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Answers
Teacher Copy
Plan
Pacing: 1 class period
Chunking the Lesson
Example A #1 Example B
Example C #2
p. 79
Check Your Understanding
Lesson Practice
Teach
Bell-Ringer Activity
Students should recall that an absolute value of a number is its distance from zero on a number line.
Have students evaluate the following:
1. |6| [6]
2. |–6| [6]
Then have students solve the following equation.
3. |x|= 6 [x = 6 or x = –6]
Example A Marking the Text, Interactive Word Wall
Point out the Math Tip to reinforce why two solutions exist. Work through the solutions to the equation algebraically. Remind students that
solutions to an equation make the equation a true statement. This mathematical understanding is necessary for students to be able to check
their results.
© 2014 College Board. All rights reserved.
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Developing Math Language
An absolute value equation is an equation involving an absolute value of an expression containing a variable. Just like when solving
algebraic equations without absolute value bars, the goal is to isolate the variable. In this case, isolate the absolute value bars because they
contain the variable. It should be emphasized that when solving absolute value equations, students must think of two cases, as there are two
numbers that have a specific distance from zero on a number line.
1 Identify a Subtask, Quickwrite
When solving absolute value equations, students may not see the purpose in creating two equations. Reviewing the definition of the absolute
value function as a piecewise-defined function with two rules may enable students to see the reason why two equations are necessary.
Have students look back at Try These A, parts c and d. Have volunteers construct a graph of the two piecewise-defined functions used to
write each equation and then discuss how the solution set is represented by the graph.
Example B Marking the Text, Simplify the Problem, Critique Reasoning, Group Presentation
Start with emphasizing the word vary in the Example, discussing what it means when something varies. You may wish to present a simpler
example such as: The average cost of a pound of coffee is $8. However, the cost sometimes varies by $1. This means that the coffee could
cost as little as $7 per pound or as much as $9 per pound. Now have students work in small groups to examine and solve Example B by
implementing an absolute value equation. Additionally, ask them to take the problem a step further and graph its solution on a number line.
Have groups present their findings to the class.
ELL Support
For those students for whom English is a second language, explain that the word varies in mathematics means changes. There are different
ways of thinking about how values can vary. Values can vary upward or downward, less than or greater than, in a positive direction or a
negative direction, and so on. However, the importance comes in realizing that there are two different directions, regardless of how you think
of it.
Also address the word extremes as it pertains to mathematics. An extreme value is a maximum value if it is the largest possible amount
(greatest value), and an extreme value is a minimum value if it is the smallest possible amount (least value).
Developing Math Language
An absolute value inequality is basically the same as an absolute value equation, except that the equal sign is now an inequality symbol: <,
>, ≤, ≥, or ≠. It still involves an absolute value expression that contains a variable, just like before. Use graphs on a number line of the
solutions of simple equations and inequalities and absolute value equations and inequalities to show how these are all related.
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Example C Simplify the Problem, Debriefing
Before addressing Example C, discuss the following: Inequalities with |A| > b, where b is a positive number, are known as disjunctions
and are written as A < –b or A > b.
For example, |x| > 5 means the value of the variable x is more than 5 units away from the origin (zero) on a number line. The solution is x
< – 5 or x > 5.
See graph A.
This also holds true for |A| ≥ b.
Inequalities with |A| < b, where b is a positive number, are known as conjunctions and are written as –b < A < b, or as –b < A and A < b.
For example: |x| < 5; this means the value of the variable x is less than 5 units away from the origin (zero) on a number line. The solution
is –5 < x < 5.
See graph B.
This also holds true for |A| ≤ b.
Students can apply these generalizations to Example C. Point out that they should proceed to solve these just as they would an algebraic
equation, except in two parts, as shown above. After they have some time to work through parts a and b, discuss the solutions with the
whole class.
Teacher to Teacher
Another method for solving inequalities relies on the geometric definition of absolute value |x – a| as the distance from x to a. Here’s how
you can solve the inequality in the example:
|2x + 3 | + 1 > 6
|2x + 3| > 5
|
2 | x − −3
2
>
5
|x − −3
| >
2
5
2
Thus, the solution set is all values of x whose distance from − 32 is greater than 52 . The solution can be represented on a number line and
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written as x < –4 or x > 1.
2 Quickwrite, Self Revision/Peer Revision, Debriefing
Use the investigation regarding the restriction c > 0 as an opportunity to discuss the need to identify impossible situations involving
inequalities.
