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Answers
Teacher Copy
Plan
Pacing: 1 class period
Chunking the Lesson
Example A #1 Example B
Example C #2
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).
Learning Targets
p. 73
Combine functions using arithmetic operations.
Build functions that model real-world scenarios.
Activating Prior Knowledge (Learning Strategy)
Definition
Recalling what is known about a concept and using that information to make a connection to a new concept
Purpose
Helps students establish connections between what they already know and how that knowledge is related to new
learning
Discussion Groups (Learning Strategy)
Definition
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Working within groups to discuss content, to create problem solutions, and to explain and justify a solution
Purpose
Aids understanding through the sharing of ideas, interpretation of concepts, and analysis of problem scenarios
Debriefing (Learning Strategy)
Definition
Discussing the understanding of a concept to lead to consensus on its meaning
Purpose
Helps clarify misconceptions and deepen understanding of content
Close Reading (Learning Strategy)
Definition
Reading text word for word, sentence by sentence, and line by line to make a detailed analysis of meaning
Purpose
Assists in developing a comprehensive understanding of the text
Think-Pair-Share (Learning Strategy)
Definition
Thinking through a problem alone, pairing with a partner to share ideas, and concluding by sharing results with
the class
Purpose
Enables the development of initial ideas that are then tested with a partner in preparation for revising ideas and
sharing them with a larger group
Summarizing (Learning Strategy)
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Definition
Giving a brief statement of the main points in a text
Purpose
Assists with comprehension and provides practice with identifying and restating key information
Paraphrasing (Learning Strategy)
Definition
Restating in your own words the essential information in a text or problem description
Purpose
Assists with comprehension, recall of information, and problem solving
Quickwrite (Learning Strategy)
Definition
Writing for a short, specific amount of time about a designated topic
Purpose
Helps generate ideas in a short time
Suggested Learning Strategies
Activating Prior Knowledge, Discussion Groups, Debriefing, Close Reading, Think-Pair-Share,
Summarizing, Paraphrasing, Quickwrite
Jim Green has a lawn service called Green’s Grass Guaranteed. Tori and Stephan are two p. 78p. 77p. 76p. 75p. 74
of his employees. Tori earns $10 per hour, and Stephan earns $8 per hour. Jim sends Tori and Stephan on a
job that takes them 4 hours.
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1. Model with mathematics. Write a function t(h) to represent Tori’s earnings in dollars for working
h hours and a function s(h) to represent Stephan’s earnings in dollars for working h hours.
t(h) = 10h; s(h) = 8h
Math Tip
Addition, subtraction, multiplication, and division are operations on real numbers. You can also
perform these operations with functions.
2. Find t(4) and s(4) and tell what these values represent in this situation.
t(4) = 40; s(4) = 32; Tori’s earnings for the 4-hour job are $40, and Stephan’s earnings for the 4-hour job
are $32.
3. Find t(4) + s(4) and tell what it represents in this situation.
t(4) + s(4) = 40 + 32 = 72; The sum of Tori and Stephan’s earnings for the 4-hour job is $72.
You can add two functions by adding their function rules.
Writing Math
The notation (f + g)(x) represents the sum of the functions f(x) and g(x). In other words, (f + g)(x) =
f(x) + g(x).
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4.
a. Add the functions t(h) and s(h) to find (t + s)(h). Then simplify the function rule.
(t + s)(h) = 10h + 8h = 18h
b. What does the function (t + s)(h) represent in this situation?
the amount in dollars Jim must spend on Tori and Stephan’s earnings for a job that takes h hours
5. Find (t + s)(4). How does the answer compare to t(4) + s(4)?
(t + s)(4) = 18(4) = 72; (t + s)(4) has the same value as t(4) + s(4).
6. How much will Jim spend on Tori and Stephan’s earnings for the 4-hour job?
$72
7. How much would Jim spend on Tori and Stephan’s earnings for a job that takes 6 hours? Explain
how you determined your answer.
$108; I evaluated (t + s)(h) for h = 6: (t + s)(6) = 18(6) = 108.
For a basic tree-trimming job, Jim charges customers a fixed $25 fee plus $150 per tree. One of Jim’s
competitors, Vista Lawn & Garden, charges customers a fixed fee of $75 plus $175 per tree for the same
service.
8. Write a function j(t) to represent the total charge in dollars for trimming t trees by Jim’s company
and a function v(t) to represent the total charge in dollars for trimming t trees by Vista.
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j(t) = 25 + 150t; v(t) = 75 + 175t
Math Tip
When subtracting an algebraic expression, remember to subtract each term of the expression. For
example, subtract 6x − 2 from 10x as follows.
10x − (6x + 2)
=
10x − 6x − 2
=
4x − 2
9.
a. Subtract j(t) from v(t) to find (v − j)(t). Then simplify the function rule.
