs of a function and its inverse function are reflections across the line y = x.
• Exponential functions can be used to model situations of growth, including the growth of an investment through compound interest.
KEY STANDARDS ADDRESSED:
MM2A2. Students will explore exponential functions.
a. Extend properties of exponents to include all integer exponents.
b. Investigate and explain characteristics of exponential functions, including domain and range, asymptotes, zeros, intercepts, intervals of increase and decrease, rates of change, and end behavior.
c. Graph functions as transformations of [pic].
d. Solve simple exponential equations and inequalities analytically, graphically, and by using appropriate technology.
e. Understand and use basic exponential functions as models of real phenomena.
f. Understand and recognize geometric sequences as exponential functions with domains that are whole numbers.
g. Interpret the constant ratio in a geometric sequence as the base of the associated exponential function.
MM2A5. Students will explore inverses of functions.
a. Discuss the characteristics of functions and their inverses, including one-to-oneness, domain, and range.
b. Determine inverses of linear, quadratic, and power functions and functions of the form [pic], including the use of restricted domains.
c. Explore the graphs of functions and their inverses.
d. Use composition to verify that functions are inverses of each other.
RELATED STANDARDS ADDRESSED:
MM2P1. Students will solve problems (using appropriate technology).
a. Build new mathematical knowledge through problem solving.
b. Solve problems that arise in mathematics and in other contexts.
c. Apply and adapt a variety of appropriate strategies to solve problems.
d. Monitor and reflect on the process of mathematical problem solving.
MM2P2. Students will reason and evaluate mathematical arguments.
a. Recognize reasoning and proof as fundamental aspects of mathematics.
b. Make and investigate mathematical conjectures.
c. Develop and evaluate mathematical arguments and proofs.
d. Select and use various types of reasoning and methods of proof.
MM2P3. Students will communicate mathematically.
a. Organize and consolidate their mathematical thinking through communication.
b. Communicate their mathematical thinking coherently and clearly to peers, teachers, and others.
c. Analyze and evaluate the mathematical thinking and strategies of others.
d. Use the language of mathematics to express mathematical ideas precisely.
MM2P4. Students will make connections among mathematical ideas and to other disciplines.
a. Recognize and use connections among mathematical ideas.
b. Understand how mathematical ideas interconnect and build on one another to produce a coherent whole.
c. Recognize and apply mathematics in contexts outside of mathematics.
MM2P5. Students will represent mathematics in multiple ways.
a. Create and use representations to organize, record, and communicate mathematical ideas.
b. Select, apply, and translate among mathematical representations to solve problems.
c. Use representations to model and interpret physical, social, and mathematical phenomena.
Unit Overview:
The unit begins with tasks focus on exploration of inverse functions. In the first task of the unit, conversions of temperatures among Fahrenheit, Celsius, and Kelvin scales and currency conversions among yen, pesos, Euros, and US dollars provide a context for introducing the concept of composition of functions. Reversing conversions is used as the context for introducing the concept of inverse function. Students explore finding inverses from verbal statements, tables of values, algebraic formulas, and graphs. In the second task, students explore one-to-oneness as the property necessary for a function to have an inverse and see how restricting the domain of a non-invertible function can create a related function that is invertible.
The next task introduces exponential functions and explores them through several applications to situations of growth: the spread of a rumor, compound interest, and continuously compounded interest. Students explore the graphs of exponential functions and apply transformations involving reflections, stretches, and shifts. The students will finish up the unit in applying exponential functions to geometric sequences.
TASKS:
The remaining content of this framework consists of student tasks or activities. The first task works with inverses and the second task concentrates on the introduction of exponential functions. Each activity is designed to allow students to build their own algebraic understanding through exploration. There is a student version, as well as a Teacher Edition version that includes notes for teachers and solutions.
Notes on Please Tell Me in Dollar and Cents Learning Task
This learning task addresses standard MM2A5 involving exploration of inverses of functions. The task introduces inverse functions through explorations of functions that convert from one quantity to another. Composition of functions is also introduced here so that students can use composition to verify that functions are inverses of each other (MM2A5, part d). Thus, the discussion of composition in GPS Geometry should be limited to that needed for the study of inverse functions. Students will investigate functions built through operations on functions, including composition, in GPS Pre-Calculus (MM4A4).
Students are introduced to composition of functions through conversion formulas for changing among Fahrenheit, Celsius, and Kelvin temperature scales and through currency conversions involving Japanese yen, Mexican pesos, Euros, and US dollar (Items 1 – 3). The familiar task of temperature conversion allows students to focus on the concept and meaning of function composition; currency conversions (the impetus for the title of the task) give students other examples to motivate the concept of function composition.
The need to convert temperatures, currency, and other quantities in the reverse direction is used to introduce the concept of inverse function. Students first see inverse functions through an example of a function that reverses, or undoes, the action of an original function (Item 4). Students are given a formal definition of inverse functions as functions whose compositions result in an identity function and then explore the meaning of the definition by starting from a verbal statement of the action of a function and producing a verbal statement of the action of its inverse to algebraic formulas. (Item 5) Students are asked to translate the verbal statements of functions and their inverses into algebraic formulas as an introduction to the idea of finding the formula for the inverse function (Item 6). Next, students apply the definition of inverse function to tables giving values for a function and its inverse to develop the Inverse Function Property (Items 7 – 8).
The Inverse Function Property is used as the guiding principle for an algebraic process to find a formula for the inverse given a formula for the original function. This algebraic process is developed by working with the inverse of the function that converts Fahrenheit to Celsius and then applied currency conversions and then to a function of the form[pic] which has a restricted domain (Items 9 – 11). In the later part of the task, students explore the relationship between the graph of an original function and its inverse function (Items 12 – 14).
Determining whether a given function has an inverse by considering whether it is one-to-one, finding inverses for quadratic and power functions, and future exploration of the characteristics of functions and their inverses are addressed in the next task.
Definitions and properties:
• Composition of functions: If f and g are functions, the composite function [pic] (read this notation as “f composed with g) is the function with the formula[pic], where x is in the domain of g and g(x) is in the domain of f.
• Inverse functions: If f and h are two functions such that [pic] for each input x in the domain of f, and [pic] for each input x in the domain of h, then h is the inverse of the function f, and we write h = [pic]. Also, f is the inverse of the function h, and we can write f = [pic].
• A function that has an inverse function is called invertible.
• Inverse Function Property: For any invertible function f and any real numbers a and b in the domain and range of f, respectively,[pic] if and only if [pic]. In terms of points on the graphs, (a, b) is a point on the graph of f if and only if (b, a) is a point on the graph of [pic].
Supplies needed:
• Calculator
• Graphing utility
• Graph paper and colored pencils
Please Tell Me in Dollars and Cents Learning Task
1. Aisha made a chart of the experimental data for her science project and showed it to her science teacher. The teacher was complimentary of Aisha’s work but suggested that, for a science project, it would be better to list the temperature data in degrees Celsius rather than degrees Fahrenheit.
a. Aisha found the formula for converting from degrees Fahrenheit to degrees Celsius: [pic].
Use this formula to convert freezing (32°F) and boiling (212°F) to degrees Celsius.
Comment(s):
Students are asked to do these calculations as introductory exploration of this function. Many students likely know the corresponding Celsius temperatures; this part asks them to use the formula to find the values.
Solution(s):
Freezing: [pic], or freezing is 0°C.
Boiling: [pic], or boiling is 100°C.
b. Later Aisha found a scientific journal article related to her project and planned to use information from the article on her poster for the school science fair. The article included temperature data in degrees Kelvin. Aisha talked to her science teacher again, and they concluded that she should convert her temperature data again – this time to degrees Kelvin. The formula for converting degrees Celsius to degrees Kelvin is
[pic].
Use this formula and the results of part a to express freezing and boiling in degrees Kelvin.
Comment(s):
This part is about giving some sense of Kelvin temperatures and comprehending the meaning of the conversion formula.
Solution(s):
Freezing: [pic], or freezing is 273°K.
Boiling: [pic], or boiling is 373°K.
c. Use the formulas from part a and part b to convert the following to °K: – 238°F, 5000°F .
