The Irrational (or Radical)
Functions Unit
(Level IV Pre-Calculus)
NSSAL
(Draft)
C. David Pilmer
2014
(Last Updated: October 2014)
This resource is the intellectual property of the Adult Education Division of the Nova Scotia
Department of Labour and Advanced Education.
The following are permitted to use and reproduce this resource for classroom purposes.
• Nova Scotia instructors delivering the Nova Scotia Adult Learning Program
• Canadian public school teachers delivering public school curriculum
• Canadian non-profit tuition-free adult basic education programs
• Nova Scotia Community College instructors delivering the Academic Careers and
Connections mathematics curriculum or core college programs
The following are not permitted to use or reproduce this resource without the written
authorization of the Adult Education Division of the Nova Scotia Department of Labour and
Advanced Education.
• Upgrading programs at post-secondary institutions (exception: NSCC)
• Core programs at post-secondary institutions (exception: NSCC)
• Public or private schools outside of Canada
• Basic adult education programs outside of Canada
Individuals, not including teachers or instructors, are permitted to use this resource for their own
learning. They are not permitted to make multiple copies of the resource for distribution. Nor
are they permitted to use this resource under the direction of a teacher or instructor at a learning
institution.
Acknowledgments
The Adult Education Division would like to thank the following university professors for
reviewing this resource to ensure all mathematical concepts were presented correctly and in a
manner that supported our learners.
Dr. Robert Dawson (Saint Mary’s University)
Dr. Genevieve Boulet (Mount Saint Vincent University)
Dr. Jeff Hooper (Acadia University)
The Adult Education Division would also like to thank the following NSCC instructors for
piloting this resource and offering suggestions during its development.
John Archibald (Truro Campus)
Barbara Gillis (Burridge Campus)
Alice Veenema (Kingstec Campus)
The Adult Education Division would also like to thank Antoine Jarjoura, CSAP Math and
Science Curriculum Consultant, for reviewing this resource and offering feedback.
Table of Contents
Introduction ………………………………………………………………………………..
Prerequisite Knowledge ……………………………………………………………………
Negotiated Completion Date ………………………………………………………………
The Big Picture and Suggested Timelines ………………………………………………….
Pre-Calculus Relations Concept Map .……………………………………………………..
Common Errors Seen in First Year University Math Classes ………….………………….
ii
iii
iv
v
vi
vii
Introduction to Irrational (or Radical) Functions …………………………………………..
Examining Domains of Irrational Functions ……………………………………………….
Solving Irrational Equations ……………………………………………………………….
More Irrational Equations ………………………………………………………………….
A Special Type of Irrational Function; Absolute Value Functions ………………………..
More Absolute Value Functions ……………………………………………………………
Absolute Value Equations ………………………………………………………………….
Absolute Value Inequalities ………………………………………………………………..
1
4
15
25
29
36
40
52
Post-Unit Reflections ………………………………………………………………………
Soft Skills Rubric …………………………………………………………………………..
Answers …………………………………………………………………………………….
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63
64
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Introduction
Level IV Pre-Calculus
Pre-Calculus is strictly designed for learners who are intending on enrolling in science or
engineering programs at the university level. The prerequisite for this course is ALP Level IV
Academic Math with a minimum mark of 85%, PSP Advanced Math 11 with a minimum mark
of 80%, or PSP Pre-Calculus Math 11 with a minimum mark of 80%. We also recommend that
the learner have a strong work ethic and the ability to work independently.
The delivery of the Pre-Calculus curriculum will likely vary from campus to campus. Many
campuses will not have the resources to offer this course on their schedule. In these cases, the
learner will have to take it as a correspondence course or as an online course. Larger campuses
may be able to offer the course, but it is likely that the learner will be placed in a class that is
predominately populated with Level IV Academic Math learners. There is a possibility that
Academic Careers and Connections (ACC) will use the ALP Pre-Calculus curriculum within
their own program. If this is the case, ACC and ALP learners may be found in the same
classroom. Ultimately, the delivery model used with Pre-Calculus will be at the discretion of the
Nova Scotia Community College and the Academic Chairs at the various campuses.
There is a strong likelihood that you may be the only Pre-Calculus learner in the class or at your
campus. For this reason, it may be a challenge to obtain assistance for extended periods of time.
This being said, we have made every effort to make the material found in this resource as
accessible as possible. One of the major differences between the Pre-Calculus resources and the
Academic Math resources can be seen in the answer key. In the Academic Math resources, the
answer keys only contained the final answers. The answer key in this resource and other PreCalculus resources is far more extensive. For many of the questions we have also provided hints,
and with the more challenging questions, we have provided complete solutions. Use this feature
judiciously; do not refer to the answer key before making a concerted effort to answer the
question on your own. Neither do we want learners to seek assistance from an instructor without
first attempting to use the answer key to resolve their problem. In the near future, we hope to
create online instructional videos to support our Pre-Calculus learners. Ultimately, learners with
a solid understanding of Academic Math concepts, a great work ethic, and the ability to work
independently, will flourish in this course. Although some of the topics can be challenging,
remember that you can do it.
Most of you will have taken ALP Level IV Academic Mathematics prior to taking this course.
In that academic course, there was a heavy emphasis on real-world application questions, and the
use of the graphing calculator. In Pre-Calculus, one will encounter application questions and
have to use graphing calculators, but to a far lesser degree. There is a greater emphasis on the
pure mathematics and the automaticity to recall relevant mathematical facts and concepts. This
parallels the expectations that one will encounter in a first year university calculus course.
Regardless of this, the material in this resource is presented in a manner to foster understanding
(opposed to encouraging memorization).
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The NSSAL Pre-Calculus Print Resources
If you are not affiliated with the Nova Scotia School for Adult Learning (NSSAL) and choosing
to use this resource, you may be wondering why some of the early topics that appear in our PreCalculus resources seem to be more introductory in nature compared to topics found in
traditional Pre-Calculus programs and textbooks. The reason stems from the fact that the
prerequisite within the ALP system for our Pre-Calculus course is Level IV Academic Math.
Traditional Pre-Calculus courses usually have an advanced level mathematics course as their
prerequisite. NSSAL does not have the resources or the number of learners necessary to offer an
advanced math prerequisite. For this reason our Pre-Calculus includes introductory material that
is typically found in an advanced math prerequisite.
Prerequisite Knowledge
Learners who are starting the Level IV Pre-Calculus Rational Functions Unit should be familiar
with the concepts examined in the Level IV Academic Rational Expressions and Radicals Unit
and the Level IV Pre-Calculus Polynomial Functions Unit. These include:
• Adding and subtracting rational expressions
9
2
e.g.
− 2
2
x + 11x + 28 3 x + 13 x + 4
9
2
=
−
( x + 7 )( x + 4 ) ( x + 4 )( 3x + 1)
9 ( 3 x + 1)
2 ( x + 7)
−
( x + 7 )( x + 4 )( 3x + 1) ( x + 4 )( 3x + 1)( x + 7 )
=
•
=
9 ( 3 x + 1) − 2 ( x + 7 )
( x + 7 )( x + 4 )( 3x + 1)
=
27 x + 9 − 2 x − 14
( x + 7 )( x + 4 )( 3x + 1)
=
25 x − 5
( x + 7 )( x + 4 )( 3x + 1)
Multiplying and dividing rational expressions
6 p 3 + 15 p 2
p2 − 4 p
e.g.
÷
p 2 + 4 p − 32 p 2 − 8 p + 16
6 p 3 + 15 p 2 p 2 − 8 p + 16
×
p 2 + 4 p − 32
p2 − 4 p
=
3 p 2 ( 2 p + 5) ( p − 4 )
=
×
( p + 8)( p − 4 ) p ( p − 4 )
2
=
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3 p ( 2 p + 5)
p +8
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•
Identifying polynomial functions
e.g.
y = 4 x3 + 5 x − 1 polynomial, specifically cubic or third order polynomial
e.g. =
y 4 x −2 + 3 x not polynomial
e.g.
•
y= 5 − 2 x 4 polynomial, specifically quartic or fourth order polynomial
Dividing two polynomial expressions using long division or synthetic division.
e.g. ( 2 x3 + 7 x 2 − 9 x − 20 ) ÷ ( x + 4 )
Long Division:
2x 2 − x − 5
Synthetic Division:
-4
2
7
-9
-8
4
2
-1
-5
x + 4 2 x3 + 7 x 2 − 9 x − 20
-20
20
0
2 x3 + 8 x 2
− x2 − 9x
− x2 − 4 x
− 5 x − 20
−5 x − 20
0
Therefore: ( 2 x3 + 7 x 2 − 9 x − 20 ) ÷ ( x + 4 )= 2 x 2 − x − 5
Negotiated Completion Date
After working for a few days on this unit, sit down with your instructor and negotiate a
completion date for this unit.
Start Date:
_________________
Completion Date:
_________________
Instructor Signature: __________________________
Student Signature:
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__________________________
