contentsfaculty.essex.edu/~bannon/documents/mth.119.bannon.f.2015.pdf · pre-calculus i mth-119...

Pre-calculus IMTH-119

Essex County CollegeDivision of Mathematics

Ron Bannon’sClass Notes

Contents

0 Introductory Remarks 60.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60.2 This Guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1 Preliminaries 7

2 Polynomials 8

3 Exponents 9

4 Rational and Radical Expressions 10

5 Linear and Nonlinear Equations 11

6 Linear and Nonlinear Inequalities 126.1 Linear Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

6.1.1 Example Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126.2 Polynomial Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

6.2.1 Example Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146.3 Rational Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

6.3.1 Example Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176.4 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196.5 Homework: MTH 119 HW Sec 6 . . . . . . . . . . . . . . . . . . . . . . . . . 19

7 Absolute Value Inequalities 237.1 Absolute Value of a Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237.2 Solving Absolute Value Inequalities . . . . . . . . . . . . . . . . . . . . . . . 23

7.2.1 Example Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237.3 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257.4 Homework: MTH 119 HW Sec 7 . . . . . . . . . . . . . . . . . . . . . . . . . 25

8 Analytic Geometry 288.1 Some Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288.2 Example Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298.3 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318.4 Homework: MTH 119 HW Sec 8 . . . . . . . . . . . . . . . . . . . . . . . . . 31

9 Functions 339.1 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339.2 Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349.3 Graphing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359.4 Example Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369.5 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

Typeset in LATEX 2ε by [email protected]. Last updated on September 2, 2015.

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9.6 Homework: MTH 119 HW Sec 9 . . . . . . . . . . . . . . . . . . . . . . . . . 45

10 Linear Functions 55

11 Transformations and Graphs 5611.1 Shifting, Reflecting, and Stretching Functions . . . . . . . . . . . . . . . . . . 56

11.1.1 Vertical and Horizontal Shifts . . . . . . . . . . . . . . . . . . . . . . . 5611.1.2 Reflections in the Coordinate Axes . . . . . . . . . . . . . . . . . . . . 5611.1.3 Non-Rigid Transformations . . . . . . . . . . . . . . . . . . . . . . . . 57

11.2 Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6011.3 Parent Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6011.4 Example Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6311.5 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6711.6 Homework: MTH 119 HW Sec 11 . . . . . . . . . . . . . . . . . . . . . . . . . 67

12 Quadratic Functions 7012.1 Polynomial Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

12.1.1 Parabolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7012.2 Example Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7112.3 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7312.4 Homework: MTH 119 HW Sec 12 . . . . . . . . . . . . . . . . . . . . . . . . . 74

13 Algebra of Functions; Inverse Functions 7613.1 Combinations and Compositions of Functions . . . . . . . . . . . . . . . . . 7613.2 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7613.3 Example Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7813.4 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8513.5 Homework: MTH 119 HW Sec 13 . . . . . . . . . . . . . . . . . . . . . . . . . 85

14 Polynomial Functions 9014.1 Overview of Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

14.1.1 Rational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9014.1.2 Irrational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9014.1.3 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9014.1.4 Operations on Complex Numbers . . . . . . . . . . . . . . . . . . . . 91

14.2 Remainder Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9214.3 Fundamental Theorem of Algebra . . . . . . . . . . . . . . . . . . . . . . . . 9214.4 Intermediate Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9214.5 Rational Root Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9214.6 An Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9314.7 Example Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9514.8 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10714.9 Homework: MTH 119 HW Sec 14 . . . . . . . . . . . . . . . . . . . . . . . . . 107

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15 Rational Functions 11315.1 Definition of a Rational Function . . . . . . . . . . . . . . . . . . . . . . . . . 11315.2 Vertical Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11315.3 Horizontal Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11315.4 Oblique Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11315.5 An Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

15.5.1 First View . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11415.5.2 Second View . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

15.6 Example Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11815.7 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12715.8 Homework: MTH 119 HW Sec 15 . . . . . . . . . . . . . . . . . . . . . . . . . 130

16 Algebraic Functions; Variation 141

17 Exponential Functions and Their Graphs 14217.1 Exponentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142

17.1.1 Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14217.1.2 Properties of Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . 14217.1.3 Euler’s Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14317.1.4 Discrete Compound Interest . . . . . . . . . . . . . . . . . . . . . . . 14317.1.5 Continuous Compound Interest . . . . . . . . . . . . . . . . . . . . . 14417.1.6 Application Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

17.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14417.3 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14817.4 Homework: MTH 119 HW Sec 17 . . . . . . . . . . . . . . . . . . . . . . . . . 148

18 Logarithmic Functions 15418.0.1 Inverses of Exponentials . . . . . . . . . . . . . . . . . . . . . . . . . . 15418.0.2 Special Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15418.0.3 Graphing Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . 156

18.1 Properties of Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15718.1.1 Basic Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15718.1.2 Changing Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158


19 Exponential and Logarithmic Equations 17119.1 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

19.1.1 Exact and Approximate . . . . . . . . . . . . . . . . . . . . . . . . . . 17119.1.2 Solving Exponential Equations . . . . . . . . . . . . . . . . . . . . . . 17119.1.3 Solving Logarithmic Equations . . . . . . . . . . . . . . . . . . . . . . 171

19.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17219.3 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177

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19.4 Homework: MTH 119 HW Sec 19 . . . . . . . . . . . . . . . . . . . . . . . . . 177

30 Systems of Linear Equations 18230.1 Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18230.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18330.3 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18430.4 Homework: MTH 119 HW Sec 30 . . . . . . . . . . . . . . . . . . . . . . . . . 185

31 Gaussian and Gauss-Jordan Elimination 18731.0.1 Introduction to Linear Algebra: Linear 2× 2 System . . . . . . . . . . 187

31.1 Linear 3× 3 System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18831.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18931.3 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19231.4 Homework: MTH 119 HW Sec 31 . . . . . . . . . . . . . . . . . . . . . . . . . 193