(v − j)(t) =
75 + 175t − (25 + 150t)
=
75 + 175t − 25 − 150t
=
50 + 25t
b. What does the function (v − j)(t) represent in this situation?
the amount in dollars a customer will save by choosing Jim’s company to trim t trees rather
than Vista
Writing Math
The notation (f − g)(x) represents the difference of the functions f(x) and g(x). In other words, (f −
g)(x) = f(x) − g(x).
10. Find (v − j)(5). What does this value represent in this situation?
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(v − j)(5) = 50 + 25(5) = 175; A customer will save $175 by choosing Jim’s company rather than
Vista to trim 5 trees.
11. How much will a customer save by choosing Jim’s company to trim 8 trees rather than
choosing Vista? Explain how you determined your answer.
$250; I evaluated (v − j)(t) for t = 8: (v − j)(8) = 50 + 25(8) = 250.
12. Look for and make use of structure. Given f(x) = 3x + 2, g(x) = 2x − 1, and h(x) = x2 − 2x
+ 8, find each function and simplify the function rule.
a. (f + g)(x)
= 5x + 1
b. (g + h)(x)
2
=x +7
c. (h + f)(x)
2
= x + x + 10
d. (f − g)(x)
=x+3
e. (g − f)(x)
= −x − 3
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f. (h − g)(x)
2
= x − 4x + 9
= 5x + 1
= x2 + 7
= x2 + x + 10
=x+3
= −x − 3
= x2 − 4x + 9
Jim has been asked to make a bid for installing the shrubs around a new office building. In the
bid, he needs to include the number of shrubs he can install in an 8-hour day, the cost per shrub
including installation, and the total cost of his services for an 8-hour day.
Math Tip
When considering restrictions on the domain of a real-world function, consider both values
of the domain for which the function would be undefined and values of the domain that
would not make sense in the situation.
13.
a. Write a function n(h) to represent the number of shrubs Jim can install in an 8-hour
day when it takes him h hours to install one shrub.
n(h) =
8
h
b. What are the restrictions on the domain of n(h)? Explain.
The value of h cannot be 0, or the function would be undefined. Also, because h
represents a number of hours, its value cannot be negative.
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14. Jim will charge $16 for each shrub. He will also charge $65 per hour for installation
services. Write a function c(h) to represent the amount Jim will charge for a shrub that
takes h hours to install.
c(h) = 16 + 65h
The total cost of Jim’s services for an 8-hour day is equal to the number of shrubs he can
install times the charge for each shrub.
Writing Math
The notation (f · g)(x) represents the product of the functions f(x) and g(x). In other words,
(f · g)(x) = f(x) · g(x).
15.
a. Find the total cost of Jim’s services using the functions n(h) and c(h) to find (n ·
c)(h). Then simplify the function rule.
(n · c)(h) = 8h (16 + 65h) =
128
h
+ 520
b. Attend to precision. What are the restrictions on the domain of (n · c)(h)?
The value of h must be positive.
Connect to Business
When a company makes a bid on a job, the company states the price at which it is willing
to do the job. The company must make its bid high enough to cover all of its expenses. If it
bids too high, however, the job may be offered to one of its competitors.
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16. Reason quantitatively. Jim estimates that it will take 0.5 hour to install each
shrub. Use the functions n(h), c(h), and (n · c)(h) to determine the following values
for Jim’s bid, and explain how you determined your answers.
a. the number of shrubs Jim can install in an 8-hour day
16 shrubs; I evaluated n(h) for h = 0.5: n(0.5) =
8
0.5
= 16.
b. the cost per shrub, including installation
$48.50; I evaluated c(h) for h = 0.5: c(0.5) = 16 + 65(0.5) = 48.50.
c. the total cost of Jim’s services for an 8-hour day
$776; I evaluated (n · c)(h) for h = 0.5: (n · c)(0.5) =
128
0.5
+ 520 = 776
17. Explain how you could check your answer to Item 16c.
Sample answer: Multiply the number of shrubs Jim can install in an 8-hour day by the cost
per shrub, including installation: 16($48.50) = $776.
Jim offers two lawn improvement services, as described in the table.
Lawn Improvement Services
Service
Compost
Hourly Charge ($)
40
Material Cost for Average
Yard ($)
140
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18.
a. Write a function c(h) to represent the total charge for applying compost to a
lawn, where h is the number of hours the job takes.
c(h) = 40h + 140
b. Write a function f(h) to represent the total charge for applying fertilizer to a
lawn, where h is the number of hours the job takes.
f(h) = 30h + 30
Writing Math
The notation (f ÷ g)(x), g(x) ≠ 0 represents the quotient of the functions f(x) and g(x)
given that g(x) ≠ 0. In other words, (f ÷ g)(x) = f(x) ÷ g(x), g(x) ≠ 0.
19.
a. Divide c(h) by f(h) to find (c ÷ f)(h) given that f(h) ≠ 0.