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The Big Picture and Suggested Timelines
The following flow chart shows the six required units for the 100 hour Level IV Pre-Calculus
course. These are presented in a suggested order.
Sequences and Series (14 hours)
• Functional Form, Recursive Form, Arithmetic Sequences and
Series, Geometric Sequences and Series
Trigonometric Equations and Identities Unit, Part 1 (18 hours)
• Unit Circle, Special Rotations, Evaluating Trigonometric
Expressions, Trigonometric Equations, Radian Measure,
Trigonometric Identities
Trigonometric Equations and Identities Unit, Part 2 (15 hours)
• Trigonometric Equations Involving Tangents, Principal
Solutions, More Trigonometric Identities
The Euler Number Unit (10 hours)
• The Euler Number, Natural Logarithms, Exponential Functions
Using Base e, Logistic Functions
Interval Notation Unit (5 hours)
• Domain, Range, Increasing, Decreasing, Concave Upwards,
Concave Downwards
Polynomial Functions Unit (12 hours)
• Graphs of Polynomial Functions, Determining the Equation of a
Polynomial Function, Polynomial Equations and Inequalities
Rational Functions Unit (9 hours)
• Graphs of Rational Functions, Rational Equations and
Inequalities, Partial Fractions
Irrational Functions Unit (9 hours)
• Domains of Irrational Functions, Irrational Equations, Absolute
Value Functions, Absolute Value Equations and Inequalities
Odds and Ends (8 hours)
• Binomial Expansion, Composite Functions, Combinations of
Functions, Inverse Functions, and Imaginary and Complex
Numbers
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The Pre-Calculus Relations Concept Map
Relations
A relation shows the relationship
between two sets of numbers
Conics
Circles, ellipses, hyperbolas,
and parabolas
Function
A relation is a function when for every
member of the first set (typically x's),
there is only one corresponding member
in the second set (typically y's).
Algebraic Functions
These functions can be described
using algebraic operations (addition,
subtraction, multiplication, division,
exponentiation).
Rational Functions
The variable has an
integral constant
exponent.
e.g. y = x −1 or y =
1
x
Transcendental Functions
These functions cannot be described
using algebraic operations.
Irrational Functions
(or Radical Functions)
The variable has a
non-integral constant
exponent.
e.g. y = x
1
2
or y = x
Polynomial Functions
The variable has a
positive integral
constant exponent.
e.g. y = x 3
Absolute Value
Functions
Trigonometric Functions
e.g. y = sin x , y = cot x
Logarithmic Functions
e.g. y = log x , y = ln x
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e.g. y = x or y = x 2
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Exponential Functions
e.g. y = 2 x , y = e x
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Common Errors Seen in First Year University Math Classes
Do not walk into a first year university mathematics classroom without being aware of the
following common errors. At this point in your Pre-Calculus course, you may not have
encountered all of the concepts associated with the errors listed below. However, by the end of
the course, you should be familiar with all of the concepts.
•
( x + y)
•
( x + y)
•
1
1 1
is not equal to + .
x+ y
x y
•
2
is not equal to x 2 + y 2 ; rather, it is equal to x 2 + 2 xy + y 2 .
3
is not equal to x3 + y 3 ; rather, it is equal to x 3 + 3 x 2 y + 3 xy 2 + y 3 .
x 2 + y 2 is not equal to x + y .
• The expression
•
1
y
cannot be simplified to
(i.e. we cannot cancel out the y's.)
x +1
x+ y
2
is not equal to 0; rather, it is undefined.
0
• We know that sin 2 x + cos 2 x =
1 , but that does not mean that sin x + cos x =
1.
• cos ( x + y ) is not equal to cos x + cos y ; rather, it is equal to cos x cos y − sin x sin y .
• sin 2x is not equal to 2sin x ; rather, it equal to 2sin x cos x .
• cos −1 x is not equal to
1
: rather, it is another way to say arccos x .
cos x
• b0 is not equal to 0, it is equal to 1 (as long as b is not equal to 0; when b = 0 , the
expression is undefined).
• The solutions to the equations x = 9 and x 2 = 9 are different. The solution to the
equation x = 9 is 3. The solution to the equation x 2 = 9 is 3 or -3.
• −52 and ( −5 ) are different expressions; −52 =
25 .
−25 while ( −5 ) =
2
2
• log a ( b + c ) is not equal to log a b + log a c ; rather, the appropriate law of logarithms is
log a=
( bc ) log a b + log a c .
If by the end of your Pre-Calculus course, you are still unsure why some of these common errors
are incorrect, sit down with your instructor and go over them. Understanding these errors is far
more important than merely memorizing them.
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Introduction to Irrational (or Radical) Functions
Application 1:
The heart rates of mammals vary greatly based on the size of the mammal. For example, a
mouse’s heart beats approximately 600 beats per minute, while an adult horse’s heart beats an
average of 35 beats per minute. The following function describes the relationship between a
mammal’s heart rate, R, and its body mass, m. The heart rate is measured in beats per minute
and the body mass is measured in kilograms.
H=
240
4
m
H = 240m−0.25
or
Application 2:
The length of a pendulum has a direct bearing on how much time it takes for that pendulum to
complete one full cycle (i.e. swing back and forth). For example a pendulum that is 0.5 metres
long takes 1.4 seconds to complete a full cycle, while a 1 metre pendulum takes 2.0 seconds.
The following function describes the relationship between the length, L, of a pendulum and the
time, T, to complete a full cycle. The length is measured in metres and the time is measured in
seconds.
1
 L 2
L
or T = 2π  
T = 2π
 9.8 
9.8
Application 3:
The critical height of a tree (i.e. maximum sustainable height of a tree) is dependent upon the
base diameter of the tree trunk. This should seem obvious; taller trees require larger tree trunks.
There are specific functions that describe this relationship for specific species of trees. However,
we work with the following generic formula that describes the relationship between critical
height, h, and base diameter, d, for most species of trees. Both the height and diameter are
measured in metres.
h = 68 d
3
2
or h = 68d
2
3
In this unit, we examine irrational or radical functions. The three functions shown above are
all examples of irrational or radical functions. We classify them as this type of function because
they possess non-integral constant exponents. These functions should not be confused with
rational functions. As we learned in the previous unit, rational functions are functions that can
be expressed as the ratio of two polynomial functions. We also stated that they can be viewed as
functions that possess integral constant exponents.
1
1
or y = x −3
3
x
integral constant exponent of -3
y=
y = x or y = x 2
1
2
Therefore irrational (or radical) function
non-integral constant exponent of
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Therefore rational function
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All irrational (or radical) functions, like rational functions, are classified as algebraic functions
because they can be described using algebraic operations (addition, subtraction, multiplication,
division and exponentiation).
Example
Classify each of the following functions using the appropriate terminology (Algebraic, Irrational,
Rational, Polynomial, Transcendental, Trigonometric, Logarithmic, Exponential).
(a) y = 3 2x − 7
(b) g(x ) = 3sec2x
2
5
x − 7x
(d) y =
(c) y =
2
x3
x−7
(e) A = 3e−0.72t
(f) y =
x (2x + 1)
Answers:
This type of question is easy to answer if you regularly refer to the Pre-Calculus Relations
Concept Map found in the beginning of this resource and all the other Pre-Calculus resources
that preceded it.
1
(a) y = 3 2x − 7 or y = (2x − 7)3
(b) g(x ) = 3sec2x
Transcendental, Trigonometric
7
1
x 2 − 7x
(c) y =
or y = x 2 − x
2
2
2
(d) y =
5
x3
or y = 5x
(e) A = 3e−0.72t
(f) y =
x−7
x (2x + 1)
Algebraic, Irrational
−
3
2
Algebraic, Rational, Polynomial
Algebraic, Irrational
Transcendental, Exponential
Algebraic, Rational
Questions
1. Classify each of the following functions using the appropriate terminology (Algebraic,
Irrational, Rational, Polynomial, Transcendental, Trigonometric, Logarithmic, Exponential).
(a) y = 3log(x −1)
(b) y = x 2 − 2x − 3
(c) A = πr 2
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3x 2
(e) p(x ) =
x+2
(g) y = cos
3π
4
(i) C = 2πr
500
(k) N =
1+ 20e−0.9t
x
 13
(f) y = 9 
 2
(h) y =
3
9
5− x
(j) y = 4 x 0.25
(l) g(t ) = −6.75t
−
4
3
4
2. (a) The formula for the volume of a sphere is V = π r 3 . What type of function is this?
3
(b) Rearrange the formula for the volume of sphere such that r is expressed in terms of V.
What type of function are we addressing now?
3. Explain the following analogy. Irrational functions are to algebraic functions as squares are
to quadrilaterals.
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Examining Domains of Irrational Functions
In this course and in Level IV Academic Math, many of the functions we have examined did not
have restrictions on their domains.
e.g. y = 3 x 2 + 7 x − 2
Domain: { x ∈ R} or x ∈ ( −∞, ∞ )
x
e.g.
y = 3( 2) 4
100
1 + 20e −0.8t
e.g.=
y 3sin x + 2
e.g.
y=
Domain:
{ x ∈ R}
or x ∈ ( −∞, ∞ )
Domain:
{ x ∈ R}
or x ∈ ( −∞, ∞ )
Domain:
{ x ∈ R}
or x ∈ ( −∞, ∞ )
However, there were other functions that did have restrictions on their domains.