32 Partial Fraction Decomposition 19532.0.1 The Pre-Calculus of Partial Fraction Decomposition . . . . . . . . . . 19532.0.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19632.0.3 Greek To Me! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198

32.1 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19832.2 Homework: MTH 119 HW Sec 32 . . . . . . . . . . . . . . . . . . . . . . . . . 198

33 Nonlinear Systems of Equations 20233.1 Introduction to Non-linear Systems . . . . . . . . . . . . . . . . . . . . . . . . 20233.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20233.3 Supplemental Readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20533.4 Homework: MTH119 HW Sec 33 . . . . . . . . . . . . . . . . . . . . . . . . . 205

34 Introduction to Matrix Alegebra 21034.1 Equal Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21034.2 Matrix Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210

34.2.1 Matrix Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21034.2.2 Scalar Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21034.2.3 Simple Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211


35 Matrix Multiplication and Inverses 21435.0.1 Something Old . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21435.0.2 Something New . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21435.0.3 Inverse of a Square Matrix . . . . . . . . . . . . . . . . . . . . . . . . . 217


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36 Determinants and Cramer’s Rule 22536.1 Determinants of a Square Matrix . . . . . . . . . . . . . . . . . . . . . . . . . 22536.2 Application of Matrices and Determinants . . . . . . . . . . . . . . . . . . . . 225

36.2.1 Invertibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22536.2.2 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22536.2.3 Cramer’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226


37 Final Exam Review 233

38 Technology 254

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0 Introductory Remarks

Errors in this document should be reported to Ron Bannon.

[email protected]

This document may be shared with MTH 119 students and instructors only.

0.1 Review

It is absolutely imperative that you have a fairly good grasp of the mathematics coveredin prior courses including College Algebra (MTH 100). You may visit

http://mth100.mathography.org/

to see what materials are covered in MTH 100. Although I may spend some time in classdiscussing material covered in prior mathematics courses, you can not expect that thispre-requisite materials will be covered sufficiently well enough to remediate inadequatepreparation. Being prepared is the first step!

0.2 This Guide

This guide is being provided as a way to structure MTH 119 across sections and semesters.This guide is not a textbook and it should not be used as such—it just provides structurefor teachers and students.

1. Please come to all classes, these notes are not to be used as an excuse not to attend.You should make every effort to attend all scheduled classes. The notes that followare not self-contained and you will need to attend class in order to make sense outof the content that follows. The notes just give structure to the classroom lectures,and will keep you on track with your assignments.

2. These notes will not be read as a script, but teachers/students are encouraged tofollow the content of these notes. Your teacher may or may not do all the problemsthat are in the notes. You should make every attempt to follow what your teacher isdoing in each and every class.

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http://mth100.mathography.org/




1 Preliminaries

There is no need to do this section of the textbook because it was extensively covered inmore basic courses. However, students may decide on their own to review Chapter 1 ofSchaum’s Precalculus.

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2 Polynomials


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3 Exponents


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4 Rational and Radical Expressions


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5 Linear and Nonlinear Equations


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6 Linear and Nonlinear Inequalities

6.1 Linear Inequalities

The following properties were covered in MTH 100 and are presented here as a review.Let a, b, c, and d be real numbers.

1. Transitive Property.

a < b and b < c ⇒ a < c

2. Addition of Inequalities.

a < b and c < d ⇒ a + c < b + d

3. Addition of Constants.

a < b ⇒ a + c < b + c

4. Multiplying by Constants.

(a) If c > 0.

a < b ⇒ ac < bc

(b) If c < 0. This is where most students mess-up. Think!

a < b ⇒ ac > bc

6.1.1 Example Questions

The examples that follow are similar to problems covered in MTH 100.

1. Solve for the variable, graph, and express the solution set using interval notation.

6x ≤ 19

Solution: Work will be done in class. Make sure you take notes, or get notes from areliable classmate if you are not present during lecture.

x ≤ 196(

−∞, 196

]

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2x + 7 > 3 (1− 2x)


x > −12(

−12 , ∞

)


3 +27

x ≤ 2x− 5


x ≥ 143[

143 , ∞

)


0 < 2− 3 (x + 1) ≤ 20


−7 ≤ x < −13[

−7, −13

)

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6.2 Polynomial Inequalities

Typically we will only be dealing with easily analyzable polynomials, and you will needto understand that linear factors change signs at their zeros. For example, the factor(x− 3) changes sign at x = 3; that is, to the right of three the factor is always positive, andto the left of three the factor is always negative. This, in fact, is the central idea of linearfactor analysis. What is key here, is that our inequalities will now be easy to analyze if wecan factor it, and we need to invariably have them compared to zero for this factor analysisto work.

Here’s the method for solving polynomial inequalities.

• First and foremost you will need to have one side of your inequality set equal tozero.

• Factor the polynomial completely—this may require the use of the quadratic for-mula. Irreducible factors1 may be left un-factored.

• Find all zeros of the polynomial, and arrange the zeros in increasing order on anumber line. Draw a small circle at each zero directly on the number line.

• Find a test value on each of the intervals to test in the factored form of the polyno-mial. It will either be positive or negative, and you’ll need to indicate this directlyon the number line using either ± sign.

• Shade the region(s) that satisfy the inequality/equality.

• Write the final answer in interval notation.


1. Solve and express the solution set using interval notation.

x2 − x− 6 ≤ 0


[−2, 3]

1For example, x2 + 1 is irreducible, and is always positive.