(c ÷ f )(h) =
40h + 140
30h + 30
b. What does the function (c ÷ f)(h) represent in this situation?
the ratio of the cost in dollars of applying compost to the cost in dollars of
applying fertilizer for a job that takes h hours
20. Find (c ÷ f)(4). What does this value represent in this situation?
(c ÷ f )(4) =
40(4) + 140
30(4) + 30
= 2; For a job that takes 4 hours, the cost of applying compost is 2
times the cost of applying fertilizer.
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Math Tip
You may be able to simplify the function rules in Items 21c, d, and e by factoring
the expression’s numerator and denominator and dividing out common factors.
21. Look for and make use of structure. Given f(x) = 2x, g(x) = x + 3, and h(x) =
2x + 6, find each function and simplify the function rule. Note any values that
must be excluded from the domain.
a. (f · g)(x)
2
(f · g)(x) = 2x(x + 3) = 2x + 6x
b. (g · h)(x)
2
(g · h)(x) = (x + 3)(2x + 6) = 2x + 12x + 18
c. (f ÷ h)(x), h(x) ≠ 0
( f ÷ h)(x) =
2x
2x + 6
=
x
,x
x+3
≠ −3
d. (h ÷ g)(x), g(x) ≠ 0
(h ÷ g)(x) =
2x + 6
x+3
= 2, x ≠ − 3
e. (g ÷ f)(x)
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(g ÷ f )(x) =
x+3
,x
2x
≠0
Discussion Group Tips
As you listen to your group members’ discussions, you may hear math terms or
other words that you do not know. Use your math notebook to record words that
are frequently used. Ask for clarification of their meaning, and make notes to
help you remember and use those words in your own communications.
22. Discuss and then answer this question with your group. How are operations
on functions similar to and different from operations on real numbers?
Sample answer: Operations on real numbers involve only numbers. Operations on
functions involve function rules. Otherwise, the processes of addition, subtraction,
multiplication, and division are essentially the same. For division of real numbers,
the divisor cannot be 0, and for division of functions, the function rule that is the
divisor cannot be equal to 0.
Check Your Understanding
23. Given that f(x) = 2x + 1 and g(x) = 3x − 2, what value(s) of x are excluded from
the domain of (f ÷ g)(x)? Explain your answer.
x ≠ 23 ; The function (f ÷ g)(x) is undefined when g(x) = 0. Because g(x) = 3x − 2,
g(x) = 0 when x = 23 . So,
2
3
is excluded from the domain of (f ÷ g)(x).
24. Make a conjecture about whether addition of functions is commutative. Give
an example that supports your conjecture.
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Addition of functions is commutative. Sample example: Given that f(x) = 4x + 2
and g(x) = −2x + 5, (f + g)(x) = (g + f)(x) = 2x + 7.
25. Given that h(x) = 4x + 5 and (h − j)(x) = x − 2, find j(x). Explain how you
determined your answer.
j(x) = 3x + 7; Sample explanation: I know that h(x) − j(x) = (h − j)(x), so j(x) =
h(x) − (h − j)(x) = 4x + 5 − (x − 2) = 3x + 7.
Lesson 5-1 Practice
For Items 26–30, use the following functions.
f (x) = 5x + 1
g(x) = 3x − 4
Find each function and simplify the function rule. Note any values that must be
excluded from the domain.
26. (f + g)(x)
(f + g)(x) = 5x + 1 + 3x − 4 = 8x − 3
27. (f − g)(x)
(f − g)(x) = 5x + 1 − (3x − 4) = 2x + 5
28. (f · g)(x)
(f • g)(x) = (5x + 1)(3x − 4) = 15x2 − 17x − 4
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29. (f ÷ g)(x), g(x) ≠ 0
( f ÷ g) =
5x + 1
4
,x ≠
3x − 4
3
30. A student incorrectly found (g − f)(x) as follows. What mistake did the
student make, and what is the correct answer?
(g − f )(x) = 3x − 4 − 5x + 1 = − 2x − 3
When subtracting the rule for f(x) from the rule for g(x), the student should
have written the rule for f(x) in parentheses so that both terms of the rule would
be subtracted. The correct answer is (g − f)(x) = 3x − 4 − (5x + 1) = −2x − 5.
31. Make sense of problems and persevere in solving them. Jim
plans to make a radio ad for his lawn company. The function a(t) = 800 +
84t gives the cost of making the ad and running it t times on an AM
station. The function f(t) = 264t gives the cost of running the ad t times
on a more popular FM station.
a. Find (a + f)(t) and tell what it represents in this situation.
(a + f)(t) = 800 + 84t + 264t = 800 + 348t; The function (a + f)(t) represents
the cost of making the ad and running it on both stations t times.
b. Find (a + f)(12) and tell what it represents in this situation.
(a + f)(12) = 800 + 348(12) = 4976; It will cost $4976 to make the ad and
run it on both stations 12 times.
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