π

e.g. y = tan x
Domain:  x ∈ R x ≠ + kπ , k ∈ I  or
2


 3π π   π π   π 3π 
x ∈ ... ∪  − , −  ∪  − ,  ∪  ,  ∪ ...
2  2 2 2 2 
 2
Domain: { x ∈ R x > 0} or x ∈ ( 0, ∞ )
e.g. y = ln x
e.g. y =
5
( x + 1)( x − 4 )
Domain: { x ∈ R x ≠ −1 or 4} or x ∈ ( −∞, −1) ∪ ( −1, 4 ) ∪ ( 4, ∞ )
There are some irrational functions that do not have restrictions on their domains.
=
y
3
x −1
y=
− 3 x2 − 9
Domain: x ∈ ( −∞, ∞ )
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Domain: x ∈ ( −∞, ∞ )
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There are other irrational functions that do have restrictions on their domains.
=
y
x+3
y =− x 2 − x − 2
Domain: x ∈ ( −∞,1] ∪ [ 2, ∞ )
Domain: x ∈ [ −3, ∞ )
Irrational functions that involve a variable within an even root (i.e. square roots, fourth roots,
sixth roots, etc.) will likely have a restriction on their domains. This stems from the fact that we
cannot take the square root, fourth root, or sixth root of a negative number (if we are restricting
ourselves to real numbers).
However, irrational functions that involve a variable within an odd root (cube root, fifth root,
seventh root, etc.) will likely have first set values from negative infinity to infinity (i.e. no
restrictions on the domain). This stems from the fact that you can take the cube root, fifth root,
or seventh root of any number.
There are exceptions to these generalizations that will be more clearly seen by examining the
following worked examples.
Example 1
Determine the domain of each of these irrational functions.
(a)=
y
(c)=
y
3
2x − 6
(b) y =
x2 − 4
(d) b =
x 2 + 5 x − 14 + 2
4
a 3 + 4a 2 − 4a − 16
1
 2 5
(e) y = 
2 
9− x 
(f) h ( x ) =( x 4 − 13 x 2 + 36 )
−
1
2
−7
Answers:
(a)=
y
2x − 6
The 2 x − 6 must be greater than or equal to zero because we cannot take the square root
of negative numbers (if we are restricting ourselves to real numbers).
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2x − 6 ≥ 0
2x ≥ 6
2x 6
≥
2 2
x≥3
Then the domain is x ∈ [3, ∞ ) .
In the second last step, we did not have to "flip" the inequality sign because we divided
both sides of the inequality by a positive number (i.e. +2). If, however, we did multiply
or divide both sides of an inequality by a negative number, then the inequality sign must
be flipped.
(b) y =
x 2 + 5 x − 14 + 2
The expression x 2 + 5 x − 14 must be greater than or equal to zero. We will now be
solving a polynomial inequality, a topic we studied in the Polynomial Functions Unit.
x 2 + 5 x − 14 ≥ 0
Change the inequality to an equation, find the roots of the equation, and then use a
number line to test the regions between those roots.
0
x 2 + 5 x − 14 =
0
( x + 7 )( x − 2 ) =
2
x=
−7 or x =
We need the positive regions.
-
+
Test x = −8 {10}
Test x = 0 {-14}
Test x = 3 {10}
-7
+
2
Domain: x ∈ ( −∞, 7 ] ∪ [ 2, ∞ )
(c)=
y 3 x2 − 4
Since we are dealing with a cube root, then the expression x 2 − 4 can generate positive or
negative values.
Domain: x ∈ ( −∞, ∞ )
(d) b = 4 a 3 + 4a 2 − 4a − 16
The a 3 + 4a 2 − 4a − 16 must be greater than or equal to zero.
a 3 + 4a 2 − 4a − 16 ≥ 0
Change to an equation, find the roots, and then use a number line to test regions between
the roots.
0
a 3 + 4a 2 − 4a − 16 =
(a
3
+ 4a 2 ) + ( −4a − 16 ) =0
Factoring by Grouping
0
a2 ( a + 4) − 4 ( a + 4) =
0
( a + 4) ( a2 − 4) =
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0
( a + 4 )( a − 2 )( a + 2 ) =
−4 or a =
−2
a=
2 or a =
Test
Test
Test
Test
x = −5
x = −3
x=0
x=3
-
{-21}
{5}
{-16}
{35}
-4
We need the positive regions.
-
+
-2
+
2
Domain: x ∈ [ −4, −2] ∪ [ 2, ∞ )
1
2
 2 5
(e) y = 
or y = 5
2 
9 − x2
9− x 
At first we might assume that there are no restrictions on the domain because we are
2
, we realize
dealing with a fifth root. However, when we consider the expression
9 − x2
that 9 − x 2 cannot be equal to zero (because we cannot divide by zero).
9 − x2 ≠ 0
( 3 − x )( 3 + x ) ≠ 0
Therefore the domain is x ∈ ( −∞, −3) ∪ ( −3,3) ∪ ( 3, ∞ )
x ≠ 3 or x ≠ −3
1
2 − 7 or h x
(f) h ( x ) =( x 4 − 13 x 2 + 36 ) =
( )
1
−7
x − 13 x 2 + 36
In this case the expression x 4 − 13 x 2 + 36 cannot be negative (because you cannot take the
square root of a negative number) and cannot be equal to zero (because you cannot divide
by zero).
x 4 − 13 x 2 + 36 > 0
As before, change the polynomial inequality to an equation, find the roots, and then use a
number line to test values between those roots.
x 4 − 13x 2 + 36 =
0
−
(x
2
4
− 9 )( x 2 − 4 ) =
0
0
( x − 3)( x + 3)( x − 2 )( x + 2 ) =
x=
3 or x =
−3 or x =
2 or x =
−2
Test
Test
Test
Test
Test
x = −4
x = −2.5
x=0
x = 2.5
x=4
{84}
{-6.2}
{36}
{-6.2}
{84}
+
+
-3
-2
+
2
3
Therefore the domain is x ∈ ( −∞, −3) ∪ ( −2, 2 ) ∪ ( 3, ∞ ) .
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Questions
1. Determine the domain of each of these functions in set notation. No work needs to be
shown.
(a) y = x
(b) y = 3 x
(c) y =
(d) y =
1
x
1
x
3
(e) y = x 2
2. Determine the domain of each of these functions in set notation. Show your work.
(a)=
y
12 − 4 x
(b) f (=
x)
(c) y =
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5 x + 35
x 2 − 3 x − 10
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1
(d) h ( x ) =−
( 2x2 + 9x − 4) 2
(e) g ( =
t)
(f) y=
4
( t 2 − 25)
1
3
x3 − 3x 2 − x + 3
(g) p ( a ) =
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−
5
a+2
a − 3a − 4
2
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(h) f ( t=
)
(i) y=
1
2 −4
( 9 − 4t )
2 x3 + x 2 − 18 x − 9
(j) q ( t ) =( t 4 − 17t 2 + 16 )
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−
1
2
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3. Match each graph of y = f ( x ) to the corresponding graph of y =
f ( x ) . (Hint: Is there
any useful information you can obtain from the graph of y = f ( x ) that would help you
understand the domain of the function y =
f ( x ) ?)
Graphs of y = f ( x )
Graphs of y =
f ( x)
(a)
(b)
(c)
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4. Match each graph of y = f ( x ) to the corresponding graph of y =
Graphs of y = f ( x )
Graphs of y =
f ( x) .
f ( x)
(a)
(b)
(c)
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5. In each case, you are supplied with the graph of y = f ( x ) . Based on each of these, find the
domain of y =
(a)
f ( x) .
Graph of y = f ( x ) :
Domain of y =
(c)
(b)
f ( x) :
Domain of y =
Graph of y = f ( x ) :
Domain of y =
Graph of y = f ( x ) :
(d)
f ( x) :
f ( x) :
Graph of y = f ( x ) :
Domain of y =
f ( x) :
6. Open-ended Questions (i.e. more than one acceptable answer)
(a) Provide the equation of an irrational function whose domain is x ∈ [ −9, ∞ ) .
(b) Provide the equation of an irrational function whose domain is x ∈ ( −∞, 6 ) ∪ ( 6, ∞ ) .
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7. (a) Determine the domain of the function of f ( x ) =
1
( x3 + x 2 − 6 x ) 2 .
(b) Create a table of values, plot the points and sketch the graph of this function.
8. (a) Determine the domain of the function of g ( x ) =− ( 3 x 2 − x − 2 )
−
1
2
.
(b) Create a table of values, plot the points and sketch the graph of this function.
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Solving Irrational Equations
With irrational equations, those involving roots (square roots, cube roots, etc.), we often have to
raise both sides of the equation to a specific exponent to eliminate the roots. However, this can
create a problem that we have not encountered in the prior mathematics courses. That is, we
create an extraneous root (also called an extraneous solution). This is a root that seems to be
appropriate based on the algebraic procedure that we have followed, yet is not a solution because
it does not satisfy the original equation. Consider the very simple irrational equation y = 9 ; we
know that the only solution to this question is x = 3 . Now look at the lengthy algebraic
procedure used to solve this simple equation; notice how many solutions we obtain.
x= 9
x2 = 9
( )
2
Square both sides of the equation to eliminate the square root.
x2 = 9
x2 − 9 = 0
(x − 3)(x + 3) = 0
x = 3 or x = −3
To solve a quadratic equation, first set it equal to zero.
Factor
The answer x = 3 is the correct solution. However, the answer x = −3 is the extraneous solution.
In the case of irrational equations, extraneous solutions results when we raise both sides of the
equation to an even exponent (e.g. square both sides). To deal with this issue, it is important to
check all solutions by substituting them back into the original equation.
Let’s look at another simple equation. Consider x = −7 . We know there is no real solution to
this question; it is impossible to square root a real number such that it generates a negative
number. Now look at the solution provided below.
x = −7
2
2
x = (−7)
( )
Square both sides
x = 49
The answer x = 49 is obviously extraneous because we know that the principal square root of 49
is 7, not -7. Again, the squaring of both sides created a situation in which an extraneous root
could be generated.
Example 1
Solve the equation
7x − 3 = 5.
Answer:
7x − 3 = 5
( 7x − 3) = 5
2
2
Square both sides.
7x − 3 = 25
7x = 28
x=4
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We still need to check the answer to make sure we are not dealing with an extraneous root.
7x − 3 = 5
7(4 ) − 3 = 5
Check:
28 − 3 = 5
25 = 5
It works. Therefore the solution is x = 4 .
5=5
Example 2
Solve 3 5x − 7 = 2.
Answer:
3
5x − 7 = 2
( 5x − 7 ) = (2)
3
3
3
Cube both sides of the equation.
5x − 7 = 8
5x = 15
x=3
Since we did not have to raise both sides to an even exponent, we do not have to check for
extraneous roots. The solution is x = 3 .
Example 3
Solve 3x + 10 = x + 4 .
Answer:
3x + 10 = x + 4
( 3x + 10 ) = (x + 4)
2
2
3x + 10 = x 2 + 8x + 16
0 = x 2 + 5x + 6
0 = (x + 3)(x + 2)
x = −3 or x = −2
Checks:
3x + 10 = x + 4
3(−3) + 10 = (−3) + 4
3x + 10 = x + 4
3(−2) + 10 = (−2) + 4
−9 + 10 = 1
−6 + 10 = 2
1 =1
1=1
4 =2
2=2
Both work so we have two solutions; x = −3 and x = −2 .
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Example 4
Solve 3 x + 6 + 2x = 2.
Answer:
3 x + 6 + 2x = 2
3 x + 6 = 2 − 2x
(3
x+6
) = (2 − 2x )
2
2
9(x + 6) = 4 − 8x + 4 x 2
9x + 54 = 4 − 8x + 4 x 2
0 = 4 x 2 −17x − 50
0 = (4 x − 25)(x + 2)
25
or x = −2
x=
4
Check:
3 x + 6 + 2x = 2
 25 
25
+ 6 + 2  = 2
3
4
4
3 −2 + 6 + 2(−2) = 2
25 24 25
3
=2
+
+
4
2
4
6 + (−4 ) = 2
49 25
=2
+
4
2
 7  25
=2
3  +
 2 2
21 25
=2
+
2
2
46
=2
2
23 ≠ 2
3 x + 6 + 2x = 2
3 4 + (−4 ) = 2
2=2
3
The solution is x = −2 .
Example 5
Solve 3 x 3 + 2x 2 + 5 −1 = x .
Answer:
3
x 3 + 2x 2 + 5 −1 = x
x 3 + 2x 2 + 5 = x + 1
3
(
3
x 3 + 2x 2 + 5
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) = (x + 1)
3
3
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x 3 + 2x 2 + 5 = (x + 1)(x + 1)(x + 1)
x 3 + 2x 2 + 5 = (x 2 + 2x + 1)(x + 1)
x 3 + 2x 2 + 5 = x 3 + x 2 + 2x 2 + 2x + x + 1
x 3 + 2x 2 + 5 = x 3 + 3x 2 + 3x + 1
0 = x 2 + 3x − 4
0 = (x + 4 )(x −1)
x = −4 or x = 1
Example 6
Solve x 2 = 3 − 2x 2 − 3 .
Answer:
x 2 = 3 − 2x 2 − 3
x 2 − 3 = − 2x 2 − 3
(x
2
(
− 3) = − 2x 2 − 3
2
)
2
x 4 − 6x 2 + 9 = 2x 2 − 3
x 4 − 8x 2 + 12 = 0
(x 2 − 2)(x 2 − 6)= 0
x2 = 2
x =± 2
or
Factoring by Inspection
x2 = 6
x =± 6
We need to check all four of these roots to see if any are extraneous. We chose not to show
this work. Suffice to say that the solutions are x = 2 and x = − 2 .
Example 7
If f (x ) = 2x − 2 and g(x ) = x + 1, find the values of x for which f (x ) = g(x ).
Answer:
f (x ) = g(x )
2x − 2 = x + 1
2
2
2x − 2 = x + 1
) (
(
)
Square both sides.
2x − 2 = x + 2 x + 1
x −3=2 x
(x − 3)
2
( )
= 2 x
2
x 2 − 6x + 9 = 4 x
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x 2 −10x + 9 = 0
(x − 9)(x −1) = 0
x = 9 or x = 1
We chose not to show the necessary check. However, we would learn that x = 1 is an
extraneous root. The only solution is x = 9 .
Example 8
Find the solution to
2x + 25 − 2 x + 4 = 1.
Answer:
2x + 25 − 2 x + 4 = 1
2x + 25 = 2 x + 4 + 1
( 2x + 25 ) = (2
2
)
x + 4 +1
2
Square both sides.
2x + 25 = 4 (x + 4 ) + 4 x + 4 + 1
2x + 25 = 4 x + 17 + 4 x + 4
−2x + 8 = 4 x + 4
−x + 4 = 2 x + 4
(−x + 4)
2
(
= 2 x+4
Divided both sides by 2.
)
2
Square both sides again.
x 2 − 8x + 16 = 4 (x + 4 )
x 2 − 8x + 16 = 4 x + 16
x 2 −12x = 0
x (x −12) = 0
x = 0 or x = 12
We chose not to show the necessary check. However, we would learn that x = 12 is an
extraneous root. The only solution is x = 0 .
Example 9
Solve 2 x − 4 x − 3 =
1
4x − 3
Answer:
Start by eliminating the denominator by multiplying everything on both sides of the equation
by 4 x − 3 .
 1 
4x − 3 2 x − 4x − 3 4x − 3 = 4x − 3 