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2x2 + 5x− 12 > 0


(−∞, −4) ∪(3

2 , ∞)


2x3 − 3x2 − 32x + 48 ≥ 0

Solution: Work will be done in class. Make sure you take notes, or get notes from areliable classmate if you are not present during lecture.[

−4, 32

]∪ [4, ∞)


(x− 3)2 ≤ 1


[2, 4]


x2 − 6x + 9 > 16

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(−∞, −1) ∪ (7, ∞)


x4 (x− 2) ≥ 0


[0, 0] ∪ [2, ∞)

0 ∪ [2, ∞)

6.3 Rational Inequalities

You need to have a good understanding of polynomial inequalities before attemptingrational inequalities. Typically we will only be dealing with easily factorable ratios ofpolynomials, and once again you will need to understand that linear factors change signsat their zeros. Central to solving rational inequalities, is that our inequalities will nowbe completely factorable ratio of polynomials, and we need to invariably have them com-pared to zero for this factor analysis to work. Big idea here is that we will only have onerational expression, completely factored, and compared to zero.

Here’s the method for solving rational inequalities.

• First and foremost you will need to have one side of your inequality set equal tozero.

• If you have more than one rational expression, you will need to combine them intoone rational expression.

• Factor the polynomials, both numerator and denominator, completely.

• Find all zeros of the numerator and denominator, and arrange the zeros in increas-ing order on a number line. Draw a small circle at each zero directly on the numberline.

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• Find a test value on each of the intervals to test in the factored form of the rationalexpression. It will either be positive or negative, and you’ll need to indicate thisdirectly on the number line using either ± sign.

• Shade the region(s) that satisfy the inequality/equality. Be very careful that youdon’t divide by zero.

• Write the final answer in interval notation.



(x− 1) (x + 2)(x− 2) (x + 1)

≤ 0


[−2, −1) ∪ [1, 2)


(2x− 1) (3x + 5)(x− 1) (x + 1)

≥ 0


(−∞, −5

3

]∪(−1, 1

2

]∪ (1, ∞)


1x≥ 1

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(0, 1]


1x≤ x


[−1, 0) ∪ [1, ∞)


1x− 2

≥ 1x + 3


(−∞, −3) ∪ (2, ∞)


1x≤ 4

x3


(−∞, −2] ∪ (0, 2]

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x + 4x− 2

≤ −2


[0, 2)

6.4 Supplemental Readings

The material discussed in class covers parts of Chapter 6 of Schaum’s Precalculus. If,for whatever reason, you missed class or just need additional time with this material,I suggest that you read over Chapter 6 of Schaum’s Precalculus, and work the SolvedProblems before attempting the homework assignment.

6.5 Homework: MTH 119 HW Sec 6

Again, you may need to review Chapter 6 of Schaum’s Precalculus if you are having trou-ble following what we are doing in class, or you have missed the lecture material. It isyour responsibility to master this material, and you will need to solve all problems in thissection. Get help if you can not do these problems.

Directions: Solve each inequality algebraically and express your solution set twoways: as a graph on a number line2 and in interval notation. Partial answers are provided,but no work is being provided as a guide. Yes, you need to be able to do these problemsand you will be tested. You need to get help if you cannot do these problems exactly aswas done in class!

1. −5 < 3x− 2 < 1

Solution: (−1, 1)

2.x + 12x + 2

< 3

2Please go to http://www.wolframalpha.com/ and graph these inequalities. A video is available at:http://screencast.com/t/vrcXRWtt. These URLs are ‘clickable’!

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http://www.wolframalpha.com/

http://screencast.com/t/vrcXRWtt




Solution: (−∞, −2) ∪ (3, ∞)

3.x3≥ 2 +

x6

Solution: [12, ∞)

4. x2 − 1 ≥ 0

Solution: (−∞, −1] ∪ [1, ∞)

5. 4 ≤ 2x + 2 ≤ 10

Solution: [1, 4]

6.1x≤ 4

Solution: (−∞, 0) ∪[

14 , ∞

)7. 1− 2x ≤ 3

Solution: [−1, ∞)

8. x (x− 7) > 8

Solution: (−∞, −1) ∪ (8, ∞)

9. −8 ≤ 1− 3 (x− 2) ≤ 13

Solution: [−2, 5]

10.1x− x ≥ 0

Solution: (−∞, −1] ∪ (0, 1]

11.x2≤ 1− x

4

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Solution: (−∞, 4/3]

12. x2 + x > 12

Solution: (−∞, −4) ∪ (3, ∞)

13.3x− 5x + 2

< 2

Solution: (−2, 9)

14. x2 + 7x ≤ −12

Solution: [−4, −3]

15.x + 3x− 2

< 0

Solution: (−3, 2)

16.12≤ x + 1

3<

34

Solution: [1/2, 5/4)

17. x2 + 4x + 4 ≥ 9

Solution: (−∞, −5] ∪ [1, ∞)

18.x− 5x− 2

≥ 0

Solution: (−∞, 2) ∪ [5, ∞)

19.x + 6x + 1

≥ 2

Solution: (−1, 4]

20. 3x2 − 11x− 4 ≤ 0

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Solution:[−1

3 , 4]

21. x3 < x2

Solution: (−∞, 0) ∪ (0, 1)

22. 3.6 + 1.5x > 2.3x− 5.2

Solution: (−∞, 11)

23.x2 − 9x− 3

≥ 0

Solution: [−3, 3) ∪ (3, ∞)

24. x2 + 8x ≤ 0

Solution: [−8, 0]

25. 4− 3 (1− x) < 3

Solution: (−∞, 2/3)

26. 2x2 < 5x + 3

Solution: (−1/2, 3)

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7 Absolute Value Inequalities

7.1 Absolute Value of a Number

The absolute value of a real number a, written |a|, is defined as follows:

|a| =

a if a ≥ 0−a if a < 0

7.2 Solving Absolute Value Inequalities

Let x be a variable of an algebraic expression and let a be a real number such that a ≥ 0.

1. The solution of |x| < a are the values of x that lie between −a and a.

|x| < a if and only if − a < x < a.