Comment(s):
This part is designed to prepare students for idea of composing the two functions by having them do the two step process.
Solution(s):
Converting – 238°F to °K : [pic]; [pic] Thus, – 238°F is 123°K.
Converting 5000°F to °K : [pic]
[pic] Thus, 5000°F is 3303°K.
In converting from degrees Fahrenheit to degrees Kelvin, you used two functions, the function for converting from degrees Fahrenheit to degrees Celsius and the function for converting from degrees Celsius to degrees Kelvin, and a procedure that is the key idea in an operation on functions called composition of functions.
Composition of functions is defined as follows: If f and g are functions, the composite function [pic] (read this notation as “f composed with g) is the function with the formula
[pic],
where x is in the domain of g and g(x) is in the domain of f.
2. We now explore how the temperature conversions from Item 1, part c, provide an example of a composite function.
a. The definition of composition of functions indicates that we start with a value, x, and first use this value as input to the function g. In our temperature conversion, we started with a temperature in degrees Fahrenheit and used the formula to convert to degrees Celsius, so the function g should convert from Fahrenheit to Celsius: [pic]. What is the meaning of x and what is the meaning of g(x) when we use this notation?
Comment(s):
This part and the next involve the key conceptual step of moving to function notation where all of the inputs are expressed with the variable x. This is the notation they will use when they work with the inverse of a function.
Solution(s):
Here “x” is a temperature in degrees Fahrenheit and “g(x)” is the corresponding temperature in degrees Celsius.
b. In converting temperature from degrees Fahrenheit to degrees Kelvin, the second step is converting a Celsius temperature to a Kelvin temperature. The function f should give us this conversion; thus,[pic]. What is the meaning of x and what is the meaning of f (x) when we use this notation?
Here “x” is a temperature in degrees Celsius and “f(x)” is the corresponding temperature in degrees Kelvin.
c. Calculate [pic]. What is the meaning of this number?
Comment(s):
This part is designed to help students build familiarity with the notation. Students will organize their work in a variety of ways; the solution below provides one example.
Solution(s):
[pic]; [pic]
The value of [pic] is the temperature in °K that corresponds to 45°F.
d. Calculate [pic], and simplify the result. What is the meaning of x and what is the meaning of [pic]?
Comment(s):
This part is designed to introduce students to the types of calculations that they will need to do in verifying inverse functions using composition. Extensive practice in simplifying formulas created using composition is reserved for GPS Pre-Calculus, standard MM4A4, part c.
Solution(s):
[pic]
Here “x” is a temperature in degrees Fahrenheit and “[pic])” is the corresponding temperature in degrees Kelvin.
e. Calculate [pic] using the formula from part d. Does your answer agree with your calculation from part c?
Comment(s):
This part insures that students realize that the simplified version of the composition formula eliminates the need for a two-step process and helps verify their calculations from part d.
Solution(s):
[pic]
Yes, this computation gives the same answer.
f. Calculate [pic], and simplify the result. What is the meaning of x? What meaning, if any, relative to temperature conversion can be associated with the value of [pic]?
Comment(s):
This part is designed to make students notice the domain requirements for function composition. Here the issue is not whether the value can be calculated but instead is the issue of meaning. It brings home the point that, when we are working with functions that have real-world contexts, the contexts must “match-up” in order to form meaningful compositions.
Solution(s):
[pic]
Here “x” is a temperature in degrees Celsius. We cannot associate a meaning to “[pic])” relative to temperature conversion since “f(x)” is a temperature in degrees Kelvin, but an input to the function g should be a temperature in degrees Fahrenheit.
We now explore function composition further using the context of converting from one type of currency to another.
3. On the afternoon of May 3, 2009, each Japanese yen (JPY) was worth 0.138616 Mexican pesos (MXN), each Mexican peso was worth 0.0547265 Euro (EUR), and each Euro was worth 1.32615 US dollars (USD).[1]
a. Using the rates above, write a function P such that P(x) is the number of Mexican pesos equivalent to x Japanese yen.
Comment(s):
Students may need to experiment with some specific values, such as starting with 1000 yen, in order to decide how to write the formula. Alternately, writing a proportion may help:
[pic].
Writing conversion functions sometimes seems backwards to students. For example, they know that 1 foot = 12 inches, but, to write a function that converts from feet to inches, the formula would be I = 12F, where I is the number of inches and F is the number of feet. Thus, in this situation, function notation may help avoid confusion.
If students are bothered by the formula, it can also be explained with the unit factors method used extensively in science classes, as indicated in the solution below.
Solution(s):
[pic]; thus, [pic], where x is a number of Japanese yen and P(x) is the corresponding number of Mexican pesos.
b. Using the rates above, write a function E that converts from Mexican pesos to Euros.
Comment(s):
This part is similar to part a, but here students must determine the meaning of x and E(x).
Solution(s):
[pic], where x is a number of Mexican pesos and E(x) is the corresponding number of Euros.
c. Using the rates above, write a function D that converts from Euros to US dollars.
[pic], where x is a number of Euros and D(x) is the corresponding number of US dollars.
d. Using functions as needed from parts a – c above, what is the name of the composite function that converts Japanese yen to Euros? Find a formula for this function. (Original values have six significant digits; use six significant digits in the answer.)
Comment(s):
This part and the two that follow are designed to reinforce student understanding of the notation for function composition as well as the concept. If students use current data from the internet and compare their numbers to the stated rate for conversion from yen to Euros, they may find differences in later digits because these values change over very short time intervals.
Unless students consistently round to the same number of significant digits, they will get different answers in part f below due to rounding error. If students do not follow rounding instructions and get different answers, there may need to be a review of scientific notation and the concept of significant digits. Scientific notation is included in the Grade 8 curriculum (standard M8N1, part j). This real-world application of currency conversion gives an opportunity to show that it is the consistent use of significant digits, and not the use of the same number of decimal places, that leads to reliable results.
Solution(s):
The function is [pic]. [pic]
e. Using functions as needed from parts a – c above, what is the name of the composite function that converts Mexican pesos to US dollars? Find a formula for this function. (Use six significant digits in the answer.)
The function is [pic]. [pic]
f. Using functions as needed from parts a – c above, what is the name of the composite function that converts Japanese yen to US dollars? Find a formula for this function. (Use six significant digits in the answer.)
Comment(s):
Some students may write the name of the function as [pic]. Others may write [pic]. Students are accustomed to having the Associative Law because all of the algebraic properties they have studied so far are associative. That either calculation gives the same result follows from the fact that function composition is an associative operation. The associativity of function composition should be mentioned, but intensive study of the operation of composition is reserved for GPS Pre-Calculus.
Solution(s):
Function name: [pic] = [pic] = [pic]
[pic] or
[pic] or
[pic]
g. Use the appropriate function(s) from parts a - f to find the value, in US dollars, of the following: 10,000 Japanese yen; 10,000 Mexican pesos; 10,000 Euros.
Comment(s):
This part of the item brings closure to this aspect of the discussion by having students select the correct function for a variety of conversions to US dollars.
Solution(s):
10,000 yen to $: [pic]
10,000 Mexican pesos to $: [pic]
10,000 Euros to $: [pic]
Returning to the story of Aisha and her science project: it turned out that Aisha’s project was selected to compete at the science fair for the school district. However, the judges made one suggestion – that Aisha express temperatures in degrees Celsius rather than degrees Kelvin. For her project data, Aisha just returned to the values she had calculated when she first converted from Fahrenheit to Celsius. However, she still needed to convert the temperatures in the scientific journal article from Kevin to Celsius. The next item explores the formula for converting from Kelvin back to Celsius.
4. Remember that the formula for converting from degrees Celsius to degrees Kelvin is
[pic].
In Item 2, part b, we wrote this same formula by using the function f where [pic]represents the Kelvin temperature corresponding to a temperature of x degrees Celsius.
a. Find a formula for C in terms of K, that is, give a conversion formula for going from °K to °C.
Comment(s):
This item begins the students’ introduction to the concept of inverse function. This first example deliberately uses a very simple formula so that students can initially focus on the concept of reversing the action of a given function.