Check Your Understanding
Debrief students’ answers to these items to ensure that they understand concepts related to absolute value equations. Have groups of
students present their solutions to Item 4.
Assess
Students’ answers to Lesson Practice problems will provide you with a formative assessment of their understanding of the lesson
concepts and their ability to apply their learning.
See the Activity Practice for additional problems for this lesson. You may assign the problems here or use them as a culmination for the
activity.
Adapt
Check students’ answers to the Lesson Practice to ensure that they understand basic concepts related to writing and solving absolute
value equations and inequalities and graphing the solutions of absolute value equations and inequalities. If students are still having
difficulty, review the process of rewriting an absolute value equation or inequality as two equations or inequalities.
Activity Standards Focus
In Activity 5, students perform operations on functions. Students then write composite functions. Throughout this activity, emphasize
that when evaluating functions combined with operations, the value of an input evaluated first in the separate functions and then
operated is equal to the value of the combined function with that input. Combining functions ahead of time is efficient when evaluating
many input values.
Plan
Pacing: 1 class period
Chunking the Lesson
#1–3 #4–7 #8–11
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#12 #13–14 #15
#16–17 #18–20
#21 #22
Check Your Understanding
Lesson Practice
Teach
Bell-Ringer Activity
Have students review some of the concepts they will need to apply in this lesson. Ask students to complete the following exercises.
1. Simplify 5a2 + 2a − 4a − 6a2. [−a2 − 2a]
2. Evaluate (m + 2)2 for m = −5. [9]
Ask students to share their responses, and answer any questions they may have prior to moving forward with the lesson.
1–3 Activating Prior Knowledge, Chunking the Activity, Paraphrasing
Ask students questions like: What would the graph of these functions look like? [t(h) would be linear, with a y-intercept at the origin
and a slope of 10
, and s(h) would be linear, with a y-intercept at the origin and a slope of 81 .] What does the y-intercept represent?
1
[zero pay for zero hours worked] What does the slope represent? [the rate of pay per hour] Would you be interested in looking at the
entire coordinate plane? [No; Quadrant I only, because Tori and Stephan will not receive pay for negative hours]
4–7 Discussion Groups, Group Presentation
Ensure students understand this application by placing them in small groups and having each group create a scenario with adding two
real-world functions. Encourage them to use the hourly earnings functions as a template but also to feel free to use a variable other
than hours.
Common Core State Standards for Activity 5
HSF-BF.A.1 Write a function that describes a relationship between two quantities.
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HSF-BF.A.1b Combine standard function types using arithmetic operations.
Technology Tip
You can use the table feature of a graphing calculator to find specific function values. You could do this for the function addition
represented in the previous items by following these steps:
1. Press the [y=] key.
2. Beside the function, type in 10x + 8x.
3. Press [2nd][WINDOW] to look at the table setup. Set TblStart to =0; Set the change in the table, or ΔTbl, to 1; Set the
Indpnt: to Ask.
4. Access the table by pressing [2nd][GRAPH].
5. Notice the table is blank. The calculator is waiting for you to enter the x-value (in this case, the number of hours) for which
you would like to know the corresponding y-value, or cost.
6. At x=, key in [4] [ENTER]. This should give the corresponding y-value of $72.
7. At x=, key in [6] [ENTER]. This should give the corresponding y-value of $108.
8. Now you can continue this by trying other numbers of hours that were not already in the examples.
For additional technology resources, visit SpringBoard Digital.
8–11 Activating Prior Knowledge, Chunking the Activity, Predict and Confirm
Lead a discussion about these items by asking the following:
How do these functions differ from those presented in Items 1–7? [These functions have a y-intercept other than (0, 0).]
What makes these functions have these y-intercepts? [the fixed fees charged by each company]
Why is subtraction being used rather than addition? [It is basically an example of comparison shopping, where one wants to
know how much will be saved by using one company instead of the other, in terms of a given number of trees.]
What is one thing you have to be cautious about when subtracting expressions? [Subtract each term of the expression, not
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just the first term; in other words, the subtraction sign is distributed throughout the subtrahend expression.]
12 Activating Prior Knowledge, Debriefing
Once students know how to add and subtract linear functions, they apply this knowledge to adding and/or subtracting a linear
function to a quadratic function. The same function rules apply, and students will use the same structure to combine like terms.
Differentiating Instruction
For those students who need additional explanation of the functions used in Item 12, explain the following:
A linear function is an algebraic equation in which the greatest degree of a variable term is 1. In other words, the greatest
exponent of a variable term is 1.