2. The solution of |x| > a are the values of x that are less than −a or greater than a.

|x| > a if and only if x < −a or x > a.

We will also discuss, by example, the similar case when we have ≤ or ≥.



|x| ≤ 7


[−7, 7]


|x| > 7

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(−∞, −7) ∪ (7, ∞)


|x− 10| ≤ 7


[3, 17]


|x− 10| > 7


(−∞, 3) ∪ (17, ∞)


|3− 2x| ≤ 4

Solution: Work will be done in class. Make sure you take notes, or get notes from areliable classmate if you are not present during lecture.[

−12 , 7

2

]

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|3− 2x| − 5 > −1

Solution: Work will be done in class. Make sure you take notes, or get notes from areliable classmate if you are not present during lecture.(

−∞, −12

)∪(7

2 , ∞)


5 |2x− 3|+ 3 < 23


−12 , 7

2

)





Directions: Solve each inequality algebraically and express your solution set twoways: as a graph on a number line and and in interval notation. Partial answers are pro-vided, but no work is being provided as a guide. Yes, you need to be able to do theseproblems and you will be tested. You need to get help if you cannot do these problemsexactly as was done in class!

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1. |5x− 1| < 9


2. 9 + |5x| < 24

Solution: (−3, 3)

3. |6 + 4x| ≥ 10

Solution: (−∞, −4] ∪ [1, ∞)

4. |10− x| ≥ 1

Solution: (−∞, 9] ∪ [11, ∞)

5. 3 |4− 5x| > 12

Solution: (−∞, 0) ∪(

85

, ∞)

6. |2x + 11| − 3 ≤ 4

Solution: [−9, −2]

7.∣∣∣∣3x +

25

∣∣∣∣ ≥ 4

Solution: (−∞, −22/15] ∪ [6/5, ∞)

8.∣∣∣∣4x− 3

5

∣∣∣∣ < 1


9.∣∣∣x2− 3∣∣∣ > 7

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Solution: (−∞, −8) ∪ (20, ∞)

10. |1 + 3x|+ 4 < 5


11.∣∣∣∣x− 3

2

∣∣∣∣ ≤ 5

Solution: [−7, 13]

12. |3x| − 8 > 1

Solution: (−∞, −3) ∪ (3, ∞)

13. 8−∣∣∣x3+ 6∣∣∣ < 5

Solution: (−∞, −27) ∪ (−9, ∞)

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8 Analytic Geometry

8.1 Some Review

In MTH 100 you were presented with three formulas related to two points in a plane. Itis typical to label these two points P1 or (x1, y1) and P2 or (x2, y2).

Distance: The distance between P1 and P2 is given by:

d =

√(x2 − x1)

2 + (y2 − y1)2

Midpoint: The midpoint between P1 and P2 is given by:(x1 + x2

2,

y1 + y2

2

)Slope: The slope of a line through P1 and P2 is given by:

y2 − y1

x2 − x1

You should already be familiar with the Cartesian coordinate system and plottingpoints, and graphing parabolas, lines, and circles. On some of these graphs you wereexpected to find the key points, among them are the intercepts.

x-intercept: The point at which the graph crosses the x-axis. At the x-intercept the ycoordinate is zero. To find the x-intercept, just set y = 0 and solve for x.

y-intercept: The point at which the graph crosses the y-axis. At the y-intercept the x coor-dinate is zero. To find the y-intercept, just set x = 0 and solve for y.

Other key points/characteristics were discussed in MTH 100 (axis-of-symmetry, ver-tex, center, radius, slope, etc.) and you may find it helpful to review MTH 100 to makesure these concepts/terms are understood.

Although symmetry was discussed in MTH 100, it was not formally presented. Fourmain symmetries are presented here, and you should make every effort to visualize thesesymmetries to understand them.

1. Symmetry with respect to the y-axis. This is often referred to as even symmetry. Ifsubstituting −x for x leads to the same equation we have even symmetry.

2. Symmetry with respect to the x-axis. If substituting −y for y leads to the sameequation we have symmetry with respect to the x-axis.

3. Symmetry with respect to the origin.This is often referred to as odd symmetry. Ifsubstituting−x for x and−y for y leads to the same equation we have odd symmetry.

4. Symmetry with respect to a line y = x. This will be discussed further when we getto inverses.

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8.2 Example Questions

Answer each of the following questions.

1. Find the x-intercept of x2 + y = 3.

Solution: Work will be done in class. Make sure you take notes, or get notes from areliable classmate if you are not present during lecture.(√

3, 0)

;(−√

3, 0)

2. Find the distance between the points(−3√

3, −3)

and(

3√

3, 3)


12

3. Find the y-intercept of x2 + xy + y2 = 9.


(0, 3) ; (0, −3)

4. Find the midpoint of the line segment with the endpoints(

3,√

2)

and(−1, 5

√2)


1, 3√

2)

5. Analyze xy2 = 1 for symmetry

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x-axis symmetry

6. Analyze x3y = 7 for symmetry.


origin symmetry, also called odd symmetry

7. Analyze x2 + y = 3 for symmetry.


y-axis symmetry, also called even

8. Analyze x2 + xy + y2 = 9 for symmetry.


origin symmetry, also called odd symmetry

9. Find the equation of the circle with center (−1, 2) and radius 2


(x + 1)2 + (y− 2)2 = 4

10. Find the equation of the circle that contains the point (−2, −2) and center is (1, −2)


(x− 1)2 + (y + 2)2 = 9

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Directions: Answer each of the following questions. Partial answers are provided,but no work is being provided as a guide. Yes, you need to be able to do these problemsand you will be tested. You need to get help if you cannot do these problems exactly aswas done in class!