Solution(s):
We subtract 273 from both sides of the given equation to obtain: [pic]. Thus, the conversion formula is: [pic].
b. Write a function h such that [pic]is the Celsius temperature corresponding to a temperature of x degrees Kelvin.
Comment(s):
One of the challenges of working with inverse functions comes from the fact that it is standard to write all the functions with the same variable to represent the input. This part has students make this change in notation for this simple example. Removing the “C” and “K” should help students focus on the action of the function as “subtract 273 from the input.”
Solution(s):
[pic]
c. Explain in words the process for converting from degrees Celsius to degrees Kelvin. Do the equation [pic] and the function f from Item 2, part b both express this idea?
Comment(s):
Throughout this task there is an underlying goal of helping students see the variable in a function formula as an algebraic representative of the input and to read formulas as statements of operations to perform on the input in a particular order (as specified by the rules for order of operations). This part of the item and the next ask students to verbalize such understanding for the particular example of conversions between Celsius and Kelvin temperatures.
Solution(s):
To convert from degrees Celsius to degrees Kelvin, we add 273 to the Celsius temperature. Yes, both functions use degrees Celsius as input and indicate to add 273 to the input.
d. Explain verbally the process for converting form degrees Kelvin to degrees Celsius. Do your formula from part a above and your function h from part b both express this idea?
To convert from degrees Kelvin to degrees Celsius, we subtract 273 from the Kelvin temperature. Yes, both functions use degrees Kelvin as input and indicate to subtract 273 from the input.
e. Calculate the composite function [pic], and simplify your answer. What is the meaning of x when we use x as input to this function?
Comment(s):
This part and the next one continue the process of leading up to the formal concept of inverse function by illustrating that composing inverse functions leaves us with the identity function.
Solution(s):
[pic]
In this function, since x is an input to the function f, x represents a temperature in degrees Celsius.
f. Calculate the composite function [pic], and simplify your answer. What is the meaning of x when we use x as input to this function?
[pic]
In this function, since x is an input to the function h, x represents a temperature in degrees Kelvin.
In working with the functions f and h in Item 4, when we start with an input number, apply one function, and then use the output from the first function as the input to the other function, the final output is the starting input number. Your calculations of [pic] and [pic] show that this happens for any choice for the number x. Because of this special relationship between f and h , the function h is called the inverse of the function f and we use the notation [pic] (read this as “f inverse”) as another name for the function h.
The precise definition for inverse functions is: If f and h are two functions such that
[pic] for each input x in the domain of f,
and
[pic] for each input x in the domain of h,
then h is the inverse of the function f, and we write h = [pic]. Also, f is the inverse of the function h, and we can write f = [pic].
Note that the notation for inverse functions looks like the notation for reciprocals, but in the inverse function notation, the exponent of “–1” does not indicate a reciprocal.
5. Each of the following describes the action of a function f on any real number input. For each part, describe in words the action of the inverse function, [pic], on any real number input. Remember that the composite action of the two functions should get us back to the original input.
a. Action of the function f : subtract ten from each input
Action of the function [pic]:
b. Action of the function f : add two-thirds to each input
Action of the function [pic]:
c. Action of the function f : multiply each input by one-half
Action of the function [pic]:
d. Action of the function f : multiply each input by three-fifths and add eight
Action of the function [pic]:
Comment(s):
This item is included to reinforce conceptual understanding of the inverse function concept as “undoing” the action of the original function. Students may need to try out some specific numbers to determine the answer for part d.
Once students have verbalized their answers, it may be helpful to discuss that the functions in parts a and b involve additive inverses, since subtraction can be viewed as adding the additive inverse of a number, and that part c involves multiplication by a number and its multiplicative inverse. These examples show that the “inverse” in inverse functions is related to inverse number operations with which students are very familiar.
Solution(s):
a. Action of [pic]: add ten to each input
b. Action of [pic]: subtract two-thirds from each input
c. Action of [pic]: multiply each input by two
d. Action of [pic]: subtract eight from each input and then multiply the result by five-thirds
6. For each part of Item 5 above, write an algebraic rule for the function and then verify that the rules give the correct inverse relationship by showing that [pic] and [pic] for any real number x.
Comment(s):
By approaching these inverses with verbal statements first, students get several examples of algebraic formulas for inverse functions before discussing a process for finding a formula for the inverse of a function given by its formula. In this way, students can understand that process rather than memorize an algorithm.
In part d, students who leave off the parentheses should be encouraged to check some specific number to make sure that they have expressed the function correctly.
Solution(s):
a. Formulas for f and [pic]: [pic], [pic]
[pic], [pic]
b. Formulas for f and [pic]: [pic], [pic]
[pic], [pic]
c. Formulas for f and [pic]: [pic], [pic]
[pic], [pic]
d. Formulas for f and [pic]: [pic], [pic]
[pic] , [pic]
Before proceeding any further, we need to point out that there are many functions that do not have an inverse function. We’ll learn how to test functions to see if they have an inverse in the next task. The remainder of this task focuses on functions that have inverses. A function that has an inverse function is called invertible.
7. The tables below give selected values for a function f and its inverse function [pic].
a. Use the given values and the definition of inverse function to complete both tables.
|x |[pic] |
|3 |15 |
|5 |10 |
|7 |6 |
|9 |3 |
|11 |1 |
|x |f (x) |
|1 |11 |
|3 |9 |
|6 |7 |
|10 |5 |
|15 |3 |
Comment(s):
Students will need to start by translating needed table entries into statements of the form [pic] or [pic]. Then, they can apply the inverse function to both names for the number to determine how to find the desired value in the other table. The number 3 is included as both an input and an output to emphasize the care students need to take in distinguishing inputs and outputs.
Solution(s):
[pic]
[pic]
[pic]. We see from the table for values of f that [pic].
[pic]. We see from the table for values of f that [pic].
[pic]. We see from the table for values of f that [pic].
b. For any point (a, b) on the graph of f, what is the corresponding point on the graph of [pic]?
Comment(s):
In determining the answers to complete the tables, students will have observed that [pic] if and only if [pic]. This part asks students to express this idea in terms of points on the graph as a simple way for them to express the idea that inputs and outputs reverse between a function and its inverse.
Solution(s):
If (a, b) is a point on the graph of f, then (b, a) is a point on the graph of [pic].
c. For any point (b, a) on the graph of [pic], what is the corresponding point on the graph of f ? Justify your answer.
Comment(s):
This part requires students to verbalize the converse of the relationship stated in part b. Asking both questions should remind students that the converse is not logically equivalent to the original statement.
Solution(s):
If (b, a) is a point on the graph of [pic], (a, b) is a point on the graph of f .
As you have seen in working through Item 7, if f is an invertible function and a is the input for function f that gives b as output, then b is the input to the function [pic] that gives a as output. Conversely, if f is an invertible function and b is the input to the function [pic] that gives a as output, then a is the input for function f that gives b as output. Stated more formally with function notation we have the following property:
Inverse Function Property: For any invertible function f and any real numbers a and b in the domain and range of f, respectively,
[pic] if and only if [pic].
8. Explain why the Inverse Function Property holds, and express the idea in terms of points on the graphs of f and [pic].
Comment(s):
Students should be presented this part of the task after they have worked on Item 7.
This item requires them to see that the computations done for Item 7 can be done in general. Students may need to review the meaning of the biconditional “if and only if” to remember that such a statement is equivalent to two separate implications.
Solution(s):
The following arguments explain why the Inverse Function Property holds.
i) Suppose that f is an invertible function and that a and b are numbers so that [pic]. Using the equal expressions as inputs to the inverse function gives equals outputs; hence, [pic]. However, [pic], so [pic], or [pic].
ii) Suppose that we start by knowing that f is an invertible function and that [pic].
Using the equal expressions as inputs to the inverse function gives equals outputs; hence, [pic]. However, [pic], so [pic], or [pic].
Alternate version of the Inverse Function Property: The point (a, b) is on the graph of an invertible function f if and only if the point (b, a) is on the graph of [pic].
9. After Aisha had converted the temperatures in the scientific journal article from Kelvin to Celsius, she decided, just for her own information, to calculate the corresponding Fahrenheit temperature for each Celsius temperature.
a. Use the formula [pic] to find a formula for converting temperatures in the other direction, from a temperature in degrees Celsius to the corresponding temperature in degrees Fahrenheit.