The standard form of a linear function is Ax + By = C, where A, B, and C are constants.
The y-intercept form of a linear equation is y = mx + b, where m is the slope and b is the y-intercept.
The point-slope form of a linear equation is y − y1 = m(x − x1), where m is the slope and (x1, y1) are the coordinates of a
point through which the line passes.
A quadratic function is an algebraic equation in which one or more of the variable terms is squared, giving the function a degree
of 2. However, a squared power is the greatest degree a quadratic function can have.
The general form is ax2 + bx + c = 0, where a, b, and c are constants.
13–14 Close Reading, Marking the Text, Differentiating Instruction, Simplify the Problem
To help students understand the function in Item 13, have them construct a table of values for h and n(h). Ask: If it takes Jim
one hour to install one shrub, how many shrubs can Jim install in an 8-hour day? What if it takes Jim 2 hours to install one
shrub? 3 hours? 4 hours? Elicit from students the operation of division between 8, the total number of hours in the workday, and
h, the number of hours it takes Jim to install one shrub. In Item 13b, highlight restrictions and domain. Support students whose
first language is not English by further explaining the word restriction.
For Item 14, tell students who are struggling to refer back to either function from Item 8 because they are the same type.
15 Activating Prior Knowledge
Explain to the students that multiplying functions will require them to multiply polynomials, which they learned in Algebra 1. In
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Item 15a, a monomial is being multiplied by a binomial. The types of polynomials being multiplied will obviously vary with
the functions. Be sure to use the Distributive Property. In Item 15b, students again encounter the topic of domain restrictions.
To ensure that students understand the correct response to 15b, engage them by asking them why the h-value must be
positive.
16–17 Create Representations, Debriefing
In Item 16, ensure students are not confused by the solution to 16a being 16 shrubs and the predetermined cost per shrub
(listed in Item 14) being the same. The answer of 16 in Item 16b represents that predetermined cost per shrub. The fact that
the number of shrubs and the cost per shrub are the same is a mere coincidence.
Present a table with four column titles—h, n(h), c(h), and (n • c)(h)—and place values for h = 0.5 in each column. Then have
students make a conjecture as to whether they think it will cost more or less if Jim estimates that it will take him 40 minutes
to install each shrub. After discussing, have students try h = 23 in the functions. Fill in values for h = 23 in a new row of the
table. [Answer should be approximately $712.]
Differentiating Instruction
Ask students to discuss whether their conjectures were correct or incorrect when they altered the value of h in Items 16 and
17 from h = 0.5 to h = 23 (a longer amount of time per shrub). Why is the total cost of Jim′s services for an 8-hour day less?
[because he is getting less work done per hour]
18–20 Predict and Confirm, Activating Prior Knowledge
Ask students to make a conjecture as to the number of hours (if any) that it would take for the total charge of applying
compost to equal the total charge of applying fertilizer. [Students will hopefully realize the impossibility of this because both
the hourly charge and material cost are greater for the compost service.]
21 Activating Prior Knowledge, Debriefing
Note that the Math Tip refers to factoring expressions in the numerator and denominator in Items 21c, 21d, and 21e. Since
factoring has not been covered in Algebra 2 at this point, you may wish to review with students the following:
In Item 21c,
Discuss that the two terms in the denominator have a common factor of 2. The coefficient 2 of 2x in the numerator cancels
with the common factor of 2 in the denominator. Furthermore, some students are going to want to cancel out the x′s. Be
prepared to explain that this is not possible because the x in the denominator is part of the term (x + 3), and the only way to
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cancel would be if there were a term of (x + 3) in the numerator.
In Item 21d,
In the numerator, there is a common factor of 2 that can be factored out. After doing so, the (x + 3)′s in the numerator and
denominator can be entirely canceled.
In Item 21e, there is no common factor.
Differentiating Instruction
For struggling students, it may be helpful to take some extra time to review factoring out a common factor before moving
forward. Here are some suggestions of samples you might use.
1. x2 + 5x = x(x + 5)
2. 10y + 15 = 5(2y + 3)
3.
3p + 12
6p + 24
=
1 3(p + 4)
2 6 (p + 4)
= 12 ,
p≠ −4
Mini-Lesson: Function Operations
If students need additional help with adding, subtracting, multiplying, or dividing functions, a mini-lesson is available to
provide practice.
See the Teacher Resources at SpringBoard Digital for a student page for this mini-lesson.