1. Find the distance between the points(√

2, −√

3)

and(−√

2,√

3)

Solution:

2√

5

2. Find the center and radius of the circle with equation x2 + y2 − 4x− 12y− 9 = 0

Solution:

(2, 6) ; 7

3. Analyze 2y = x3 for symmetry

Solution: origin symmetry, also called odd symmetry

4. Analyze −xy2 = 4 for symmetry and intercepts. Sketch the graph by plotting pointsand using symmetry.3

3You may want to try solving for y and then choosing simple values for x to evaluate.

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Solution: x-axis symmetry; no intercepts. Your graph (Figure 1, page 32) shouldlook similar to mine.

-15 -10 -5 0 5

-5

-4

-3

-2

-1

1

2

3

4

5

Figure 1: Partial graph of −xy2 = 4.

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9 Functions

9.1 Equations

In algebra we usually use equations to represent relationships. For example, you alreadyknow many such relationships. Here’s a partial list.

Lines: y = mx + b

Parabolas: y = Ax2 + Bx + C

Circles: (x− h)2 + (y− k)2 = r2

If you can4 solve for y in terms of x, you most likely have a function where the elementsin x is the domain and the elements in y is the range. Here we often say that y dependson x or is a function of x.

Notationally, we often write y = f (x), which is read as, “y is a function of x.” We areoften asked to evaluate functions for given values, for example, if f (x) = 3x2 − 2x + 1,then f (1) = 3− 2 + 1 = 2, and f (2 + s) = 3 (2 + s)2 − 2 (2 + s) + 1 = 3s2 + 10s + 9.We’re just “plugging in” and then simplifying—some students have trouble with this,but it’s really not hard once you try! Yes, trying is critical to your success.

Definition of a Function: A function f from a set A to a set B is a relation that assignsto each element x in the set A exactly one element y in the set B. The set A is the domainof the function f , and the set B contains the range. You should note that the set B is notnecessarily the range, it just contains the range.

Characteristics of a function from set A to set B5

• Each element of A must be matched with an element from B.

• An element from A cannot be matched with two different elements of B.

Not all relationships are functional and you’ll need to evaluate if a relationship isfunctional.6 We’ll keep it very simple!

Once again, you may need to be reminded that you should never divide by zero! So ifyou’re asked to find the domain of function

f (x) =5

x− 3,

4This is not entirely true, for example

y (x− 1) = x2 − 1

can easily be solved for y, but y is not a function of x. However, examples presented here don’t go anywherenear this level of understanding.

5We’ll discuss using graphs (vertical line test) and diagrams to determine if we have a function or not.6Vertical Line Test: If a vertical line crosses a graph more than once, the graph is not the graph of a

function.

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you’re basically being asked what values of x are allowed? I hope you can see that x = 3will result in a division by zero—so the domain of this function is all real numbers exceptthree. This is often written: R, x 6= 3, but you should also be able to write the domain ininterval notation.7

Also, the radicands of even roots, when dealing with real numbers, need to be non-negative. For example if f (x) =

√4− x2, then the domain is: [−2, 2]. Being able to solve

an inequality is essential in many domain problems.Don’t get your brain into a bunch, we’ll do this material by example—there’s very little

to remember and some of this was already covered in MTH 100.

9.2 Terminology

• Increasing Function: A function f is increasing on an interval if, for any x1 and x2in the interval, x1 < x2 implies that f (x1) < f (x2).

• Decreasing Function: A function f is decresing on an interval if, for any x1 and x2in the interval, x1 < x2 implies that f (x1) > f (x2).

• Constant Function: A function f is constant on an interval if, for any x1 and x2 inthe interval, that f (x1) = f (x2).

• Relative Minimum of a Function: A function value f (a) is called a relative mini-mum of f if there exists an interval (x1, x2) that contains a such that

f (a) ≤ f (x) , ∀ x ∈ (x1, x2) .

• Relative Maximum of a Function: A function value f (a) is called a relative maxi-mum of f if there exists an interval (x1, x2) that contains a such that

f (a) ≥ f (x) , ∀ x ∈ (x1, x2) .

• A Function with Even Symmetry: A function f (x) is said to have even symmetry(y-axis symmetry) if for each x in f ’s domain, f (x) = f (−x).

• A Function with Odd Symmetry: A function f (x) is said to have odd symmetry(origin symmetry) if for each x in f ’s domain, f (x) = − f (−x).

• Special Points: A function f (x) has a y-intercept if 0 is in its domain, and the y-intercept is (0, f (0)). A function f (x) has an x-intercept if 0 is in its range, and thex-intercept is (a, f (a) = 0). An x-intercept is often referred to as a root or a zero ofthe function.

7(−∞, 3) ∪ (3, ∞)

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• Average Rate of Change: The average rate of change of a function, f (x), with re-spect to x over the interval [a, b] is given by:

f (b)− f (a)b− a

If we instead look at the interval [x, x + h], we get:

f (x + h)− f (x)h

which is referred to as a difference quotient. Here h 6= 0. Both of these formulasare related to the slope of a secant line through f .

9.3 Graphing

Make sure you have access to a graphic calculator. You may need to read your calculator’smanual, especially as it relates to identifying relative extrema and roots/zeros on a graph.It’s important that you can graph simple functions and identify intervals and points asthey relate to the terminology above. For example, supposed you are asked to graph

f (x) = x3 − 3x2 − x + 3

and discuss this graph as it relates to the terminology above. Using your own calculatoryou should be able to get a graph (Figure 2, page 36) similar to mine.8 And, also us-ing your calculator you should be able to identify—with a good deal of accurarcy—therelative minimum and maximum on this graph. Here’s what I’m getting.9

• Increasing: On the interval (−∞, −0.15) ∪ (2.15, ∞)

• Decreasing: On the interval (−0.15, 2.15)

• Constant: Never constant, it is always changing as we move left to right.

• Relative Minimum of a Function: The point is (2.15, −3.08)

• Relative Maximum of a Function: The point is (−0.15, 3.08)

• Even Symmetry: Not even.