Comment(s):
This part starts the process for finding the formula for the inverse of a function. By asking students to find the inverse of the formula given with the letters “C” and “F”, students should easily conclude that they need to solve for F in terms of C. This sets the stage for the general process.
Solution(s):
[pic]
b. Now let [pic], as in Item 2, so that [pic] is the temperature in degrees Celsius corresponding to a temperature of x degrees Fahrenheit. Then [pic] is the function that converts Celsius temperatures to Fahrenheit. Find a formula for [pic].
Comment(s):
Students should realize that they have already found the formula in part a and just need to convert to using x for inputs and [pic] for outputs.
Solution(s):
[pic]
c. Check that, for the functions g and [pic] from part b, [pic] and [pic] for any real number x.
Comment(s):
This part gives students important practice with algebraic skills and verifies that students have the correct formula. Some students may know that zero degrees Kelvin is absolute zero so that there is a minimum temperature of – 273 degrees Celsius, which is [pic] degrees Fahrenheit. (This site provides a good discussion of the concept of absolute zero including some neat graphics .) Teachers should decide whether to discuss appropriate restrictions on the domains of g and [pic]depending on the scientific knowledge and mathematical maturity of the students in the class. Note that whether there is a hottest temperature is a hot topic (pun intended) among physicists, so there is no agreed upon value to use as an upper limit for the domains and ranges of g and [pic]. For these reasons, the instructions here ask students to verify the inverse properties for the unrestricted domain of all real numbers.
Solution(s):
[pic], and
[pic]
Our next goal is to develop a general algebraic process for finding the formula for the inverse function when we are given the formula for the original function. This process focuses on the idea that we usually represent functions using x for inputs and y for outputs and applies the inverse function property.
10. We now find inverses for some of the currency conversion functions of Item 3.
a. Return to the function P from Item 3, part a, that converts Japanese yen to Mexican pesos. Rewrite the formula replacing [pic] with y and then solve for x in terms of y.
Comment(s):
This part starts teaching a process of finding the inverse function by focusing on the fact that [pic] if and only if [pic]. Once students have a formula for [pic] in terms of y, they can simply use x in place of y to write the formula.
Note that we round [pic] to 7.21417 in order to be consistent with the currency conversion sites where the conversion factor is always written as a number in decimal format. Consistent with previous computations, we round the number to six significant digits.
Solution(s):
Replacing: [pic]
Solving for x in terms of y:[pic]
b. The function [pic]converts Mexican pesos back to Japanese yen. By the inverse function property, if [pic], then [pic]. Use the formula for x, from part a, to write a formula for [pic] in terms of y.
Comment(s):
This part explains the next step in the process of finding the inverse. In parts d and e below, students will follow the process without step by step guidance.
Solution(s):
[pic]
c. Write a formula for [pic].
Comment(s):
This part explains the last step in the process of finding the inverse. It requires that students comprehend that changing the variable used to write a function formula does not change the function. Earlier work in this task has been designed to help students gain, or deepen, this understanding before the need to apply it here.
Solution(s):
[pic] Hence, [pic].
d. Find a formula for [pic], where E is the function that converts Mexican pesos to Euros from Item 3, part b.
Comment(s):
Students need to copy the process shown in parts a – c.
Some authors outline this process as one in which the first step is to swap x and y. The advantage of this process is that students are solving for y to find the formula for the inverse function. The disadvantage is that it turns the process into more of an algorithm to be memorized than one to be understood. The process described here is designed to be a natural progression based on the inverse function property.
Solution(s):
Replacing: [pic]
Solving for x in terms of y:[pic]
[pic]
Formula for the inverse function: [pic]
e. Find a formula for [pic], where D is the function that converts Euros to US dollars from Item 3, part c.
Comment(s):
Students need to copy the process shown in parts a – c.
Solution(s):
Replacing: [pic]
Solving for x in terms of y:[pic]
Formula for the inverse function: [pic]
11. Aisha plans to include several digital photos on her poster for the school-district science fair. Her teacher gave her guidelines recommending an area of 2.25 square feet for photographs. Based on the size of her tri-fold poster, the area of photographs can be at most 2.5 ft high. Aisha thinks that the area should be at least 1.6 feet high to be in balance with the other items on the poster.
Comment(s):
This item is designed to introduce inverses of “functions of the form [pic], including the use of restricted domains” (see part b of MM2A5) and to help students understand that the range of the original function gives the domain of the inverse function.
a. Aisha needs to decide on the dimensions for the area for photographs in order to complete her plans for poster layout. Define a function W such that W(x) gives the width, in feet, of the photographic area when the height is x feet.
Comment(s):
At this point in their study of functions, students should realize that the definition will require a formula and a statement of restrictions on the domain.
Solution(s):
The width of the poster area for photographs is given by
[pic], [pic], where x is the height in feet of the poster area for photographs.
b. Write a definition for the inverse function, [pic].
Comment(s):
In finding the formula for W, students use the relationship that area = width*height. Thus, it is likely that many will immediately write the formula for [pic] by using the symmetry in the relationship. Those who follow the process outlined in Item 10 may be surprised that the inverse has the same formula as the original function. Students may think that [pic] is the same function as W and need to remember that, to be equal, functions must have the same rule and the same domain.
The possible values for the width of the area for photographs constitute the domain of [pic]. These values also make up the range of W. Students need to consider the graph of W to determine that the function is decreasing and, hence, that the range is the interval of values (from largest to smallest) of [pic] to [pic]. Students may determine the shape of the graph of W without actually graphing the function because the formula yields a function that is a vertical stretch of the basic reciprocal function, [pic]. Others may need to see the graph so a graph of W is shown at the right.
Solution(s):
[pic], [pic]
In the remaining items you will explore the geometric interpretation of this relationship between points on the graph of a function and its inverse.
12. We start the exploration with the function W from Item 11.
a. Use technology to graph the functions W and [pic] on the same coordinate axes. Use a square viewing window.
Comment(s):
Students need to use the techniques learned earlier in this unit for using technology to graph functions with restricted domains. It is important that students use a square viewing window so that they begin to see that the graphs of a function and its inverse are reflections across the line y = x.
For TI calculators, students should graph [pic] and [pic]. A window like the one in the graph at the right is recommended; set the window and then zoom to a square window.
Solution(s):
The graph is shown at the right below. Endpoints of each graph are labeled in anticipation of the next part of the item.
b. State the domain and range of the function W.
Domain of W: [pic]; Range of W: [pic]
c. State the domain and range of the function [pic].
Domain of [pic]: [pic]; Range of [pic]: [pic]
d. In general, what are the relationships between the domains and ranges of an invertible function and its inverse? Explain your reasoning.
Comment(s):
The explanation below shows set equality by establishing that each set is a subset of the other.
Solution(s):
If f is an invertible function, then the domain of f is equal to the range of [pic] and the domain of [pic] is equal to the range of f. This happens because of the inverse function property.
Each number in the domain of f is the first coordinate of a point on the graph of f so it becomes the second coordinate of a point on the graph of [pic]and hence is a number in the range of [pic]. Thus, the domain of f is a subset of the range of [pic]. Similarly, any number in the range of [pic] is the second coordinate of a point on the graph of [pic] so it becomes the first coordinate of a point on the graph of f and hence is a number in the domain of f. Thus, the range of [pic] is a subset of the domain of f. This explains why the domain of f is equal to the range of [pic].
Each number in the domain of [pic] is the first coordinate of a point on the graph of [pic] so it becomes the second coordinate of a point on the graph of f and hence is a number in the range of f. Thus, the domain of [pic] is a subset of the range of f. Similarly, any number in the range of f is the second coordinate of a point on the graph of f so it becomes the first coordinate of a point on the graph of [pic]and hence is a number in the domain of [pic]. Thus, the range of f is a subset of the domain of [pic]. This explains why the domain of [pic] is equal to the range of f.
13. Explore the relationship between the graph of a function and the graph of its inverse function. For each part below, use a standard, square graphing window with [pic] and [pic].