 4x − 3 
(
)
(
)
2 4 x 2 − 3 x − ( 4 x − 3) =
1
2 4 x 2 − 3x − 4 x + 3 =
1
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2 4 x 2 − 3x = 4 x − 2
4 x 2 − 3x = 2 x − 1
(
4 x 2 − 3x
)
2
=( 2 x − 1)
Divided both sides by 2.
2
4 x 2 − 3x = 4 x 2 − 4 x + 1
x =1
The answer x = 1 checks out when substituted back into the original equation.
Questions
1. A variety of graphs are provided below. Use these graphs to solve each of the following
equations. No work needs to be shown.
y=
−2 3 x + 3 and other functions
(a) 4 =
15 + 2x − x 2
y=
x
(b) −2 3 x + 3 =
15 + 2 x − x 2 =− x − 3
(c) −4 =−2 3 x + 3
(d)
(e) 2 =
−2 3 x + 3
(f) −5=
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15 + 2 x − x 2 and other functions
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15 + 2x − x 2
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2. Solve each of the following radical equations.
(a)
(c)
3
2 x + 10 =
2
(b)
3x + 4 =
−2
(d) x − 5=
(e) 2 3 x + 6 − 5 =
x
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©2014
3 x + 1 =−5
(f) x −=
2
21
x −3
3
39 x − 8
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C. D. Pilmer
(g)
x 3 − 2 x 2 + 24 x + 1 − 2 x =
5
(i) 3 + x + 4 =
NSSAL
©2014
6− x +5
(h)
x +=
1
(j)
2 x − 5 +=
1 2 x −3
22
3x − 1
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C. D. Pilmer
(k)
x + x −5 =
10
x −5
(l) 3 x − 2 x + 1 =
9
2x +1
3. Find the point(s) of intersection between the linear function f ( x )= x + 6 and the radical
function g ( x=
)
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©2014
6 − x algebraically. Confirm your answer using graphing technology.
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4. The surface area of a cone=
is A π r r 2 + h 2 . Solve the equation for h.
5. The speed of sound changes slightly based on the air temperature. Warmer temperatures
result in slightly higher speeds of sound. This relationship between the speed of sound, s, in
metres per second and the air temperature, T, in degrees Celsius is described by the following
function.
T
=
s 331.3 1 +
273.15
(a) What restriction is placed on T given this function?
(b) Determine the difference between the speed of sound at -20oC and to 30oC.
(c) What air temperature would result in a speed of sound of 337.3 m/s?
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More Irrational Equations
In the previous section we primarily dealt with irrational equations that were simply the non1
1
integral constant exponents of (square roots) and (cube roots). In this section, we examine
3
2
irrational equations with a greater variety of non-integral constant exponents.
Example
Solve each of the following.
3
(b) 5.6x −0.75 = 2.9
(a) 40 = 16x 2
2
1
(c) x 3 + 2 x 3 − 8 =
0
Answers:
In both cases, we raise both sides of the equation to a value that will ultimately change the
exponent of the x to that of one.
3
(a) 40 = 16x 2
3
40 16x 2
=
16
16
2.5 = x
Divide both sides of the equation by 16.
3
2
2
 3 3
(2.5) =  x 2 
 
1.84 = x
2
3
Raise both sides of the equation to the exponent
2
.
3
Calculator: 2.5^(2/3) is equal to 1.84
3
Check:
40 = 16(1.84 )2
40 = 16(2.5)
It works.
40 = 40
(b) 5.6x −0.75 = 2.9
5.6x
−
3
4
−
3
4
5.6x
5.6
−
= 2.9
=
Express the exponent as a fraction, rather than a decimal.
2.9
5.6
Divide both sides of the equation by 5.6.
3
x 4 = 0.5179
4
4
 − 3 − 3
−
 x 4  = (0.5179) 3
 
x = 2.40
Check:
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4
Raise both sides of the equation to the exponent − .
3
Calculator: 0.5179^(-4/3) is equal to 2.40
5.6(2.40)
= 2.9
5.6(0.5186) = 2.9
It works.
2.9 = 2.9
−0.75
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C. D. Pilmer
2
3
1
3
(c) x + 2 x − 8 =
0
1
3
Let a = x
a 2 + 2a − 8 =
0
0
( a + 4 )( a − 2 ) =
−4 or a =
a=
2
1
1
−4 or x 3 =
x3 =
2
−64 or x =
x=
8 (Both work in the original equation.)
Questions
1. Solve each of the following.
4
−
2
3
= 10.8
(a) 17.6 = 0.9x 3
(b) 15x
(c) 8.4 x1.25 = 21.6
(d) 1.5 = 2.4 x −0.4
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3
2. Solve ( x 2 + x − 4 ) 4 =
8.
2
1
3. Solve x 3 − x 3 − 6 =.
0
4. The heart rates of mammals vary greatly based on the size of the mammal. The function
240
describes the relationship between a mammal’s heart rate, R, and it’s body mass, m.
H=4
m
The heart rate is measured in beats per minute and the body mass is measured in kilograms.
(a) What is the anticipated heart rate of a species of mammal that has a typical body mass of
150 kg?
(b) What is the typical body mass for a species of mammal that has a heart rate of 100 beats
per minute?
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5. The critical height of a tree (i.e. maximum sustainable height of a tree) is dependent upon the
2
base diameter of the tree trunk. The function h = 68d 3 describes the relationship between
critical height, h, and base diameter, d, for most species of trees. Both the height and
diameter are measured in metres.
(a) What is the anticipated base diameter for a tree that has a critical height of 37 metres?
(b) What is the anticipated critical height for a tree that has a base diameter of 0.23 metres?
6. The Beaufort scale was created in 1805 by Sir Francis Beaufort and was used to describe
wind speed based mainly on observed sea conditions. For example, a Beaufort number of 2 is
given if the sea is comprised of small non-breaking wavelets (0.2 to 0.5 metres high) and
winds can just be felt on exposed skin. By comparison, a Beaufort number of 8 is given for
gale force winds where the sea is comprised of moderately high breaking waves (5.5 to 7.5
metres high) which create considerable spray. At this level, the winds are strong enough to
seriously impede walking. The highest number on the Beaufort scale is 12, which is applied
to hurricane conditions. Wind speed, s, in metres per second in relation to the Beaufort
number, B, can be described by the following function.
3
s = 0.837 B 2
(a) Knowing that the Beaufort scale goes from 0 to 12, determine the domain and range for
the above function.
(b) What Beaufort number would be applied if wind speeds were 26 metres per second?
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A Special Type of Irrational Function; Absolute Value Functions
Absolute Values
Before we look at absolute value functions, we should learn about absolute values. The absolute
value of a number describes the distance a number is from zero on a number line without
considering the direction from zero the number lies (i.e. we don't care if it is to the right or to the
left of zero).
• The absolute value of 5 is 5 because we are 5 units from zero. This can be written as
5 = 5.
Distance =
5
-6
•
-5
-4
-3
-2
-1
0
1
2
3
4
The absolute value of -5 is 5 because we are 5 units from zero. This can be written as
−5 =
5.
Distance =
5
-6
-5
-4
-3
-2
-1
0
1
2
3
4
•
The absolute value of 0 is 0. Do not fall into the trap of saying that the absolute value of
a number is a positive. Zero is the case in point because it is neither negative nor
positive.
•
Additional Examples:
e.g. −9 =
9
e.g. −8.3 =
8.3
e.g. 6 = 6
e.g.
3 3
=
4 4
In conclusion,
1. When x is a positive number or zero, then the value will not change when you take its
absolute value.
x.
For x ≥ 0, x =
e.g. 7 = 7
2. When x is a negative number, then the number will change its sign when you take its absolute
value.
For x < 0, x =
e.g. −3 =− ( −3) =3
−x .
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When would we use absolute values in the real world? Consider the following examples.
• Suppose we have a person standing on a cliff and their altitude is +15 metres. Another
individual is submerged below the water and has an altitude of -20 metres. If we are
asked to determine, who is closest to sea level (i.e. altitude of 0 metres), then we
intuitively use absolute values. Since the +15 =
15 and −20 =
20 , then we conclude
that person standing on the cliff is closest to sea level.
• Suppose a person walks forward 6 metres and then backwards 4 metres. We would say
that their displacement is 2 metres. This is obtained by taking the displacement of +6
metres (forward movement) and adding it to the displacement of -4 metres (backwards
movement). However, if we are asked to state the distance the person has traveled, we
would state 10 metres. Distance, unlike displacement, does not include a directional
component; it is merely how much ground an object has covered. So in this particular
case, to calculate the distance using the displacements provided we added 6 and −4 to
obtain the distance 10 metres.
Questions
1. Evaluate each of the following. No work needs to be shown.
(a) 12 =
(b) −9 =
(c) −3.45 =
(d) 6.5 =
(e) 2
3
=
8
(f) −3
2
=
5
2. If a = −4 , b = −1 , and c = 2 , find the value of each expression. Show your work.
(a) 5 a + b
(b) 2 a − bc
(c) 4 a + 3b − 2 c
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©2014
(d) ab + 3 ac
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(e) a b + b − c
(f) 2 b + c − 3 a − b
Absolute Value Functions
Absolute value functions come in numerous forms; however, for this course, we narrow our
y ax + b and y = ax 2 + bx + c , in which a, b,
focus to absolute value functions of the form =
and c are real numbers. Rather than explicitly telling you how to graph such functions, we ask
you to discover this for yourself by completing the following investigation.
Investigation
In parts 1 and 2 of the investigation, you compare linear functions of the form f ( x=
) ax + b to
their corresponding absolute value functions of the form g ( x=
) ax + b . In parts 3 and 4 of the
investigation, you will compare quadratic functions of the form f ( x ) = ax 2 + bx + c to their
corresponding absolute value functions of the form g ( x ) = ax 2 + bx + c .
Part 1
Complete the table of values for f ( x )= x + 2 and g ( x )= x + 2 . Graph both functions on the
same coordinate system. You may wish to use color pencils to distinguish the two functions.
x
f ( x)
g ( x)
1
3
3
-3
3
0
-1
-2
-3
-4
-5
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Part 2
Complete the table of values for f ( x )= 2 − 2 x and g ( x )= 2 − 2 x . Graph both functions on the
same coordinate system.
x
f ( x)
g ( x)
4
-6
6
3
2
1
0
-1
-2
Part 3
Complete the table of values for f ( x ) = x 2 + 2 x − 3 and g ( x ) = x 2 + 2 x − 3 . Graph both
functions on the same coordinate system.
x
f ( x)
g ( x)
2
1
0
-1
-2
-3
-4
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Part 4
1
1
Complete the table of values for f ( x ) =
− x 2 + 2 and g ( x ) =
− x 2 + 2 . Graph both
2
2
functions on the same coordinate system.
x
f ( x)
g ( x)
3
-2.5
2.5
2
1
0
-1
-2
-3
Investigation Questions
(a) Look at the work you completed in Parts 1 through 4 of this investigation. Describe how the
graph of a linear or quadratic function is related to its corresponding absolute value function.
(b) Would the graph of =
y x 2 + 1 look any different from the graph of =
y x 2 + 1 ? Why or
why not?
(c) What can you say about the domain and range of absolute functions of the form =
y ax + b ?
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(d) Using what you learned when answering investigation questions (a) through (c), use the
graph on the left to determine the graph of the corresponding absolute value function on the
right for each of the following. No work needs to be shown.
(i)
(ii)
NSSAL
©2014
y
=
1
x +1
2
=
y
3
y=
− x+6
2
1
x +1
2
3
y=
− x+6
2
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(iii)
(iv)
NSSAL
©2014
=
y 3x 2 − 6 x
=
y 3x 2 − 6 x
y=
−2 x 2 − 8 x − 6
y=
−2 x 2 − 8 x − 6
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More Absolute Value Functions
In the previous section we examined linear and quadratic functions and their corresponding
absolute value functions. Although we are now able to visualise what the graphs of these types
of absolute value functions look like, you are probably asking yourself
why absolute value functions are considered irrational (or radical)
functions. To answer this, let's consider the absolute value function
y = x whose graph is shown on the right. There are three different
ways of writing this equation.
 x, if x ≥ 0
y= x
y=
y = x2
− x, if x < 0
In the second version, we have written y = x as a piecewise function, in which the function is
composed of two or more separate functions, or pieces, each with its specific domain. In the
third version, we write y = x using a radical, specifically a square root. If a function has a nonintegral constant exponent, then it is classified as irrational. In this case, y = x or y = ( x
2
1
2 2
)
1
. This is why we classify this absolute value function as an
2
irrational function. This reasoning applies to all absolute value functions.
has the non-integral exponent of
Consider the absolute value function =
y 2 x + 6 whose graph is shown
on the right. The two other ways of writing this equation are as follows.
 2 x + 6, if x ≥ −3
y=
− ( 2 x + 6 ) , if x < −3
( 2x + 6)
=
y
2
Again, the second version is a piecewise function, and the third version clearly explains why
absolute value functions are classified as irrational functions.
Consider the absolute value function =
y x 2 − 6 x whose graph is shown
on the right. The two other ways of writing this equation are as follows.
 x 2 − 6 x, if x ≤ 0 or x ≥ 6
y=
2
 − ( x − 6 x ) , if 0 < x < 6
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Questions
1. Express each using an absolute value.
(a)
=
y
( 3x − 2 )
(b) y =
2
 8 − 2 x, if x ≤ 4
(c) y = 
− ( 8 − 2 x ) , if x > 4
(e) y =
−5
(x
2
− 2 x − 8)
2
(−x
2
+ x + 12 )
2
9 − x 2 , if − 3 ≤ x ≤ 3
(d) y = 
2
 − ( 9 − x ) , if x < −3 or x > 3
 5 x + 10, if x ≥ −2
(f) y = 
− ( 5 x + 10 ) , if x < −2
2. In each case, an absolute value function and its graph are supplied. Write the equation of the
function as a piecewise function. The scale on both axes of each graph goes from -10 to 10.
(a)
=
y 2 x − 10
(b)
y= 3 − x
(c)
=
y 3 x + 12
(d)
y = x2 + x − 6
(e)
y= 4 − x 2
(f) =
y 0.5 x 2 + 3.5 x
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3. Use the graph on the left to determine the graph of the corresponding absolute value function
on the right.
(a)
2
y=
− x+2
3
2
y=
− x+2
3
(b)
1
y =− x 2 − x + 4
2
1
y =− x 2 − x + 4
2
4. Write each of the absolute functions in question 3 as a piecewise function.
2
(a) y =
− x+2
3
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(b) y =− x 2 − x + 4
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5. Two learners were asked to determine the equation
of the function shown on the right but to express it
as a piecewise function. Their answers are
supplied below.
First Learner's Answer:
 2 x − 6, if x ≥ 3
y=
− ( 2 x − 6 ) , if x < 3
Second Learner's Answer:
−2 x + 6, if x ≤ 3
y=
− ( −2 x + 6 ) , if x > 3
Look at both answers. Who is correct? Explain.
6. In a layperson's terms, a cusp is a point on a graph where the curve makes an abrupt change
in direction. The formal definition of a cusp involves derivatives, a topic that will be studied
extensively in a calculus course.
(a) Look at the graphs you generated for the absolute value functions in question 3. Provide
the coordinates of the cusps in each case.
2
− x+2
y=
3
1
y =− x 2 − x + 4
2
(b) Absolute value functions of the form =
y ax + b always have a cusp at the point
 b 
 − , 0  . Why is this so?
 a 
(c) Under what circumstances would an absolute value function of the form y = ax 2 + bx + c
not have cusps?
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Absolute Value Equations
 x, if x ≥ 0
As we know, x = 
.
− x, if x < 0
This notation can be used to describe other absolute value expressions. Some examples are
provided below.
 x + 3, if x ≥ 3
x+3 =