• Odd Symmetry: Not odd.

• Special Points: The y-intercept is (0, 3);. The exact x-intercepts are (−1, 0), (1, 0),and (3, 0).

8Yes, I am using a calculator to graph.9When a decimal is given I am indicating that it’s approximate.

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-3 -2 -1 0 1 2 3 4 5 6

-3

-2

-1

1

2

3

Figure 2: f (x) = x3 − 3x2 − x + 3

Again, make sure you can use your calculator—cell phones are not allowed on exams,so please do not ask. Each calculator model varies, but I am almost 100% positive thatall graphic calculators available nowadays will be able to do these problems. And, inthe future you should be able to graph many functions without the aid of a calculator.Graphing takes time and having a good graphic calculator will certainly speed up thelearning process. Let’s get to work!



1. Given the following graph (Figure 3, page 37) is a relationship between a set A ∈[−1, 2] and a set B ∈ R. Determine if the relationship is functional, and specify itsrange?

Solution: This relationship is functional and its range is [2, 6]

2. Graph x2 + y2 − 4x + 6y + 4 = 0 and determine if y is a function of x.


Its graph (Figure 4, page 37) is a circle and it fails the vertical line test, so y is a not afunction of x.

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-1 0 1 2

1

2

3

4

5

6

Figure 3: A relationship from A to B

-3 -2 -1 0 1 2 3 4 5 6

-7

-6

-5

-4

-3

-2

-1

1

2

Figure 4: Graph is x2 + y2 − 4x + 6y + 4 = 0.

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3. Given f (x) = x2 − 3x + 1, find each of the following.

(a) f (−1)

(b) f (s)

(c) f (2x)

(d) f (a + b)

(e)f (a)− f (b)

a− b


(a) f (−1) = 5

(b) f (s) = s2 − 3s + 1

(c) f (2x) = 4x2 − 6x + 1

(d) f (a + b) = a2 + 2ab + b2 − 3a− 3b + 1

(e)f (a)− f (b)

a− b= a + b− 3

4. Given f (x) = 3x− 2x2, find the difference quotient.


f (x + h)− f (x)h

= 3− 4x− 2h

5. Find the domain of f (x) = 9− 2x3.


(−∞, ∞)

6. Find the domain of f (x) =x2 − 1

x2 − x− 6.

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(−∞, −2) ∪ (−2, 3) ∪ (3, ∞)

7. Find the domain of f (x) =√

x2 − 8x + 12.


(−∞, 2] ∪ [6, ∞)

8. Find the domain of f (x) = 3

√x2 + 8x + 12

2x− 3.


(−∞, 3/2) ∪ (3/2, ∞)

9. Write the circumference10 C of a circle as a function of its area11 A.


C = f (A) = 2√

πA

10. Determine if the following functions are even, odd, neither?

(a) f (x) =x2

x3 − 1

(b) f (x) =x3

x4 − 1(c) f (x) = x2 − 9

10C = 2πr11 A = πr2

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(a) Neither

(b) Odd

(c) Even

11. Given f (x) = x3 − 9x2 − 48x + 52 and its graph (Figure 5, page 40), answer each ofthe following questions.

I want to point out that I’ve indicated two points on this function, one point12 is(−2, 104) and the other point13 is (8, −396)—you should be able to use your cal-culator to verify these points. The y-intercept should be obvious, it’s (0, 52). Again,you should start using your graphic calculators to verify these graphs.

-10 0 10 20

-400

Figure 5: f (x) = x3 − 9x2 − 48x + 52

(a) Domain in interval notation?

(b) Range in interval notation?

(c) The interval where f is increasing?

(d) The interval where f is decreasing?

12This is the relative maximum.13This is the relative minimum.

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(e) The y-intercept.

(f) How many x intercepts are there? Can you find their exact values or approximatethem?

(g) Symmetry?

(h) Relative maximum?

(i) Relative minimum?


(a) (−∞, ∞)

(b) (−∞, ∞)

(c) (−∞, −2) ∪ (8, ∞)

(d) (−2, 8)

(e) (0, 52)

(f) Three x intercepts. Approximate x values are: −4.44, 0.94, 12.51. You’ll need touse your calculators to do this!

(g) No symmetry

(h) (−2, 104); you need to learn how to do this on your calculators.

(i) (8, −396); you need to learn how to do this on your calculators.

12. Given f (x) = x2 − 6x + 9 and its graph (Figure 6, page 42), answer each of the fol-lowing questions. You really don’t need a calculator for this one—you graphed theseby hand in MTH 100.







(g) Symmetry?



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-3 0 3 6 9

3

6

9

Figure 6: f (x) = x2 − 6x + 9


(a) (−∞, ∞)

(b) [0, ∞)

(c) (3, ∞)

(d) (−∞, 3)

(e) (0, 9)

(f) One x intercepts. Exact x value is: 3.

(g) Line symmetry, x = 3. This is what you did in MTH 100, it’s called the axis-of-symmetry. However, this function is not even, and it is not odd.

(h) None

(i) (3, 0)

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13. Given14

f (x) =

x2 + 1 x < 1−x2 − 1 x ≥ 1

,

and its graph (Figure 7, page 43), answer each of the following questions. You reallydon’t need a calculator for this one—you graphed both pieces by hand in MTH 100.

-3 -2 -1 0 1 2 3

-9

-6

-3

3

6

9

Figure 7: f (x)







(g) Symmetry?



14This is a piecewise defined function.

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(a) (−∞, ∞)

(b) (−∞, −2] ∪ [1, ∞)

(c) (0, 1)

(d) (−∞, 0) ∪ (1, ∞)

(e) (0, 1)

(f) None

(g) None

(h) None

(i) (0, 1)

14. Is 2x + 3y = 7 a function, that is, is y a function of x?


Yes, y is a function of x?