Comment(s):
By examining these graphs on a square window, in addition to having graphed W and [pic] , students have a variety of examples from which to deduce that the graph of the inverse is the reflection of the graph of the original function across the line y = x.
a. For functions in Item 6, part a, graph f, [pic], and the line y = x on the same axes.
b. For functions in Item 6, part c, graph f, [pic], and the line y = x on the same axes.
c. For functions in Item 6, part d, graph f, [pic], and the line y = x on the same axes.
Solution(s):
a. Item 6, part a b. Item 6, part c
c. Item 6, part d
d. If the graphs were drawn on paper and the paper were folded along the line y = x, what would happen?
When the graphs are drawn on paper and the paper is folded along the line y = x, the graphs of f and [pic]coincide.
e. Do you think that you would get the same result for the graph of any function f and its inverse when they are drawn on the same axes using the same scale on both axes? Explain your reasoning.
Comment(s):
It is likely that students will look at other functions and their inverses before deciding on their answers to this question. They should recognize that the question relates to the graphs being reflections across the line y = x. The restriction on the domain of the function W in Item 11 was chosen so that students would see a distinct inverse that is the reflection of the original function across the line y = x.
Once students become convinced that the graphs will always coincide, they need to develop reasoning that connects this phenomenon to the characteristics of a function and its inverse. Students may come to the conclusion that it is because of the inverse function property, that (a, b) is on the graph of a function if and only if (b, a) is on the graph of the inverse. Then, the key issue will be to realize that that they need to justify why reflecting the point (a, b) across the line y = x produces the point (b, a).
To give such a justification, students will need to remember the geometric relationships among a point, a line of reflection, and the reflected point. The reflection of a point through a line requires a line through the initial point that is perpendicular to the line of reflection. The point of reflection is on the other side of this perpendicular line from the initial point and is the same distance from the line of reflection as the original point as shown in the diagram above.
Two lines of reasoning for why (b, a) is the reflection of (a, b) through the line y = x. Many other lines of reasoning are also possible.
Solution(s):
Yes, the graph of a function and its inverse will always coincide if both functions are drawn on the same axes using the same scale on both axes and we fold the paper along the line y = x.
If we use the same scale on both axes, when we fold along the line y = x, the points that coincide are reflections of each other through the line y = x. When we have functions f and [pic], we know that for each point (a, b) on the graph of one, the point (b, a) is on the graph of the other. The points (a, b) and (b, a) are reflections of each other across the line y = x, as we explain below.
Explanation 1:
If a = b, then (a, b) = (a, a), so the point (a, b) is on the line y = x. Its reflection across the line is itself, which also be labeled (b, a).
If a < b, then we can draw a rectangle with corners (a, a), (a, b), (b, b), and (b, a) as shown in the diagram at the right. This rectangle is a square because each side has length b – a. The diagonal through (a, a) and (b, b) is the line y = x. The diagonal through (a, b) and (b, a) is the perpendicular bisector of the first diagonal. Therefore, the points (a, b) and (b, a) reflections of each other across the line y = x.
If b < a, then we can draw a rectangle with corners (a, a), (a, b), (b, b), and (b, a) as shown in the diagram at the right. This rectangle is a square because each side has length a – b. The argument proceeds as in the case for a < b.
Explanation 2:
The line through the points (a, b) and (b, a) has slope [pic]. The equation of the line is [pic] which simplifies to [pic]. This line has slope [pic], so the two lines are perpendicular. Solving the system
[pic] , we find that the two lines intersect at the point [pic]. The distance from the point (a, b) to [pic] is [pic].
The distance from the point (b, a) to [pic] is
[pic].
The distances are the same so the points are reflections of each other across the line y = x.
14. Consider the function [pic].
a. Find the inverse function algebraically.
Comment(s):
This function is included to allow an application of the reflection property for the graphs of an invertible function and its inverse and to give further exploration of inverses of functions of the form [pic].
Solution(s):
Substitute y for f(x) and solve for x: [pic].
Then, [pic] so that [pic], which is the same formula as the original function.
The graph of [pic] can be obtained from the graph of [pic]by performing a vertical stretch by a factor of 3 followed by a reflection through the x-axis. Thus, the function f has the set of all real numbers except 0 as its domain and range. Since the domain and range are the same set, when these are reversed to form [pic], we have the same domain and range as the function f. Therefore, [pic]is exactly the same function as f, that is, f is its own inverse.
b. Draw an accurate graph of the function f on graph paper and use the same scale on both axes.
The graph is shown on the next page. The line y = x is shown as a dotted line for reference in answering part c.
c. What happens when you fold the paper along the line y = x? Why does this happen?
Comment(s):
It may help students to understand what is happening here to also consider functions with the same formula but with the domain restricted to negative values of x or restricted to positive values of x so that they can see that each “half” of the function f is the inverse of the other half.
Solution(s):
When we fold the paper along the line y = x, the piece of the graph corresponding to negative values of x coincides with the piece of the graph for positive value of x. This shows us that, when we reflect the graph through the line y = x, we get exactly the same graph. This happens because f is its own inverse.
Notes on Growing by Leaps and Bounds Learning Task
Introduction:
This task introduces students to exponential functions. At this point in their study, students have extended their understanding of exponents to include all integer values but have not yet discussed rational or real number exponents. In Part 1, students investigate a mathematical model of spreading a rumor in which the domain of the function is limited to a finite set of nonnegative integers. In Part 2, students learn the definition of an exponential function and see the model from Part 1 as an example of such a function. The emphasis in Part 2 is the pattern for the formula of an exponential function and an introduction to the shape of the graph. In Part 3, students work with the compound interest formula. In Part 4, students learn about the graphs of exponential functions with domains of all real numbers and perform transformations on these functions: vertical stretches and shrinks, reflections in the x-axis, vertical shifts, and horizontal shifts. Students explore these graphs without addressing the meaning of non-integer exponents.
This task provides a guided discovery for the following:
The characteristics of the formula for and graphs of exponential functions
Application of the compound interest formula
Transformations of exponential functions
Supplies Needed:
Graph paper
Graphing utility
Optional: spreadsheet software
Part 1: Meet Linda
Linda’s lifelong dream had been to open her own business. After working and sacrificing and saving, she finally had enough money to open up an ice cream business. The grand opening of her business is scheduled for the Friday of Memorial Day weekend. She would like to have a soft opening for her business on the Tuesday before. The soft opening should give her a good idea of any supply or personnel issues she has and give her time to correct them before the big official opening.
A soft opening means that the opening of the business is not officially announced; news of its opening is just spread by word of mouth (see, not all rumors are bad!). Linda needs a good idea of when she should begin the rumor in order for it to spread reasonably well before her soft opening. She has been told that about 10% of the people who know about an event will actually attend it. Based on this assumption, if she wants to have about 50 people visit her store on the Tuesday of the soft opening, she will need 500 people to know about it.
1. Linda plans to tell one person each day and will ask that person to tell one other person each day through the day of the opening, and so on. Assume that each new person who hears about the soft opening is also asked to tell one other person each day through the day of the opening and that each one starts the process of telling their friends on the day after he or she first hears. When should Linda begin telling others about the soft opening in order to have at least 500 people know about it by the day it occurs?
Comment(s):
With the table in item 2 below, it is likely that many students will organize their work in a similar way. Whether or not they use such a table, they will need to count up from the first day that Linda begins to spread the news to find out how many days it will take for the number of people who know to reach 500 and then count backwards to determine the day Linda should start. The Memorial Day reference in the problem gives a convenient way to express the answer.
Solution(s):
Linda should tell her first person about the soft opening on Monday two weeks before Memorial Day because:
1st day: Linda tells one other person – 2 people know.
2nd day: Each of the two people who know tell another person – 4 people know.
3rd day: Each of the four people who know tell another person – 8 people know.
4th day: 16 people know
5th day: 32 people know
6th day: 64 people know
7th day: 128 people know
8th day: 256 people know
9th day: 512 people know
The 9th day corresponds to the Tuesday of the soft opening. So, the 2nd day is the Tuesday one week before, and the 1st day is the Monday that is two weeks before Memorial Day.