22 Debriefing
This item guides students toward the conclusion that operations with functions follow similar processes and rules as
operations with numbers. As with numbers, addition and multiplication of functions follow the commutative properties,
whereas subtraction and division do not.
With both real numbers and function division, the divisor cannot equal zero.
The main differences when performing function operations are the use of function notation and variables. Rather than
simple addition, function operations involve combining like terms. Lastly, function division may require knowledge of
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factoring polynomials.
Check Your Understanding
Debrief students’ answers to these items to ensure that they understand concepts related to function operations.
Assess
Students’ answers to Lesson Practice problems will provide you with a formative assessment of their understanding of
the lesson concepts and their ability to apply their learning.
See the Activity Practice for additional problems for this lesson. You may assign the problems here or use them as a
culmination for the activity.
Adapt
Check students’ answers to the Lesson Practice to ensure that they understand function operations. Sometimes students
understand the concepts but are confused by the notation. If this happens, encourage them to begin by rewriting an
expression such as (f + g)x as f(x) + g(x) and then substitute expressions for f(x) and g(x).
Plan
Pacing: 1 class period
Chunking the Lesson
#1–2 #3–5 #6–8
#9 #10 #11–14
Check Your Understanding
Lesson Practice
Teach
Bell-Ringer Activity
Write the equations y = 6x and y = 6x + 11 on the board. Have students discuss what kind of equations these represent
and their similarities and differences. Use the discussion as an opportunity to review independent and dependent
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variables in equations, and how they relate to domain and range of functions.
1–2 Create Representations, Debriefing
Item 1 serves as a concrete, numerical check for student understanding of the scenario. Item 2 asks students to write
an algebraic rule relating cost and time. Use Item 2 as a quick assessment of understanding of function notation.
3–5 Quickwrite, Identify a Subtask, Group Presentation
Obtaining a cost estimate provides a numerical example for a two-step process that involves composing one function
with another. Be sure that all students understand the numerical process before they encounter the symbolic function
notation later on. The term composition will be introduced later in the activity.
6 Think-Pair-Share
Use this item to assess students’ understanding of slope. It will also help them to focus on the quantities represented
by each variable.
Teacher to Teacher
In the linear equation y = mx + b, m is the constant rate of change of y with respect to x. When that equation is
graphed, m indicates the slope of the line. Therefore, slope is the graphical representation of rate of change. The rate
of change has two representations—a numerical one with units of measure (e.g., $20/h) and a graphical one (e.g.,
slope of 20).
7–8 Think-Pair-Share, Graphic Organizer, Debriefing
Domain and range of the functions are represented in a table and in a graphic organizer. These items set the stage for
understanding how the domain and range of a composite function are related to the domain and range of the functions
from which it is created.
Mini-Lesson: Linear Functions and Slope
If students need additional help with identifying the slope and y-intercept of a linear function, as well as interpreting
slope and rate of change, a mini-lesson is available to provide practice.
See the Teacher Resources at SpringBoard Digital for a student page for this mini-lesson.
Developing Math Language
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The word composition means the act of combining parts to form a whole. Pertaining to math functions, composition
is the act of forming a new function by involving two or more functions in succession. It involves taking the output
of the first function and using it as the input for the second function, and so on. The result of following these steps
creates an entirely new function, called a composite function. It is important to note that when working with
composite functions, students start from the inside and work their way outward, similar to order of operations with
arithmetic.
9 Think-Pair-Share, Graphic Organizer
Students will recognize that composition of the two functions creates a new function, the cost-to-mow function, and
will identify the domain and range of the composite function.
10 Marking the Text, Graphic Organizer
This item allows for a different representation to help students identify the domain and range of the composite
function. The representations in Items 9 and 10 parallel the representations in Items 7 and 8.
11–14 Quickwrite, Group Presentation, Debriefing
Work with students to analyze what happens in this situation. Students should see that the time function has been
substituted into the cost function. The resulting composite is a function of a acres. Students will practice function
composition later in this activity.
Use Item 11 to determine whether or not students understand the process of composition.
Item 12 brings students back to the context. During debriefing, make sure that the point is made that composition is
an efficient process if you need to repeatedly use the two-step process represented by composition.
Use student communication to reinforce understanding of the meaning of composition. Be sure that students see the
different meanings of the value 50 in Items 13 and 14.
Mini-Lesson: Composition Function Notation
If students need additional help with understanding the meaning of composite function notation, a mini-lesson is
available to provide practice.