− ( x + 3) , if x < 3
 x − 5, if x ≥ 5
x −5 =

− ( x − 5 ) , if x < 5
 4 − 2 x, if x ≤ 2
4 − 2x =

− ( 4 − 2 x ) , if x > 2
5

 3 x + 5, if x ≥ − 3
3x + 5 =

− ( 3 x + 5 ) , if x < − 5

3
Understanding this notation is critical to understanding how to solve absolute value equations.
Example 1
Solve x − 3 =
5.
Answer:
This question can be quickly worked out in your head. Most learners can see that x must be
equal to 8 or -2. However, we use this question as our launching point to introduce the case
method.
Case 1: If x − 3 ≥ 0 , then x − 3 = x − 3 .
So the equation changes to x − 3 =
5.
When we solve this equation, we obtain x = 8.
We have to check that this value of x satisfies the original equation. The 8 − 3
does equal 5, therefore we know that we have a correct solution.
− ( x − 3) .
Case 2: If x − 3 < 0 , then x − 3 =
So the equation changes to − ( x − 3) =
5.
When we solve this equation, we obtain x = −2.
We have to check that this value of x satisfies the original equation. The −2 − 3
does equal 5, therefore we know that we have a correct solution.
Therefore, the solutions for this equation are
=
x 8 or − 2 . This can
be confirmed by graphing y= x − 3 and y = 5 on the same
coordinate system and examining the points of intersection. On a
graphing calulator, enter Y1=abs(X-3), where abs is found in the
Catalog menu.
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Example 2
Solve x − 1 = 2 x − 7 .
Answer:
Case 1: If x − 1 ≥ 0 , then x − 1 = x − 1 .
So the equation changes to x − 1= 2 x − 7 .
x − 1= 2 x − 7
x − 2 x =−7 + 1
− x =−6
x=6
We have to check that this value of x satisfies the original equation.
6 −=
1 2 ( 6) − 7
5= 12 − 7
5=5
It satisfies the equation. We have a correct solution.
Case 2: If x − 1 < 0 , then x − 1 =− ( x − 1) .
So the equation changes to − ( x − 1) = 2 x − 7 .
− ( x − 1) = 2 x − 7
− x + 1= 2 x − 7
− x − 2 x =−7 − 1
−3 x =
−8
8
3
We have to check that this value of x satisfies the original equation.
8
8
−=
1 2  − 7
3
3
5 16 21
=
−
3 3 3
5
5
≠−
It does not satisfies the equation. This is an extraneous root.
3
3
Why does this Case 2 not work? It has to do with the assumption we make at the
beginning of Case 2 when we said x − 1 < 0 . This only applies when x < 1 . When
8
we solved the equation based on this assumption, we obtained , a number that is
3
greater than 1, rather than less than 1. The solution to this equation was contrary to
the initial assumption; that is why it did not satisfy the original equation.
x=
Therefore, the solution for this equation is x = 6 . This can be
confirmed by graphing y= x − 1 and =
y 2 x − 7 on the same
coordinate system and examining the points of intersection.
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Example 3
Solve x + 7 = x 2 + 4 x + 3 .
Answer:
Case 1: If x 2 + 4 x + 3 ≥ 0 , then x 2 + 4 x + 3 = x 2 + 4 x + 3 .
So the equation changes to x + 7 = x 2 + 4 x + 3 .
x + 7 = x2 + 4x + 3
0 = x 2 + 3x − 4
0=
( x + 4 )( x − 1)
x=
−4 or x =
1
We have to check that these values of x satisfy the original equation.
( −4 ) + 7 = ( −4 )
2
+ 4 ( −4 ) + 3
(1) + 7= (1)
3 = 16 − 16 + 3
8 = 1+ 4 + 3
3= 3
8= 8
=
3 3
← satisfies equation
=
8 8
2
+ 4 (1) + 3
← satisfies equation
Case 2: If x 2 + 4 x + 3 < 0 , then x 2 + 4 x + 3 =
− ( x 2 + 4 x + 3) .
So the equation changes to x + 7 =− ( x 2 + 4 x + 3) .
x + 7 =− ( x 2 + 4 x + 3)
0=
− x 2 − 5 x − 10
x=
5 ± −15
−2
no real solution
Therefore, the solution for this equation is x = −4 or 1 . This can be
confirmed by graphing y = x 2 + 4 x + 3 and y= x + 7 on the same
coordinate system and examining the points of intersection.
Example 4
Solve x 2 − 2 x =
1.
Answer:
Case 1: If x 2 − 2 x ≥ 0 , then x 2 − 2 x =x 2 − 2 x .
So the equation changes to x 2 − 2 x =
1.
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x2 − 2 x =
1
x 2 − 2 x − 1 =0
x= 1± 2
(Quadratic Formula)
x � 2.414 or x � −0.414
We have to check that these values of x satisfy the original absolute value equation.
We have not shown this work but suffice to say both satisfy the original equation.
Case 2: If x 2 − 2 x < 0 , then x 2 − 2 x =
− ( x2 − 2 x ) .
So the equation changes to − ( x 2 − 2 x ) =
1.
− ( x2 − 2 x ) =
1
− x 2 + 2 x − 1 =0
x2 − 2 x + 1 =
0
( x − 1)
2
=
0
x =1
Although we have not shown the work, we discover that x = 1 satisfies the original
absolute value equation.
Therefore, the solution for this equation is x =
1 − 2, 1 + 2, or 1 .
This can be confirmed by graphing =
y x 2 − 2 x and y = 1 on the
same coordinate system and examining the points of intersection.
Example 5
Solve 2 x − 4 = x − 3 + 1 .
Answer:
With two absolute value expressions in this equation, we end up with four possible cases.
Case 1: If 2 x − 4 ≥ 0 and x − 3 ≥ 0 , then 2 x − 4 = 2 x − 4 and x − 3 = x − 3 .
So the equation changes to 2 x − 4 = x − 3 + 1 .
2x − 4 = x − 3 +1
2 x − x =−2 + 4
x=2
This value does not satisfy the original absolute value equation; it is an extraneous
root.
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Case 2: If 2 x − 4 ≥ 0 and x − 3 < 0 , then 2 x − 4 = 2 x − 4 and x − 3 =
− ( x − 3) .
So the equation changes to 2 x − 4 =− ( x − 3) + 1 .
2 x − 4 =− ( x − 3) + 1
2 x − 4 =− x + 3 + 1
2x + x = 4 + 4
3x = 8
x=2
2
3
This value does satisfy the original absolute value equation.
− ( 2 x − 4 ) and x − 3 = x − 3 .
Case 3: If 2 x − 4 < 0 and x − 3 ≥ 0 , then 2 x − 4 =
So the equation changes to − ( 2 x − 4 ) = x − 3 + 1 .
− ( 2x − 4) = x − 3 + 1
−2 x + 4 = x − 2
−2 x − x =−2 − 4
−3 x =
−6
x=2
This value does not satisfy the original absolute value equation. This should make
sense given the contrary nature of the two assumptions in Case 3. The assumption
2 x − 4 < 0 can only occur for x-values less than 2. However, the assumption
x − 3 ≥ 0 can only occur for x-values greater than or equal to 3. How can a number
be simultaneously less than 2 and greater than or equal to 3?
Case 4: If 2 x − 4 < 0 and x − 3 < 0 , then 2 x − 4 =
− ( x − 3) .
− ( 2 x − 4 ) and x − 3 =
So the equation changes to − ( 2 x − 4 ) =
− ( x − 3) + 1 .
− ( 2x − 4) =
− ( x − 3) + 1
−2 x + 4 =− x + 3 + 1
−2 x + x = 4 − 4
x=0
This value does satisfy the original absolute value equation.
2
or 0 . This can be
3
confirmed by graphing =
y 2 x − 4 and y = x − 3 + 1 on the same
coordinate system and examining the points of intersection.
Therefore, the solution for this equation is x = 2
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Questions
1. Four equations and four coordinate systems (each with two functions) are provided below.
Use the appropriate graphs to solve each of the equations. No work needs to be shown.
3
1
(b) x − 3 =
(a) 2 x + 6 =
4
− x+5
2
2
Answer:
Answer:
(c) x − 3 =
2x
(d) x 2 − 1 = x + 5
Answer:
Answer:
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2. Six equations and two coordinate systems (each with four functions) are provided below.
Use the appropriate graphs to solve each of the equations. No work needs to be shown.
2
(a) 2 x − 8 = x
(b) x + 2 =− x + 6
3
Answer:
1
(c) − x − 1 = x + 2
4
Answer:
(d) =
6 2x − 8
Answer:
(e) 3= x + 2
Answer:
−2
(f) 2 x − 8 =
Answer:
Answer:
3. Solve each of the following equations. Show your work.
(a) 2 x − 3 =
15
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(b) 3 x + 2 = x + 14
(c) x 2 − 10 x =10 − x
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(d) −3 x= 2 x 2 − 2
4. In question 3, you solved four absolute value equations algebraically, specifically using the
case method. Below you have four screen shots from a graphing calculator, each
representing a solution to one of the four absolute value equations in question 3. Determine
which equation corresponds to each screen shot. No work needs to be shown. Please note
the horizontal scale on all graphs is at increments of 1; the vertical scale has not been
provided.
(i)
(ii)
(iii)
(iv)
Equation:
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Equation:
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5. A learner is solving an absolute value equation of the form 4 x − 12 = ax + b using the case
method. In case 1, the learner states, "If 4 x − 12 ≥ 0 , then 4 x − 12 = 4 x − 12 ." She correctly
solves the resulting equation and obtains the answer x = 1 . Is this root extraneous for this
case? Explain.
6. Solve each of the following equations.
(a) 3 x + 4 = x − 5 + 13
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(b) x 2 − 2 x − 16 =
8
(c) 5 − 2 x + 6 = 3 x − 7
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(d) −4 x + 6 = x 2 − 3 x
(e) x 2 + 3 x = 2 x − 5 + 9
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Absolute Value Inequalities
In the previous section, we examined a wide variety of absolute value equations. In this section,
we examine absolute value inequalities but largely restrict them to inequalities of the following
four forms.
ax + b ≤ c
ax + b ≥ c
ax + b < c
ax + b > c
Initially, we show you two methods for solving these types of inequalities. The first method
relies heavily on techniques we have used with other types of inequalities and on the case
method that we learned in the previous section. The second method, which is what you find in
most textbooks and explained on most online math videos, requires far fewer steps, but learners
are often uncertain why this technique works. We attempt to use the first method to explain why
the second method works. We do this so that learners do not think that the second method is
some kind of "mathematical trick."
Example 1
Solve 3 x − 6 < 15 .
Answer:
Method 1:
When we solved polynomial inequalities, we started by simplifying the question to an
equation. We verified the inequality in the last few steps when we tested different regions on
a number line. We do the same here.
15 , and solve the resulting equation using the case method.
Change 3 x − 6 < 15 to 3 x − 6 =
Case 1: If 3 x − 6 ≥ 0 , then 3 x − 6 = 3 x − 6 .
3x − 6 =
15
3 x = 21
x 7 ← satisfies absolute value equation
=
Case 2: If 3 x − 6 < 0 , then 3 x − 6 =
− ( 3x − 6 ) .
− ( 3x − 6 ) =
15
−3 x + 6 =
15
−3 x =
9
x=
−3 ← satisfies absolute value equation
We know that the x-values of 7 and -3 set the expression 3 x − 6 equal to 15, but the original
problem (i.e. the inequality) is looking for x-values that make that same expression less than
15. To determine this, we test different regions along a number line using numbers of our
choosing.
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Test x = −4
Test x = 0
Test x = 8
{18, which is not less than 15}
{6, which is less than 15}
{18, which is not less than 15}
Fals
True
-3
Final Answer: x ∈ ( −3, 7 )
Fals
7
Method 2:
Before we look at the alternate method, let's consider some simpler inequalities.
• Consider x < 2 . If we applied the case method to this inequality, we would get the
following.
Case 1: If x ≥ 0 , then x = x
x<2
Case 2: If x < 0 , then x = − x
−x < 2
−x 2
>
Flip the inequality sign when you divide by a negative.
−1 −1
x > −2
This can be written as −2 < x < 2 , which in interval notation is x ∈ ( −2, 2 ) .