2. Let x represent the day number and let y be the number of people who know about the soft opening on day x. Consider the day before Linda told anyone to be Day 0, so that Linda is the only person who knows about the opening on Day 0. Day 1 is the first day that Linda told someone else about the opening.
a. Complete the following table.
|Day |0 |1 |2 |3 |4 |5 |
|Number of people who know |1 |2 |4 |8 |16 |32 |
Comment(s):
The table of values is limited to fewer than that needed to answer the question in item 1 so that the graph in part b below will show the y-intercept and the shape typical of an exponential function.
b. Graph the points from the table in part a.
Comment(s):
Since this is the students’ first experience with an exponential graph, it is recommended that students draw this graph by hand on graph paper.
Solution(s):
The graph is shown at the right.
3. Write an equation that describes the relationship between x (day) and y (number of people who know) for the situation of spreading the news about the soft opening of Linda’s ice cream store.
Comment(s):
Students should easily see that the outputs of the function are powers of 2 and then note that the day number and the power of 2 are the same.
Solution(s):
[pic]
4. Does your equation describe the relationship between day and number who know about Linda’s ice cream store soft opening completely? Why or why not?
Comment(s):
The point of this question is that students realize that the domain is restricted in ways not implied by the equation. Since students have not yet studied a definition for non-integer exponents, they may believe that the equation makes sense only for integer exponents. However, they know about negative integer exponents and thus need to explicitly exclude these from the domain. They also need to exclude integers greater than 9 from the domain since Linda’s method of spreading the news of the soft opening stops on the day of the opening. If students state the correct inequalities but do not explicitly state that the exponents should be integers, teachers need to explain that this restriction must be included since other numbers can be exponents, although they will not study other exponents explicitly until a later course.
Solution(s):
No, the equation does not describe the relationship completely because the domain needs to be restricted to the integers 0, 1, 2, . . . , 9, and this information is not included in the equation.
Part 2: What if?
The spread of a rumor or the spread of a disease can be modeled by a type of function known as exponential function; in particular, an exponential growth function. An exponential function has the form
[pic],
where a is a non-zero real number and b is a positive real number other than 1. An exponential growth function has a value of b that is greater than 1.
1. In the case of Linda’s ice cream store, what values of a and b yield an exponential function to model the spread of the rumor of the soft store opening?
Comment(s):
For the rumor model, the coefficient a has a value of 1. However, other exponential functions in this task have coefficients other than 1, so students are introduced to the general definition of an exponential function from the beginning.
Solution(s):
a = 1 and b = 2
2. In this particular case, what is an appropriate domain for the exponential function? What range corresponds to this domain?
Comment(s):
Students are asked to specify the domain for this particular case. There is overlap with item 4 of part 1. The earlier question focused on whether all of the information is included in the equation. Here, the focus is explicitly to find the domain. Students may express the correct answer in a variety of ways. Two of these are shown in the solutions.
Solution(s):
The set of all nonnegative integers less than or equal to 9, or [pic]
3. In part 1, item 2, you drew a portion of the graph of this function. Does it make sense to connect the dots on the graph? Why or why not?
Comment(s):
The question of whether to connect the dots was prominent in students’ early formal study of models in middle school and functions in GPS Algebra. It reminds students to think of the meaning of points on the graph and to consider what values of the independent variable are meaningful in the situation. The description here is that the output is total number of people who know on a given day. Fractional parts of a day are not meaningful. Note: It would be impossible to draw an accurate model of this situation with a continuous time domain since we do not know when during the day each person who knows tells another person.
Solution(s):
No, it does not make sense to connect the dots. Connecting the dots would imply time passing continuously. We do not know when during the day people hear about the soft opening. We just have a count of the total number of people who know on each day.
4. How would the graph change if Linda had told two people each day rather than one and had asked that each person also tell two other people each day?
Comment(s):
This question asks students to think in terms of function values or points on the graph, but they will have to think through the situation in a similar manner to the original. If students answer more generally here without being specific about new function values, then they will have more work to do in item 5 to find the new equation.
Solution(s):
The point (0, 1) would stay the same since on Day 0 Linda would still be the only person who knows about the opening. But for the other days, more people would know so the points for the other days would be higher.
In particular, on the first day, 3 people (Linda and the two people she tells) would know giving the point (1, 3). On the second day, 9 people would know, because each of the 3 who know will tell 2 others giving a total of 3 + 2(3) =9. So the point for Day 2 is (2, 9). We can continue in this way for the other points.
5. How would the equation change if Linda had told two people each day rather than one and had asked that each person also tell two other people each day? What would be the values of a and b in this case?
Comment(s):
In addition to giving the formula, students must specify the values of a and b in order to reinforce the definition of exponential function.
Solution(s):
The equation would have a 3 as base for the exponent instead of a 2, that is, the equation would be [pic]. In this case, a = 1 and b = 3.
6. How long would it take for at least 500 people to find out about the opening if the rumor spread at this new rate?
Comment(s):
Students can use the graph or a table of values to determine the answer. If they draw a graph using a graphing utility, they should realize that the points of this function are only the integer valued points on the continuous graph shown. It is likely that most students will just count up to find the first power of 3 that is greater than 500.
Solution(s):
It would take 6 days for at least 500 people to find out.
|Day number, x |0 |1 |2 |3 |4 |5 |6 |
|No. people who know, [pic] |1 |3 |9 |27 |81 |243 |729 |
Part 3: The Beginning of a Business
How in the world did Linda ever save enough to buy the franchise to an ice cream store? Her mom used to say, “That Linda, why she could squeeze a quarter out of a nickel!” The truth is that Linda learned early in life that patience with money is a good thing. When she was just about 9 years old, she asked her dad if she could put her money in the bank. He took her to the bank and she opened her very first savings account.
Each year until Linda was 16, she deposited her birthday money into her savings account. Her grandparents (both sets) and her parents each gave her money for her birthday that was equal to twice her age; so on her ninth birthday, she deposited $54 ($18 from each couple).
Linda’s bank paid her 3% interest, compounded quarterly. The bank calculated her interest using the following standard formula:
[pic]
where A = final amount, P = principal amount, r = interest rate, n = number of times per year the interest is compounded, and t is the number of years the money is left in the account.
1. Verify the first entry in the following chart, and then complete the chart to calculate how much money Linda had on her 16th birthday. Do not round answers until the end of the computation, then give the final amount rounded to the nearest cent.
For her deposit at age 9, P = 54, r = 0.03, n = 4, t = 1.
[pic]
[pic] , as in the chart
|Age |Birthday $ |Amt from previous year plus |Total at year end |
| | |Birthday | |
|9 |54 |0 |55.63831630 |
|10 |60 |115.63832 |115.63832(1.0075)4 =119.14669 |
|11 |66 |185.14669 |185.14669(1.0075)4 = 190.76389 |
|12 |72 |262.76389 |262.76389(1.0075)4 = 270.73593 |
|13 |78 |348.73593 |348.73593(1.0075)4 = 359.31630 |
|14 |84 |443.31630 |443.31630(1.0075)4 = 456.76616 |
|15 |90 |546.76616 |546.76616(1.0075)4 = 563.35460 |
On the day before her 16th birthday, a year after her 15th, Linda had $563.35.
Comment(s):
Students will need exploration time to understand the compound interest formula. They may need to look up the term “principal amount” to understand that it refers to the amount deposited into the account. They will also need to realize that the interest rate r must be expressed in decimal form. If they enter numbers in their calculators following the formula exactly, they may need to be reminded about order of operations and that the calculator will not make the correct calculation unless the expression, nt, in the exponent is put in parentheses during calculation.