15. Is y2 − x = 5 a function, that is, is y a function of x?


y is not function of x?

16. Given that A is the set of people in this classroom, and B ∈ [0, 10] measured in feet.Is this relationship functional, and if it is, what is its range?


Yes, it is a functional relationship. The range may be too sensitive to determine, butI think we all have an idea that it would list all heights from the shortest person tothe tallest person present.

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17. Evaluate the function (this is a piecewise defined function) at each specified value ofthe independent variable and simplify.

f (x) =

4x + 5, x ≤ 04x2 + 4, x > 0

(a) f (−2)

(b) f (0)

(c) f (2)

Solution:

(a) f (−2) = −3

(b) f (0) = 5

(c) f (2) = 20

18. Find all values of x such that f (x) = 0.

f (x) =2x− 10

9

Solution: x = 5






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1. Is 5y− 2x3 − 4x2 + 7x = 9 a function, that is, is y a function of x?

Solution: Yes, y is a function of x.

2. Find the domain of the function.

g (x) = 4√

x2 + 7x

Solution: (−∞, −7] ∪ [0, ∞)

3. Consider the function below.

f (x) =1

1 + x

Find the difference quotient below (where h 6= 0) and simplify your answer.

f (x + h)− f (x)h

Solution:

f (x + h)− f (x)h

=−1

(1 + x + h) (1 + x)

4. Find the domain of the function.

f (x) =x + 2√4− x2

Solution: (−2, 2)

5. Given the following graph (Figure 8, page 47) is a relationship between a set C ∈[−2, 4] and a set D ∈ R. Determine if the relationship is functional, and specify itsrange?

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-4 -3 -2 -1 0 1 2 3 4 5

-4

-3

-2

-1

1

2

3

Figure 8: A relationship from C to D

Solution: This relationship is not functional and its range is [−4, 2]

6. Let f (x) = 2x− 3, find each of the following.

(a) f (1)

(b) f (−2)

(c) f (a)

(d) f (2x)

(e) f (x + 1)

(f) f (x + h)

(g) f (x + h)− f (x)

(h) For h 6= 0,f (x + h)− f (x)

h

Solution:

(a) f (1) = −1

(b) f (−2) = −7

(c) f (a) = 2a− 3

(d) f (2x) = 4x− 3

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(e) f (x + 1) = 2x− 1

(f) f (x + h) = 2x + 2h− 3

(g) f (x + h)− f (x) = 2h

(h) For h 6= 0,f (x + h)− f (x)

h= 2

7. Given f (x) = x4 − 5x2 + 4 and its graph (Figure 9, page 48), answer each of the

following questions. Two points are indicated here, and they occur at x = ±√

52

.

Again, your calculator can approximate these points as x ≈ ±1.58

-2 -1 0 1 2

-3

-2

-1

1

2

3

4

5

Figure 9: f (x) = x4 − 5x2 + 4

(a) Is the function even, odd, or neither?

(b) Domain in interval notation?

(c) Range in interval notation?

(d) The interval where f is increasing?

(e) The interval where f is decreasing?

(f) The y-intercept.

(g) How many x intercepts are there? Can you find their exact values or approximatethem?

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Solution:

(a) Even, because f (x) = f (−x)

(b) (−∞, ∞)

(c) [−9/4, ∞), your calculator should give the same answer, but in decimal form.

(d)(−√

5/2, 0)∪(√

5/2, ∞)

(e)(−∞, −

√5/2

)∪(0,√

5/2)

(f) (0, 4)

(g) Four x intercepts. The x-intercepts are: ±1, ±2

(h) (0, 4)

(i)(±√

5/2, −9/4)

8. Given f (x) = x√

x2 − 9 and its graph (Figure 10, page 49), answer each of the fol-lowing questions. This graph may look deceptive if you don’t yet understand domain.

-12 -9 -6 -3 0 3 6 9 12

-150

-100

-50

50

100

150

Figure 10: f (x) = x√

x2 − 9

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(h) Symmetry?

(i) Relative maximum?

(j) Relative minimum?

Solution:

(a) Odd, because f (x) = − f (−x)

(b) (−∞, −3] ∪ [3, ∞)

(c) (−∞, ∞)

(d) (−∞, −3) ∪ (3, ∞)

(e) Nowhere

(f) None

(g) Two x intercepts. The x-intercepts are: ±3

(h) Odd

(i) None

(j) None

9. Given f (x) = x√

4− x and its graph (Figure 11, page 51), answer each of the follow-ing questions. Use your calculator to verify that the maximum occurs are x = 8/3.







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-3 -2 -1 0 1 2 3 4

-3

-2

-1

1

2

3

Figure 11: f (x) = x√

4− x


(h) Symmetry?(i) Relative maximum?(j) Relative minimum?

Solution:

(a) Neither

(b) (−∞, 4]

(c)(−∞, 16

√3/9

](d) (−∞, 8/3)

(e) (8/3, 4)

(f) (0, 0)

(g) Two x intercepts. The x-intercepts are: 0, 4

(h) None

(i)(

8/3, 16√

3/9)

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(j) None

10. Given f (x) = −x |x + 1| and its graph (Figure 12, page 52), answer each of the fol-lowing questions.

-2 -1 0 1 2

-1

1

Figure 12: f (x) = −x |x + 1|

(a) Rewrite the function in terms of pieces.









Solution:

(a)

f (x) =

x2 + x x < −1−x2 − x x ≥ −1

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(b) (−∞, ∞)

(c) (−∞, ∞)

(d) (−1, −1/2)

(e) (−∞, −1) ∪ (−1/2, ∞)

(f) (0, 0)

(g) Two x intercepts. The x-intercepts are: −1, 0

(h) (−1/2, 1/4)

(i) (−1, 0)

11. Given

f (x) =

x2 + 2 x ≥ 01− 3x x < 0

and its graph (Figure 13, page 53), answer each of the following questions.