See the Teacher Resources at SpringBoard Digital for a student page for this mini-lesson.
Check Your Understanding
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Debrief students’ answers to these items to ensure that they understand concepts related to composition of
functions. Be sure students understand what the domain and range of the composition are in the context of the
situation.
Assess
Students’ answers to Lesson Practice problems will provide you with a formative assessment of their
understanding of the lesson concepts and their ability to apply their learning.
See the Activity Practice for additional problems for this lesson. You may assign the problems here or use them
as a culmination for the activity.
Adapt
Check students’ answers to the Lesson Practice to ensure that they understand basic concepts related to function
composition. If students are confused by the notation, encourage them to write each step. For example, in Item
22, they can begin by writing c(h(r)) = c(0.5r). If students have difficulty substituting algebraic expressions into
functions, encourage the use of colored pencils or highlighters. For example, students can write c(h) = 20 + 12h,
highlighting both h′s, and then write c(0.5r) and highlight 0.5r. This can help students see what should be
substituted into the function (0.5r) and where it should be substituted.
Learning Targets
p. 79
Write functions that describe the relationship between two quantities.
Explore the composition of two functions through a real-world scenario.
Create Representations (Learning Strategy)
Definition
Creating pictures, tables, graphs, lists, equations, models, and /or verbal expressions to interpret text or
data
Purpose
Helps organize information using multiple ways to present data and to answer a question or show a
problem solution
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Identify a Subtask (Learning Strategy)
Definition
Breaking a problem into smaller pieces whose outcomes lead to a solution
Purpose
Helps to organize the pieces of a complex problem and reach a complete solution
Group Presentation (Learning Strategy)
Definition
Presenting information as a collaborative group
Purpose
Allows opportunities to present collaborative solutions and to share responsibility for delivering
information to an audience
Graphic Organizer (Learning Strategy)
Definition
Arranging information into maps and charts
Purpose
Builds comprehension and facilitates discussion by representing information in visual form
Debriefing (Learning Strategy)
Definition
Discussing the understanding of a concept to lead to consensus on its meaning
Purpose
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Helps clarify misconceptions and deepen understanding of content
Suggested Learning Strategies
Create Representations, Identify a Subtask, Group Presentation, Graphic Organizer, Debriefing, Self
Revision/Peer Revision
Recall that Jim has a lawn service called Green’s Grass Guaranteed. On every mowing job, p. 82p. 81p. 80
Jim charges a fixed $30 fee to cover equipment and travel expenses plus a $20 per hour labor charge.
Work with your group on Items 1–14.
Discussion Group Tips
With your group, reread the problem scenarios in this lesson as needed. Make notes on the
information provided in the problems. Respond to questions about the meaning of key
information. Summarize the information needed to create reasonable solutions, and describe the
mathematical concepts your group uses to create its solutions.
1. On a recent mowing job, Jim worked for 6 hours. What was the total charge for this job?
$150; 30 + (20 × 6) = 150
2. Model with mathematics. If Jim works for t hours, what will he charge for a mowing job?
Write your answer as a cost function where c(t) is Jim’s charge for t hours of work.
c(t) = 30 + 20t
It takes Jim 4 hours to mow 1 acre. Jim prepares a cost estimate for each customer based on the size
(number of acres) of the property.
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3. The APCON company is one of Jim’s customers. APCON has 2 acres that need mowing.
How many hours does that job take?
8 hours
4. Another customer has a acres of property. Write the equation of a function in terms of a for
the number of hours t it will take Jim to mow the property.
t(a) = 4a
5. How much will Jim charge APCON to mow its property? Justify your answer.
$190. Sample explanation: Use the number of hours (8) to substitute into the cost function: c(t) =
30 + 20t = 30 + 20(8) = 190 dollars.
The functions in Items 2 and 4 relate three quantities that vary, based on the needs of Jim’s
customers:
The size in acres a of the property
The time in hours t needed to perform the work
The cost in dollars c of doing the work.
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Math Tip
In a linear function f(x) = mx + b, the y-intercept is b. The variable m is the rate of change in
the values of the function—the change of units of f(x) per change of unit of x. When the
function is graphed, the rate of change is interpreted as the slope. So y = mx + b is called the
slope-intercept form of a linear equation.
6. Attend to precision. Complete the table below by writing the rate of change with units
and finding the slope of the graph of the function.