Some students may benefit from verifying the meaning of the compound interest formula by stepping through the compound interest calculation as four applications of simple interest using a rate of [pic]for each quarter for four quarters of one year as shown in the table below.
|Quarter number |Amount invested at beginning of |Amount of interest paid |Amount at end of quarter |
| |quarter | | |
|1 |54 |54(.0075) = 0.405 |54.405 |
|2 |54.405 |0.4080375 |54.8130375 |
|3 |54.8130375 |0.4110977813 |55.22413528 |
|4 |55.22413528 |0.4141810146 |55.63831630 |
Advanced students may benefit from seeing how the compound interest formula is developed using calculations similar to the above but using P for the amount of money, as shown below. One quarter is one-fourth of a year, so the number of quarters is always 4 times the number of years.
|No. of yrs|No. of |Amount invested at |Amount of interest paid |Amount at end of quarter |
| |qtrs |beginning of quarter| | |
|1/4 |1 |P |P(.0075) |P(1 + 0.0075) |
|1/2 |2 |P(1.0075) |[P(1.0075)](.0075) |P(1 + 0.0075)(1 + .0075) = P(1.0075)2 |
|3/4 |3 |P(1.0075)2 |[P(1.0075) 2](.0075) |[P(1.0075) 2](1 + .0075) = P(1.0075)3 |
|1 |4 |P(1.0075)3 |[P(1.0075) 3](.0075) |[P(1.0075) 3](1 + .0075) = P(1.0075)4 |
|5/4 |5 |P(1.0075)4 |[P(1.0075) 4](.0075) |[P(1.0075) 4](1 + .0075) = P(1.0075)5 |
|3/2 |6 |P(1.0075)5 |[P(1.0075) 5](.0075) |[P(1.0075) 5](1 + .0075) = P(1.0075)6 |
2. On her 16th birthday, the budding entrepreneur asked her parents if she could invest in the stock market. She studied the newspaper, talked to her economics teacher, researched a few companies and finally settled on the stock she wanted. She invested all of her money in the stock and promptly forgot about it. When she graduated from college on her 22nd birthday, she received a statement from her stocks and realized that her stock had appreciated an average of 10% per year. How much was her stock worth on her 22nd birthday?
Comment(s):
The challenge for students here is realizing that the information that Linda’s stock had appreciated an average of 10% per year means that the money grew as if it were invested at 10% compounded annually for the 6 years.
Solution(s):
Linda’s stock was worth $998.01 by application of the compound interest formula with P = 563.35, r = 0.10, n = 1, and t = 6:
[pic]
3. When Linda graduated from college, she received an academic award that carried a $500 cash award. On her 22nd birthday, she used the money to purchase additional stock. She started her first job immediately after graduation and decided to save $50 each month. On her 23rd birthday she used the $600 (total of her monthly amount) savings to purchase new stock. Each year thereafter she increased the total of her savings by $100 and, on her birthday each year, used her savings to purchase additional stock. Linda continued to learn about stocks and managed her accounts carefully. On her 35th birthday she looked back and saw that her stock had appreciated at 11% during the first year after college and that the rate of appreciation increased by 0.25% each year thereafter. At age 34, she cashed in enough stock to make a down payment on a bank loan to purchase her business. What was her stock worth on her 34th birthday? Use a table like the one below to organize your calculations.
|Age |Amt from previous year |Amt Linda added from|Amount invested for the year |Interest rate for the |Amt at year end|
| | |savings that year | |year | |
|22 |998.01 |500 |1498.01 |11.00% |1662.79 |
|23 |1662.79 |600 | |11.25% | |
|24 | |700 | |11.50% | |
|25 | |800 | |11.75% | |
|. |. |. |. |. |. |
|. |. |. |. |. |. |
|. |. |. |. |. |. |
Comment(s):
This item brings closure to this part of the learning task. The calculations are simple applications of the compound interest formula. If students have access to a spreadsheet program, having them set up the spreadsheet formulas is a possible extension of this activity.
Solution(s):
At age 34, Linda’s stock was worth $30,133.63.
The completed table is given below as an embedded Excel file.
[pic]
The same spreadsheet with formulas turned on is pasted in below.
[pic]
Part 4: Some Important Questions
In learning about Linda’s journey, we have seen several examples of exponential growth functions… the rumor, compounding interest in a savings account, appreciation of a stock. You have already identified the exponential functions related to spreading the news of the soft opening of Linda’s ice cream store. Now we’ll consider some other exponential functions you have encountered in this task and explore some new ones.
1. The formula you used to find the value of Linda’s stocks on her 22nd birthday (item 2 of Part 3) can be considered an application of an exponential function. Think of the values of P, r, and n as constant and let the number of years vary so that the number of years is the independent variable and the value of the stocks after t years is the dependent variable.
a. Write the equation for this exponential function.
Comment(s):
Continuous compounding of interest is the typical compounding method used by financial institutions in the 21st century. Students will study the function y = ex and various applications, including continuous compounding of interest, in GPS Advanced Algebra. In preparation for this study, students need to understand calculations of compound interest as applications of exponential functions.
The compound interest formula is a function of four variables P, r, n, and t. When we hold P, r, and n fixed and study how a particular investment grows over the years, then we have an exponential function, as defined in Part 2 of this task. In the compound interest formula, it is customary to restrict values of n to positive integers since these are the values that coincide with the actual practice of financial institutions. In this example, n = 1, so it is easy to see how the function fits the definition of exponential function.
Solution(s):
For Linda’s stock investment from ages 16 to 22, P = 563.35, r = 10% = 0.1, and n = 1.
The function is [pic].
b. What are the values of a and b so that it fits the definition of exponential function?
Comment(s):
This part enhances student understanding of the form of an exponential function: a constant multiple of a fixed base to a variable power.
Solution(s):
a = 563.35, b = 1.1
c. What point on the graph of this function did you find when you calculated the value of Linda’s stock at age 22?
Comment(s):
This question helps students to see the previous calculation as an application of an exponential function.
Solution(s):
(6, 998.01)
2. The formula you used to find the amount of money in Linda’s bank account when she was 10 years old can be considered an application of an exponential function where the number of years, t¸ is the independent variable and the amount of money in the account at the end of t is the dependent variable.
a. Write the equation for this exponential function.
Comment(s):
Another example to confirm students’ understanding of the definition of exponential function.
Solution(s):
For Linda’s ban account opened at age 9, P = 54, r = 3% = 0.03, and n = 4.
The function is [pic],
which simplifies to [pic].
b. What are the values of a and b so that it fits the definition of exponential function?
Comment(s):
This question is more difficult than it appears at first glance. The above formula has the expression 4t where the definition of exponential function has a simple “t”. Rather than regard this as a transformation of an exponential function, students can use properties of exponents to rewrite the formula with an appropriate base b raised to the exponent t. Note that (1.0075)4 is the number that students used again and again in their year by year calculations for the amount of money that Linda had in the bank.
The answer for b given below is correct to the accuracy of most of the calculators that students use. The exact value is (1.0075)4 = 1.0303391906640625
Solution(s):
a = 54
Applying properties of exponents, [pic],
thus, b = 1.030339191.
c. If Linda had not added money to the account each year, how much would she have had in the account at age 16 from her original investment at age 9?
Comment(s):
This question asks students to find a value of the function for t = 7 to emphasize that they have an exponential function where t is the independent, or input, variable.
Solution(s):
16 – 9 = 7, so Linda would have left the money in the account for 7 years.
[pic]
At age 16 she would have $66.57.
3. Consider the function[pic]with an unrestricted domain.
a. Use a graphing utility to graph the function. In a future course you will learn the meaning of the values of the function when x is not at integer.
Comment(s):
Students were asked about connecting the dots for a function with the same formula in Part 1 and indicated that the inputs needed to be integers. The only exponents they have studied are integer exponents. There calculators will show a graph with domain all real numbers. Students should realize that the calculator is giving meaning to non-integer exponents. Discussing these ideas is a topic for a later course. The comment here is designed to reassure students that the rules have not changed about connecting the dots.
Students may start with a standard viewing window. Since the function has no negative values, they should be encouraged to adjust the viewing window to see more of the positive y-axis. The graph shown uses a window of x-values from – 10 to 10 and y-values from – 2 to 18.
Solution(s):
The graph is shown at the right.
b. What is the range of the function when the domain is all real numbers?
Comment(s):
Students using graphing calculators will need to explore the graph, by looking at different windows and/or tracing along the graph, to see what happens to the graph for negative values of t. This question asks them to see that negative powers of 2 are always positive. The next part asks them to think about why.
Solution(s):
Range of f: the set of all real numbers greater than 0, or
[pic].
c. Why doesn’t the graph drop below the x-axis?