Notice that x < 2 , ultimately translates into −2 < x < 2 .
If we were given x ≤ 2 , then it would ultimately translate into −2 ≤ x ≤ 2 .
• Consider x > 2 . If we applied the case method to this inequality, we would get the
following.
Case 1: If x ≥ 0 , then x = x
x>2
Case 2: If x < 0 , then x = − x
−x > 2
−x 2
<
Flip the inequality sign when you divide by a negative.
−1 −1
x < −2
Therefore x < −2 or x > 2 . In interval notation, this answer is expressed as
x ∈ ( −∞, −2 ) ∪ ( 2, ∞ ) .
Notice that x > 2 , ultimately translates into x < −2 or x > 2 .
If we were given x ≥ 2 , then it would ultimately translate into x ≤ −2 or x ≥ 2 .
The point of all this was to learn that:
(i) x < a can be expressed as −a < x < a , when a ≥ 0 .
(ii) x > a can be expressed as x < −a or x > a , when a ≥ 0 .
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Let's apply this new found knowledge to the absolute value inequality 3 x − 6 < 15 .
3 x − 6 < 15
− 15 < 3 x − 6 < 15
−15 + 6 < 3 x − 6 + 6 < 15 + 6
Add 6 to all three parts.
− 9 < 3 x < 21
−9 3 x 21
<
<
Divide all three parts by 3.
3
3
3
−3< x < 7
When expressed in interval notation, we obtain x ∈ ( −3, 7 ) .
If we wanted to check this using a graphing
calculator, we would graph =
y 3 x − 6 and
y = 15 on the same coordinate system and look
for the region(s) on the absolute value function
that are below the constant function. We have
attempted to illustrate this on the right. The
"bolded" portion of the line represents the
desired region (i.e. between -3 and 7).
Example 2
Solve 2 x + 3 ≥ 9 .
Answer:
Method 1:
9 , and solve the resulting equation using the case method.
Change 2 x + 3 ≥ 9 to 2 x + 3 =
Case 1: If 2 x + 3 ≥ 0 , then 2 x + 3 = 2 x + 3 .
2x + 3 =
9
2x = 6
x 3 ← satisfies absolute value equation
=
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Case 2: If 2 x + 3 < 0 , then 2 x + 3 =
− ( 2 x + 3) .
− ( 2 x + 3) =
9
−2 x − 3 =
9
−2 x =
12
x=
−6 ← satisfies absolute value equation
We know that the x-values of 3 and -6 set the expression 2 x + 3 equal to 9, but the original
problem (i.e. the inequality) is looking for x-values that make that same expression greater
than or equal to 15. To determine this, we test different regions along a number line using
numbers of our choosing.
Test x = −7
Test x = 0
Test x = 4
{11, which is greater than or equal to 9}
{3, which is not greater than or equal to 9}
{11, which is greater than or equal to 9}
True
-6
Final Answer: x ∈ ( −∞, −6] ∪ [3, ∞ )
Fals
True
3
Method 2:
2x + 3 ≥ 9
2 x + 3 ≤ −9
or
2x + 3 ≥ 9
2 x + 3 − 3 ≤ −9 − 3
2x + 3 − 3 ≥ 9 − 3
Subtract 3 from all three parts.
2 x ≤ −12
2x ≥ 6
2x 6
2 x −12
≥
Divide all three parts by 2.
≤
2 2
2
2
x ≤ −6
x≥3
When expressed in interval notation, we obtain x ∈ ( −∞, −6] ∪ [3, ∞ ) .
If we wanted to check this using a graphing
calculator, we would graph =
y 2 x + 3 and
y = 9 on the same coordinate system and look
for the region(s) on the absolute value function
that are above or intersecting the constant
function. We have attempted to illustrate this on
the right. The "bolded" portions of the line
represent the desired regions.
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You may be wondering why we bothered learning method 1, when method 2 is so much shorter.
The problem is that method 2 does not work with more challenging inequalities such as
3 x + 4 ≥ x − 5 + 13 , which are beyond the scope of this course. However, method 1 does work
to solve these more challenging problem.
For the rest of the examples, we only focus on method 2 because we address simpler absolute
value inequalities.
Example 3
Solve 9 − 2 x − 7 > −6 .
Answer:
9 − 2 x − 7 > −6
9 − 2x > 1
9 − 2 x < −1
or
9 − 2x > 1
9 − 2 x − 9 < −1 − 9
9 − 2x − 9 > 1− 9
− 2 x < −10
− 2 x > −8
−2 x −10
−2 x −8
>
<
−2
−2
−2 −2
x>5
x<4
When written in interval notation, we obtain x ∈ ( ∞, 4 ) ∪ ( 5, ∞ ) .
Example 4
Samir and Angela are asked to solve two similar inequalities. They show their work and supply
their final answers.
Samir's Question and Solution
x − 7 ≤ −2
Angela's Question and Solution
x − 7 ≥ −2
2 ≤ x − 7 ≤ −2
x−7 ≤ 2
x≤9
9≤ x≤5
Therefore x ∈ ( −∞,5] ∪ [9, ∞ )
or
x − 7 ≥ −2
x≥5
Therefore x ∈ [5,9]
Are their answers correct? Explain.
Answer:
Previously we learned that:
(i) x < a can be expressed as −a < x < a , when a ≥ 0 .
(ii) x > a can be expressed as x < −a or x > a , when a ≥ 0 .
Note the condition that a ≥ 0 . In Samir and Angela's inequality questions, a = −2 , a number
that is less than 0. Both of them missed this and blindly followed an inappropriate
procedure. They could have caught their mistake if they had considered the graphs of the
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absolute value function y= x − 7 and the constant function y = −2 .
These two functions never intersect and the constant function is always
below the absolute value function. That means that all values of x
make the expression x − 7 greater than -2.
If they had considered the graphs, Samir would have discovered the answer to the question
x − 7 ≤ −2 should have been x ∈ φ (i.e. empty set). Similarly, Angela would have
discovered that the answer to the question x − 7 ≥ −2 should have been x ∈ ( −∞, ∞ ) .
If they had used method 1, they would have caught their mistake as they would have
−2 . When they would
discovered that 5 and 9 were extraneous roots to the equation x − 7 =
have subsequently tested the one region on the number line, Samir would find that no value
of x satisfied his equality and Angela would have discovered that all values of x satisfy her
inequality.
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Questions
1. Eight inequalities and two coordinate systems (each with three functions) are provided
below. Use the appropriate graphs to solve each of the inequalities. Express your answers in
interval notation. No work needs to be shown.
(a) 3 x − 6 < 3
(b) 2 x + 8 > 2
Answer:
(c) 2 x + 8 ≤ 2
Answer:
(d) 3 x − 6 ≥ 3
Answer:
Answer:
(e) 2 x + 8 ≤ −1
(f) 2 x + 8 > −1
Answer:
Answer:
(g) 3 x − 6 > 6
(h) 3 x − 6 ≤ 6
Answer:
Answer:
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2. (a) 4 x + 7 ≤ 13
(b) 6 x + 13 > 1
(c) 12 − 3 x < 15
2
(d) 1 − x ≥ 5
3
(e) 0.2 x − 1.3 > 2.7
(f) −0.5 x − 3.5 ≤ 6
3. Without doing any paper-and-pencil calculations, explain what the answer to the inequality
0.7 x − 4.8 ≥ −3.7 should be?
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4. Open-ended Questions (i.e. Answers will vary.)
(a) Create an absolute value inequality whose solution is x ∈ [ −1,3] .
(b) Create an absolute value inequality whose solution is x ∈ φ .
5. Eight inequalities and two coordinate systems (each with three functions) are provided
below. Although we do not ask you to solve these inequalities or one's of a similar form
algebraically in this course, we ask you to solve them graphically; specifically using the
graphs we provide. Express your answers in interval notation. No work needs to be shown.
(a) 2 x + 6 > x + 6
(b) x 2 + 4 x + 3 ≥ 8
Answer:
(c) 2 x + 6 ≤ −3 x + 1
Answer:
(d) x 2 + 4 x + 3 ≥ − x + 3
Answer:
(e) x 2 + 4 x + 3 < 8
Answer:
(f) 2 x + 6 < x + 6
Answer:
(g) x 2 + 4 x + 3 ≤ − x + 3
Answer:
(h) 2 x + 6 ≥ −3 x + 1
Answer:
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6. Solve x 2 − 7 x − 2 ≤ 2 x − 4 . Confirm your answer with a graphing calculator.
(Hint: Change the inequality into an equation, solve for the roots using the case method, and
then test the different regions on the number line.)
7. Yasha has a cylinder with a radius of 6 cm. She wants to pour 1 litre (1000 cm3)
of hydrochloric acid into it, with a 1% or less error in volume. How accurately
must Yasha measure the height of the acid?
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Post-Unit Reflections
What is the most valuable or important
thing you learned in this unit?
What part did you find most interesting or
enjoyable?
What was the most challenging part, and
how did you respond to this challenge?
How did you feel about this math topic
when you started this unit?
How do you feel about this math topic
now?
Of the skills you used in this unit, which
is your strongest skill?
What skill(s) do you feel you need to
improve, and how will you improve them?
How does what you learned in this unit fit
with your personal goals?
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Soft Skills Rubric
Look back over the module you have just completed and assess yourself using the following
rubric. Use pencil or pen and put a checkmark in the column that you think best describes your
competency for each description. I will look at how accurately you have done this and will
discuss with you any areas for improvement.
You will be better prepared for the next step, whether it is work or further education, if you are
competent in these areas by the end of the course. Keep all of these rubrics in one place and
check for improvement as you progress through the course.
Date:
Competent
demonstrates the concept
fully and consistently
Throughout this module, I…
Approaching
Competency
Developing
Competency
demonstrates the concept
most of the time
demonstrates the concept
some of the time
• Attended every class
• Let my instructor know if not
able to attend class
• Arrived on time for class
• Took necessary materials to class
• Used appropriate language for
class
• Used class time effectively
• Sustained commitment
throughout the module
• Persevered with tasks despite
difficulties
• Asked for help when needed
• Offered support/help to others
• Helped to maintain a positive
classroom environment
• Completed the module according
to negotiated timeline
• Worked effectively without close
supervision
Comments:
(Created by Alice Veenema, Kingstec Campus)
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Answers
Introduction to Irrational (or Radical) Functions (pages 1 to 3)
1. Note: You are not expected to supply the information supplied in the brackets below.
(a) Transcendental, Logarithmic
(b) Algebraic, Irrational
(c) Algebraic, Rational, Polynomial (specifically quadratic)
(d) Transcendental, Trigonometric (specifically sinusoidal)
(e) Algebraic, Rational
(f) Transcendental, Exponential
(g) Algebraic, Rational, Polynomial (specifically constant)
(h) Algebraic, Irrational
(i) Algebraic, Rational, Polynomial (specifically linear)
(j) Algebraic, Irrational
(k) Transcendental (not exponential, rather it’s logistic)
(l) Algebraic, Irrational
2. (a) Polynomial, specifically cubic
3
V , Irrational
(b) r = 3
4π
3. Irrational functions are a special type of algebraic function, just as squares are a special type
of quadrilateral.
Examining Domains of Irrational Functions (pages 4 to 14)
1. (a) x ∈ [ 0, ∞ )
(b) x ∈ ( −∞, ∞ )
(c) x ∈ ( 0, ∞ )
(d) x ∈ ( −∞, 0 ) ∪ ( 0, ∞ )
(e) x ∈ ( −∞, ∞ )
2. (a) x ∈ ( −∞,3]
(b) x ∈ [ −7, ∞ )
(c) x ∈ ( −∞, −2] ∪ [5, ∞ )
1 
(d) x ∈  , 4 
2 
(e) t ∈ ( −∞, −5 ) ∪ ( −5,5 ) ∪ ( 5, ∞ )
(f) x ∈ [ −1,1] ∪ [3, ∞ )
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(g) a ∈ ( −∞, −1) ∪ ( −1, 4 ) ∪ ( 4, ∞ )
 3 3
(h) t ∈  − , 
 2 2
1