Rate of Change (with
units)
Function
Slope
c(t) =
30 + 20t
$20 per hour
20
t(a) =
4a
4 hours per acre
4
7. Complete the table below by naming the measurement units for the domain and range of
each function.
Function Notation
Description of
Function
Domain (units)
Range (units)
c(t)
cost for job
hours
dollars
t(a)
time to mow
acres
hours
8. Calculating the cost to mow a lawn is a two-step process. Complete the graphic organizer
below by describing the input and output, including units, for each part of the process.
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Math Terms
A composition is an operation on two functions that forms a new function. To form the
new function, the rule for the first function is used as the input for the second function.
A composite function is the function that results from the composition of two functions.
The range of the first function becomes the domain for the second function.
The graphic organizer shows an operation on two functions, called a composition. The
function that results from using the output of the first function as the input for the second
function is a composite function.
In this context, the composite function is formed by the time-to-mow function and the
cost-for-job function. Its domain is the input for the time function, and its range is the output
from the cost function.
9. Make sense of problems. The cost to mow is a composite function.
Describe its input and output as you did in Item 8.
When a composite function is formed, the function is often named to show the
functions used to create it. The cost-to-mow function, c(t(a)), is composed of the
cost-for-job and the time-to-mow functions.
The c(t(a)) notation implies that a was assigned a value t(a) by the time-to-mow
function. Then the resulting t(a) value was assigned a value c(t(a)) by the cost-to-mow
function.
10. Complete the table by writing a description for the composite function c(t(a)). Then
name the measurement units of the domain and range.
Function Notation
Description of Function
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c(t(a))
cost to mow
Domain (units)
Range (units)
acres
dollars
Jim wants to write one cost function for mowing a acres of property. To write the cost c as a
function of a acres of property, he substitutes t(a) into the cost function and simplifies.
c(t)
=
c(t(a))
Substitute t(a) for t in the cost function.
c(t(a))
=
c(4a)
t(a)=4a, so write the function in terms of a.
=
30 + 20(4a)
Substitute 4a for t in the original c(t) function.
=
30 + 80a
c(t(a))
11. Attend to precision. Write a sentence to explain what the expression c(t(2))
represents. Include appropriate units in your explanation.
It represents the cost in dollars to mow 2 acres.
12. Construct viable arguments. Why might Jim want a single function to determine
the cost of a job when he knows the total number of acres?
Sample response: If Jim needs to determine the cost for different-sized properties, this
function will be useful.
13. Explain what the expression c(t(50)) represents. Include appropriate units in your
explanation.
It represents the cost in dollars to mow 50 acres.
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14. Explain what information the equation c(t(a)) = 50 represents. Include appropriate
units in your explanation.
It represents the number of acres that can be mowed for $50.
Check Your Understanding
p. 83
15. Given the functions a(b) = b + 8 and b(c) = 5c, write the equation for the composite
function a(b(c)) and evaluate it for c = −2.
a(b(c)) = 5c + 8; a(b(−2)) = −2
16. The first function used to form a composite function has a domain of all real
numbers and a range of all real numbers greater than 0. What is the domain of the
second function in the composite function? Explain.
All real numbers greater than 0; The domain of the second function in the composite
function is the range of the first function.
17. The notation f(g(x)) represents a composite function. Explain what this notation
indicates about the composite function.
Sample answer: The composite function is composed of the functions f and g. The
variable x represents the input. The composite function first evaluates the function g for
the value of x to give a value for g(x). Then it evaluates the function f for the value of
g(x) to give a value for f(g(x)).
Lesson 5-2 Practice
Model with mathematics. Hannah’s Housekeeping charges a $20 flat fee plus $12 an
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hour to clean a house.
18. Write a function c(h) for the cost to clean a house for h hours.
c(h) = 20 + 12h
19. What are the units of the domain and range of this function?
domain: hours; range: dollars
20. What is the slope of this function? Interpret the slope as a rate of change.
The slope is 12. It means the cost will increase by $12 for each additional hour of
cleaning time.
Hannah’s Housekeeping can clean one room every half hour.
21. Write a function h(r) for the hours needed to clean r rooms.
h(r) = 0.5r
22. Write a function c(h(r)) to represent the cost of cleaning r rooms.
c(h(r)) = 20 + 12(0.5r) = 20 + 6r
23. What is the value and meaning of c(h(12))?
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$92 is the cost to clean 12 rooms.
24. Look for and make use of structure. Explain how a composition of
functions forms a new function from the old (original) functions.
Possible answer: The output of one function becomes the input of another function.
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