Comment(s):
Students know the definition of negative integer exponents( first addressed in 8th grade) and should be encouraged to reason that the definition for other negative exponents will follow the same pattern, that is, when r is a positive real number, [pic].
Solution(s):
The graph does not drop below the x-axis because all of the powers of 2, including the negative ones, are greater than 0. Consider the following negative powers of 2 as examples:
[pic]
4. Consider the function[pic].
a. Predict how the graph of g is related to the graph of f from item 3 above.
Comment(s):
Students should be able to see that the formula for the function g is obtained by adding 3 to the formula for f and apply their knowledge of vertical shifts to answer this question.
Solution(s):
The graph of g should be the same as the graph of the function f except shifted up 3 units.
b. Now use your graphing utility to graph the function g.
Comment(s):
Students may want to graph g on the same axes as the graph of f to verify their predictions although the question just asks them to graph the function g. Because of the shape of exponential functions, students are likely to recognize that g has been shifted up three units more readily just graphing g alone.
Solution(s):
The graph of g is shown at the left below. The graphs of f and g on the same axes are shown at the right below.
5. What is the range of the function g? How does this range compare to the range of the function f? Explain why the ranges are related in this way.
Comment(s):
This question helps students understand the fact that the values of the function g are all greater than 3 as an application of the vertical shift of the graph.
Solution(s):
Range of g: the set of all real numbers greater than 3, or
[pic].
The lower limit for the range of g is 3 more than the lower limit for the range of f because all the points of the graph of f are shifted up 3 units to obtain the graph of g.
6. The graph of an exponential function has a horizontal asymptote. Where is the asymptote located in the graph of f? Where is the asymptote located in the graph of g?
Comment(s):
Students learned about horizontal asymptotes in their study of the basic function [pic]. The previous questions should have prepared students to answer these questions after they review the meaning of the term “horizontal asymptote.”
Solution(s):
The horizontal asymptote occurs at y = 0 for the function f and at y = 3 for the function g.
7. Use your graphing utility to graph the following equations. Describe the graphs in parts b – e as transformations of the graph of the function in part a.
a. [pic]
b. [pic]
c. [pic]
d. [pic]
e. [pic]
Comment(s):
Students should be able to apply what they learned about transformations of other functions to this situation but working from the graphs will help them confirm that the concepts work the same way for exponential functions.
Solution(s):
a. b.
This graph in part b is a vertical stretch of the graph of the function in part a. The graph of [pic]is shown on the second coordinate system as a dotted line for reference.
c.
This graph in part c is a horizontal shift of the graph of the function in part a by 3 units to the right. The graph of [pic]is shown on the second coordinate system as a dotted line for reference.
d.
This graph in part d is a vertical shift of the graph of the function in part a by 2 units to u[ward. The graph of [pic]is shown on the second coordinate system as a dotted line for reference.
e.
This graph in part e is a reflection of the graph of the function in part a in the x-axis. The graph of [pic]is shown on the second coordinate system as a dotted line for reference. Be sure that students explore and fully understand that f(x) = - 4x is actually f(x) = -(4x) and not (-4)x. The second does NOT produce an exponential function because we do not use negative numbers as bases.
8. Make some generalizations. What impact did each of the changes you made to the equation have on the graph?
shifts how?
shifts how?
[pic]
shifts how?
Comment(s):
This question requires students to make general statements about transformations as they apply to exponential functions.
Solution(s):
The graph of an exponential function [pic]:
is a vertical stretch or shrink of the graph of [pic]by a factor of a when a > 0, and is a vertical stretch or shrink of the graph of [pic]by a factor of |a | combined with a reflection through the x-axis when a < 0. In this case, we think of the basic formula being multiplied by a constant.
The graph of a function of the form [pic] is obtained by shifting the graph of [pic]in the vertical direction by |k| units, up when k > 0 and down when k < 0. In this case, we think of the constant k added to the formula for the function.
The graph of a function of the form [pic]is obtained by shifting the graph of [pic]in the horizontal direction by |h| units, to the right when h > 0 and to the left when h < 0. In this case, we think of the variable x in the exponent replaced with an expression of the form x – h.
The above transformations can be combined. In the example above, the graph of [pic]is reflected through the x-axis since [pic], shifted to the right 1 unit since [pic] and shifted down 5 units since [pic].
Part 5: Geometric Sequences
Sequences of numbers that follow a pattern of multiplying a numbers from one term to the next are called geometric sequences. The following sequences are geometric:
A. 1, 3, 9, 27, 81, …
B. .01, .05, .25, 1.25, 6.25, …
C. 32, -16, 8, -4, 2, …
The sequence in A is generated by multiplying by 3, if you multiply the first term by 3 to get the second term. This works for any pair of consecutive terms. The third term times 3 is the fourth term: [pic], and so on.
For the sequence B, if we multiply 5 to the first number we will get the second number. This also works for any pair of consecutive numbers. The third number times 5 is the fourth number: [pic], with will work throughout the entire sequence.
The sequence in C seems a little different because it appears that we are dividing; in terms of geometric sequences we must think in terms of multiplication. In order to get the second term we must multiply the first term by -1/2. This too works for any pair of consecutive terms. The fourth term times -1/2 is the fifth term: [pic].
Because these sequences behave according to this simple rule of multiplying a constant number to one term to get to another, they are called geometric sequences. In order to examine these sequences to greater depth, we must know that the fixed numbers that generate each sequence are called the common ratios (r).
The common ratio can be calculated by dividing any two consecutive terms in a geometric sequence. The formula for calculating r is: [pic]
Take a look at the general formula for terms of a geometric sequence:
[pic]
[pic]
[pic]
[pic]
1. What is the relationship between the power of r and the number of the term, n in each case?
The power of 4 is equal to n – 1.
2. Generalize a formula for the nth term of the geometric sequence.
[pic]
3. For sequence A if we match each term with it’s corresponding term number we get:
|n |1 |2 |3 |5 |5 |… |
|Term |1 |3 |9 |27 |81 |… |
What is the general formula for sequence A?
[pic]
4. What would be the 15th term for sequence A? 4782969
5. What is the general formula for sequence B?
[pic]
6. What is the 12th term in sequence B? 488281.25
7. What is the formula for sequence C?
[pic]
8. What is the 13th term of sequence C? [pic]
9. Determine whether the following sequences are geometric. If it is find the general formula:
a. 3, 6, 9, 12, … no
b. 1, -3, 9, -27, … yes, [pic]
c. 2, 1, ½, ¼, … yes, [pic]
d. 4, 6, 9, 13.5, … yes, [pic]
e. 1, 2, 3, 4, … no
10. Let’s go back and take a look at the spreading of the “rumor” of Linda’s ice cream shop. Linda plans to tell one person each day and will ask that person to tell one other person each day through the day of the opening, and so on. Assume that each new person who hears about the soft opening is also asked to tell one other person each day through the day of the opening and that each one starts the process of telling their friends on the day after he or she first hears.
a. What is the formula for determining how many people know on each day?
[pic]
b. How many people know on days 1, 2, 3, and 4? 2, 4, 8, 16, respectively
c. Is this a geometric sequence? Explain your thinking. Yes, there is a common ratio between terms
11. If Linda had told two employees, so that three people knew on day one and each day everyone tells one person each.
a. What would be the new formula for determining how many people know on each day?
[pic]
b. How many people know on days 1, 2, 3, and 4? 3, 6, 12, and 24, respectively
12. Linda decides to spread the word via email so that more people will know. She and her two employees send out invitations to the soft opening of the store to five other people. The email asks that the receiver forward the message to 5 other people the following day. Assume this process continues each day without any repetition of recipients.
a. How many new recipients would there be for day 2, day 3, day 4, and day 5?
15, 75, 375, and 5625, respectively
b. What is the common ratio to the sequence of new recipients? 5
c. What is the general formula that can be used to determine the number of new recipients on the nth day?
[pic]
d. Use the formula to determine the number of new recipients on the 7th day.
46,875
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[1] Students may find it more interesting to look up current exchange values to use for this item and Item 9, which depends on it. There are many websites that provide rates of exchange for currency. Note that these rates change many times throughout the day, so it is impossible to do calculations with truly “current” exchange values. The values in Item 3 were found using .
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