(i) x ∈  −3, −  ∪ [3, ∞ )
2

(j) t ∈ ( −∞, −4 ) ∪ ( −1,1) ∪ ( 4, ∞ )
3. Hint: The zeros and positive regions of y = f ( x ) must correspond to the domain of
y=
f ( x ) . For example, with Graph (a), the zero occurs at x = 2 and the positive region
(i.e. the region(s) above the x-axis where y-values are positive) occurs in the interval
x ∈ ( 2, ∞ ) . That means that the corresponding graph of y = f ( x ) must have the domain
x ∈ [ 2, ∞ ) . Therefore Graph (a) matches with the third graph in the right-hand column.
(a) Matches with the third graph
(b) Matches with the first graph
(c) Matches with the second graph
4. (a) Matches with the second graph
(b) Matches with the third graph
(c) Matches with the first graph
5. (a) x ∈ ( −∞,3]
(c) x ∈ [ −2, 0]
(b) x ∈ ( −∞,1] ∪ [3, ∞ )
(d) x ∈ [ −3, −1] ∪ [ 0, ∞ )
6. There is more than one acceptable answer for each of the following. We provide the most
likely answer that learners will generate.
(a) =
y
x+9
(b) y =
3
1
x−6
7. (a) x ∈ [ −3, 0] ∪ [ 2, ∞ )
(b) Only choose x-values that are within the domain of the function.
x
y
-3
0
-2
2.83
-1
2.45
-0.5 1.77
0
0
2
0
2.5
2.62
3
4.24
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2

8. (a) x ∈  −∞,  ∪ (1, ∞ )
3

(b) You will discover when drawing this graph that there are vertical asymptotes at x = 1 and
2
x = − , and a horizontal asymptote at y = 0 . In the previous unit, we learned that many
3
rational functions possess both a vertical asymptote and horizontal asymptote. So why
are we seeing both of these asymptotes with an irrational function? This occurs for this
1
function because we have a rational expression,
, embedded within a
( 3x + 2 )( x − 1)
square root.
x
-4
-2
-1
-0.8
1.2
2
3
4
y
-0.14
-0.29
-0.71
-1.18
-0.95
-0.35
-0.21
-0.15
Solving Irrational Equations (pages 15 to 24)
1. (a) 1
(c) 5
(e) -4
2.
(b) -2
(d) -3
(f) no solution
2 x + 10 =
2
(a)
(
2 x + 10
)
2
(
=
22
2 x + 10 =
4
3x + 1
)
2
=
( −5)
2
3 x + 1 =25
3 x = 24
x = 8 (extraneous)
2 x = −6
x = −3 (correct)
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3 x + 1 =−5
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(c)
3
(
3x + 4 =
−2
3
3x + 4
)
=
( −2 )
3
(d)
x − 5=
( x − 5)
3
3 x + 4 =−8
3 x = −12
x = −4
x −3
(
=
2
x −3
)
2
x 2 − 10 x + 25 =x − 3
x 2 − 11x + 28 =
0
0
( x − 4 )( x − 7 ) =
x = 4 (extraneous)
x = 7 (correct)
(e)
x
2 3x + 6 − 5 =
(f)
x −=
2
3
39 x − 8
4 ( 3 x + 6 ) = x 2 + 10 x + 25
( x − 2 )= ( 3 39 x − 8 )
( x − 2 )( x − 2 )( x − 2 ) =39 x − 8
( x 2 − 4 x + 4 ) ( x − 2 ) = 39 x − 8
12 x + 24 = x 2 + 10 x + 25
x3 − 2 x 2 − 4 x 2 + 8 x + 4 x − 8 = 39 x − 8
0 = x2 − 2x + 1
x3 − 6 x 2 + 12 x − 8= 39 x − 8
2 3x + 6 = x + 5
(2
=
0
3x + 6
( x − 1)
)
2
3
3
=( x + 5 )
2
2
x3 − 6 x 2 − 27 x =
0
x ( x 2 − 6 x − 27 ) =
0
x = 1 (correct)
x ( x − 9 )( x + 3) =
0
x = 0, x = 9, or x = −3
(g)
(
x 3 − 2 x 2 + 24 x + 1 = 2 x + 5
(
x 3 − 2 x 2 + 24 x + 1
)
2
= ( 2 x + 5)
1
x +=
(h)
x 3 − 2 x 2 + 24 x + 1 − 2 x =
5
3x − 1
) (
2
x +1 =
)
3x − 1
2
x + 1 = 3x − 2 3x + 1
2
x3 − 2 x 2 + 24 x + 1 = 4 x 2 + 20 x + 25
−2 x =
−2 3 x
x3 − 6 x 2 + 4 x − 24 =
0
x = 3x
x2 ( x − 6) + 4 ( x − 6) =
0
x2 =
(
3x
)
2
0
( x − 6) ( x2 + 4) =
x 2 = 3x
x = 6 (correct)
0
x 2 − 3x =
0
x ( x − 3) =
x = 0 (extraneous)
x = 3 (correct)
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(i)
3+ x + 4 =
x+4=
(
6− x +5
(
6− x +2
) (
2
x+4 =
6− x +2
2 x − 5 +=
1 2 x −3
(j)
)
2
) (2
2
2 x − 5 + 1=
x−3
)
2
2x − 5 + 2 2x − 5 +=
1 4 ( x − 3)
x+4 = 6− x+4 6− x +4
2 2 x − 5 = 4 x − 12 − 2 x + 5 − 1
x + 4 − 6 + x − 4= 4 6 − x
2 2x − 5 = 2x − 8
2 x − 6= 4 6 − x
2x − 5 = x − 4
(
3 2 6− x
x −=
( x − 3) =
2
(2
6− x
)
2
2x − 5
)
2
=( x − 4 )
2
2 x − 5 = x 2 − 8 x + 16
x 2 − 6 x + 9= 4 ( 6 − x )
0 =x 2 − 10 x + 21
x 2 − 6 x + 9 = 24 − 4 x
0=
( x − 3)( x − 7 )
x − 2 x − 15 =
0
x = 3 (extraneous)
x = 7 (correct)
2
0
( x − 5)( x + 3) =
x = 5 (correct)
x = −3 (extraneous)
x + x −5 =
(k)
x −5
( x)+
10
x −5
x −5
(
)
x −5 =
 10 
x −5

 x −5 
10
x2 − 5x + x − 5 =
x 2 − 5 x =15 − x
(
x2 − 5x
)
2
=(15 − x )
2
x 2 − 5 x = 225 − 30 x + x 2
25 x = 225
x = 9 (correct)
(l)
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3 x − 2x +1 =
9
2x +1
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(
)
2x +1 3 x − 2x +1
(
)
2x +1 =
 9 
2x +1 

 2x +1 
3 2 x 2 + x − ( 2 x + 1) =
9
3 2 x 2 + x = 2 x + 10
(
3 2 x2 + x
)
2
= ( 2 x + 10 )
2
9 ( 2 x 2 + x ) = 4 x 2 + 40 x + 100
18 x 2 + 9 x = 4 x 2 + 40 x + 100
14 x 2 − 31x − 100 =
0
0
( x − 4 )(14 x + 25) =
x = 4 (correct)
x= −
25
(extraneous)
14
3. Use substitution.
f ( x )= x + 6
6− x =x+6
(
6− x
)
2
=( x + 6 )
2
6 − x = x 2 + 12 x + 36
0 =x 2 + 13 x + 30
0 =+
( x 10 )( x + 3)
x = −10 (extraneous)
x = −3 (correct)
Now find the y-coordinate. Use either of the functions.
f ( −3) =−3 + 6
=
f ( −3) 3
Point of Intersection: ( −3,3)
A2
1
to h
− r 2 which can be simplified further
=
2 2
π r
πr
4. h
=
A2 − π 2 r 4
T
≥0
273.15
Answer: T ≥ −273.15 or T ∈ [ −273.15, ∞ )
(This might sense to some of you because you know that absolute zero is -273.15oC)
5. (a) Hint: 1 +
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(b) Hints:
=
s 331.3 1 +
−20
30
, s 331.3 1 +
, then take difference
=
273.15
273.15
Answer: 30 m/s
337.3 331.3 1 +
(c) Hints:=
T
, divide both sides by 331.3, and then square both sides
273.15
Answer: 10oC
More Irrational Equations (pages 25 to 28)
1. (a) Hint: After dividing, raise both sides to the exponent
3
4
Answer: 9.30
(b) Hint: After dividing, raise both sides to the exponent −
3
2
Answer: 1.64
(c) Hints: Change 1.25 to
5
4
and after dividing, raise both sides to the exponent .
4
5
Answer: 2.13
(d) Hints: Change -0.4 to −
2
5
and after dividing, raise both sides to the exponent − .
5
2
Answer: 3.24
3
2.
( x 2 + x − 4 ) 4 =8
4
3
4
 2

(8) 3
 ( x + x − 4)  =


2
16
x + x−4=
3
4
0
x 2 + x − 20 =
0
( x + 5)( x − 4 ) =
4 (Both work when substituted back into the original equation.)
x=
−5 or x =
2
1
3. x 3 − x 3 − 6 =
0
1
Let a = x 3
a2 − a − 6 =
0
0
( a + 2 )( a − 3) =
−2 or a =
a=
3
1
3
1
3
−2 or x =
x =
3
−8 or x =
x=
27 (Both work in the original equation.)
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4. (a) 69 beats per minute
(b) Hint: 100 = 240m
Answer: 33 kg
−
1
4
5. (a) 0.4 m
(b) 25.5 m
6. (a) Domain: B ∈ [ 0,12]
(b) 10
Range: s ∈ [ 0,34.8]
A Special Type of Irrational Function; Absolute Value Functions (pages 29 to 33)
(b) −9 =
9
1. (a) 12 = 12
3.45
(c) −3.45 =
(e) 2
(d) 6.5 = 6.5
3
3
=2
8
8
(f) −3
2. (a) 25
(c) 9
(e) -1
2
2
=
3
5
5
(b) 6
(d) 28
(f) -7
Investigation
Part 1: f ( x )= x + 2 and g ( x )= x + 2
x
f ( x)
g ( x)
1
3
3
0
2
2
-1
1
1
-2
0
0
-3
-1
1
-4
-2
2
-5
-3
3
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Part 2: f ( x )= 2 − 2 x and g ( x )= 2 − 2 x
x
f ( x)
g ( x)
4
-6
6
3
-4
4
2
-2
2
1
0
0
0
2
2
-1
4
4
-2
6
6
Part 3: f ( x ) = x 2 + 2 x − 3 and g ( x ) = x 2 + 2 x − 3
x
f ( x)
g ( x)
2
5
5
1
0
0
0
-3
3
-1
-4
4
-2
-3
3
-3
0
0
-4
5
5
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1
1
Part 4: f ( x ) =
− x 2 + 2 and g ( x ) =
− x2 + 2
2
2
x
f ( x)
g ( x)
3
-2.5
2.5
2
0
0
1
1.5
1.5
0
2
2
-1
1.5
1.5
-2
0
0
-3
-2.5
2.5
Investigation Questions
(a) The ordered pairs corresponding to the positive regions (i.e. regions above the x-axis) and
zeros for a graph of a linear or quadratic function do not change in the corresponding
absolute value function. However, the negative regions (i.e. regions below the x-axis) for a
graph of a linear or quadratic function are reflected in the x-axis, changing all the negative yvalues to positive y-values.
(b) The graphs are the same because =
y x 2 + 1 is above the x-axis for all values of x.
y ax + b is x ∈ ( −∞, ∞ ) . The range of all
(c) The domain for all functions of the form =
functions of the form =
y ax + b is y ∈ [ 0, ∞ ) .
(d) (i)
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(ii)
(iii)
(iv)
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More Absolute Value Functions (pages 36 to 39)
1. (a) =
y 3x − 2
(b) y =− x 2 + x + 12
(c) y= 8 − 2 x
(d) y= 9 − x 2
(e) y =
−5 x 2 − 2 x − 8
y 5 x + 10
(f) =
 2 x − 10, if x ≥ 5
2. (a) y = 
− ( 2 x − 10 ) , if x < 5
 3 x + 12, if x ≥ −4
(c) y = 
− ( 3 x + 12 ) , if x < −4
4 − x 2 , if − 2 ≤ x ≤ 2
(e) y = 
2
 − ( 4 − x ) , if x < −2 or x > 2
 3 − x, if x ≤ 3
(b) y = 
− ( 3 − x ) , if x > 3
 x 2 + x − 6, if x ≤ −3 or x ≥ 2
(d) y = 
2
− ( x + x − 6 ) , if − 3 < x < 2
 0.5 x 2 + 3.5 x, if x ≤ −7 or x ≥ 0
(f) y = 
2
− ( 0.5 x + 3.5 x ) , if − 7 < x < 0
3. (a)
(b)
 2
 − 3 x + 2, if x ≤ 3
4. (a) y = 
−  − 2 x + 2  , if x > 3
  3

 1 2
− 2 x − x + 4, if − 4 ≤ x ≤ 2
(b) y = 
 −  − 1 x 2 − x + 4  , if x < −4 or x > 2
  2

5. Both answers are correct.
6. (a) (3,0)
(-4,0) and (2,0)
(b) The cusp always corresponds to the x-intercept of the function.
(c) No cusp occurs when the quadratic ax 2 + bx + c =
0 has one root or no roots.
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Absolute Value Equations (pages 40 to 51)
1. (a) x = -5 or -1
(b) x = -2 or 4
(c) x = -3 or 1
(d) x = -2 or 3
2. (a) x = 3 or 6
(c) no solution
(e) x = -5 or 1
(b) x = 2
(d) x = 1 or 7
(f) no solution
3. (a) Hint: no extraneous roots
Answer: x = 9 or x = −6
(b) Hint: no extraneous roots
Answer: x = 6 or x = −4
(c) Hint: extraneous root: x = 1
Answer: x =
−1 or x =
10
(d) Hint: Two extraneous roots: x =
1
and x = 2
2
1
Answer: x =
− or x =
−2
2
4. Graph (i) corresponds with 3 x + 2 = x + 14
Graph (ii) corresponds with −3 x= 2 x 2 − 2
Graph (iii) corresponds with x 2 − 10 x =10 − x
15
Graph (iv) corresponds with 2 x − 3 =
5. Yes, it is extraneous because the assumption that 4 x − 12 ≥ 0 only applies for x-values
greater than or equal to 3. The root 1 is not greater than or equal to three therefore it is
extraneous.
6. (a) Hint: Two extraneous roots: x = 2 and x = −3
Answer: x =
−11 or x =
3.5
(b) Hint: No extraneous roots
Answer: x =
−4, x =
−2, x =
4, or x =
6
18
6
(c) Hint: Two extraneous roots: x =
and x =
5
5
Answer: x =
−4 or x =
8
(d) Hint: Two extraneous roots: x = 2 and x = 6
Answer: x =
−3 or x =
1
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(e) Hints: For the case in which we assume that x 2 + 3 x < 0 and 2 x − 5 < 0 , there are no real
roots.
−1 + 17
−1 − 17
, and x =
Four extraneous roots: x = −4 , x = −1 , x =
2
2
Answer: x =
−7 or x =
2
Absolute Value Inequalities (pages 52 to 61)
1. (a) x ∈ (1,3)
(b) x ∈ ( −∞, −5 ) ∪ ( −3, ∞ )
(c) x ∈ [ −5, −3]
(d) x ∈ ( −∞,1] ∪ [3, ∞ )
(e) x ∈ φ
(f) x ∈ ( −∞, ∞ )
(g) x ∈ ( −∞, 0 ) ∪ ( 4, ∞ )
(h) x ∈ [1, 4]
2. (a) x ∈ [ −5,1.5]
(c) x ∈ ( −1,9 )
(e) x ∈ ( −∞, −7 ) ∪ ( 20, ∞ )
7

(b) x ∈  −∞, −  ∪ ( −2, ∞ )
3

(d) x ∈ ( −∞, −6] ∪ [9, ∞ )
(f) x ∈ [ −19,5]
3. x ∈ ( −∞, ∞ )
4. Answers will vary. Have your instructor check your answers.
5. (a) x ∈ ( −∞, −4 ) ∪ ( 0, ∞ )
(c) x ∈ ( −∞, −1]
(b) x ∈ ( −∞, −5] ∪ [1, ∞ )
(d) x ∈ ( −∞, −5] ∪ [ 0, ∞ )
(e) x ∈ ( −5,1)
(f) x ∈ ( −4, 0 )
(g) x ∈ [ −5, 0]
(h) x ∈ [ −1, ∞ )
6. Change to x 2 − 7 x − 2 = 2 x − 4
Case 1: If x 2 − 7 x − 2 ≥ 0 , then x 2 − 7 x − 2 = x 2 − 7 x − 2
x2 − 7 x − 2 = 2 x − 4
0
x2 − 9x + 2 =
9 + 73
9 − 73
or x
=
2
2
or x = 0.228
x � 8.772
(works)
(extraneous)
x
=
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Case 2: If x 2 − 7 x − 2 < 0 , then x 2 − 7 x − 2 =
− ( x2 − 7 x − 2)
− ( x2 − 7 x − 2) = 2 x − 4
− x2 + 7 x + 2 = 2x − 4
x2 − 5x − 6 =
0
0
( x − 6 )( x + 1) =
x= 6
(works)
Test x = 5
Test x = 7
Test x = 9
or x = −1
(extraneous)
{12 is not less than 6}
{2 is less than 10}
{16 is not less than 14}
False
True
6
 9 + 73 
Final Answer: x ∈ 6,

2 

False
8.772
7. Hints: Cylinder: V = π r 2 h
2
1000 = π ( 6 ) h
1% of 1000 is 10
36π h − 1000 ≤ 10
Answer: The height of the acid must be equal to or between 8.8 and 8.9 cm.
NSSAL
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78
Draft
C. D. Pilmer