-1 0 1

1

2

3

4

5

Figure 13: Partial graph of f (x).




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(g) Relative maximum?

(h) Relative minimum?

Solution:

(a) (−∞, ∞)

(b) (1, ∞)

(c) (0, ∞)

(d) (−∞, 0)

(e) (0, 2)

(f) There are no x intercepts.

(g) None.

(h) None.

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10 Linear Functions


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11 Transformations and Graphs

11.1 Shifting, Reflecting, and Stretching Functions

11.1.1 Vertical and Horizontal Shifts

Let c > 0 and f (x) be a function, and h (x) be a function related to f by a simple verticalor horizontal shift.

Vertical shift c units upward: h (x) is simply f (x) shifted c units upward.

h (x) = f (x) + c

Vertical shift c units downward: h (x) is simply f (x) shifted c units downward.

h (x) = f (x)− c

Horizontal shift c units right: h (x) is simply f (x) shifted c units rightward.

h (x) = f (x− c)

Horizontal shift c units left: h (x) is simply f (x) shifted c units leftward.

h (x) = f (x + c)

11.1.2 Reflections in the Coordinate Axes

Reflections in the coordinate axes of the graph of f (x), are as follows:

Reflections in the x-axis: h (x) is simply f (x) rotated about the x-axis.

h (x) = − f (x)

Reflections in the y-axis: h (x) is simply f (x) rotated about the y-axis.

h (x) = f (−x)

The rigid transformation should be somewhat reasonable, but I still think that we need todevelop a technique that depends on reasoning and not memorizing the above. More onthat later!

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11.1.3 Non-Rigid Transformations

When we’re dealing with a constant multiple of f or x we’ll get a stretching or com-pressing behavior—both vertical and horizontal. These are perhaps more difficult to un-derstand, and things can get pretty messy when we start combine rigid and non-rigidtransformations. I will essentially avoid memorizing results and will instead use a simplepoint translation method. However, it is nonetheless important that you are able to seewhat’s happening, even if it takes graphing. Let’s take a simple example to see what anon rigid transformation looks like. The parent (Figure 14, page 57) is f (x), and we willsee what happens when we do the following:

1. 2 · f (x). Give it some thought and I think you’ll see that we’ll get a vertical stretch(Figure 15, page 58) by a factor of 2.

2. 0.5 · f (x). Give it some thought and I think you’ll see that we’ll get a vertical com-pression (Figure 16, page 58) by a factor of 2.

3. f (2 · x). Give it some thought and I think you’ll see that we’ll get a horizontalcompression (Figure 17, page 59) by a factor of 2.

4. f (0.5 · x). Give it some thought and I think you’ll see that we’ll get a horizontalexpansion (Figure 18, page 59) by a factor of 2.

-10 -7.5 -5 -2.5 0 2.5 5 7.5 10

-5

-2.5

2.5

5

Figure 14: f (x)

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-10 -7.5 -5 -2.5 0 2.5 5 7.5 10

-5

-2.5

2.5

5

Figure 15: 2 f (x)

-10 -7.5 -5 -2.5 0 2.5 5 7.5 10

-5

-2.5

2.5

5

Figure 16: 0.5 f (x)

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-10 -7.5 -5 -2.5 0 2.5 5 7.5 10

-5

-2.5

2.5

5

Figure 17: f (2x)

-10 -7.5 -5 -2.5 0 2.5 5 7.5 10

-5

-2.5

2.5

5

Figure 18: f (0.5x)

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11.2 Methods

Now we will be faced with taking simple functions, often called parents, and using theabove descriptions to draw a related, often called the child, graph. It is not easy when weare faced with several transformative factors. So, my plan is to take simple characteristicpoints on a function or graph and then move them, and in the end connect these pointstogether to create a new graph that looks similar enough to the parent. Don’t worry,we’ll keep it very simple, and I will do my best to illustrate the method with examples.However, please feel free to think through the above material as we do problems.

Certainly there are many simple curves to know (often called parents) and you’ll startrecognizing them (often called families) as you do more-and-more problems—not unlikelife itself. However, you will often be given graphs that you will not recognize, but thesegraphs too will have characteristic points. If you are given a graph you should try todetermine several characteristic points on f that will be easy to shift, and once shiftedyou should be able to connect-the-dots in such a way that the graphs have similar shapes.

11.3 Parent Graphs

You should have a rough feel for how simple graphs of some functions appear. Althoughthere’s no definite list, I think everyone should know some very simple parent graphs.Here’s a very short list.

Linear Here’s the graph (Figure 19, page 60) of a very simple linear function.

-5 -4 -3 -2 -1 0 1 2 3 4 5

-2

-1

1

2

Figure 19: f (x) = x

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Quadratic Here’s the graph (Figure 20, page 61) of a very simple quadratic function.

-5 -4 -3 -2 -1 0 1 2 3 4 5

-2

-1

1

2

Figure 20: f (x) = x2

Cubic Here’s the graph (Figure 21, page 61) of a very simple cubic function.

-5 -4 -3 -2 -1 0 1 2 3 4 5

-2

-1

1

2

Figure 21: f (x) = x3

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Square Root Here’s the graph (Figure 22, page 62) of a very simple square root function.

-5 -4 -3 -2 -1 0 1 2 3 4 5

-2

-1

1

2

Figure 22: f (x) =√

x

Cubic Root Here’s the graph (Figure 23, page 62) of a very simple cube root function.

-5 -4 -3 -2 -1 0 1 2 3 4 5

-2

-1

1

2

Figure 23: f (x) = 3√

x

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Simple Wave Here’s the graph (Figure 24, page 63) of a very simple wave (sine) function.

-5 -4 -3 -2 -1 0 1 2 3 4 5

-2

-1

1

2

Figure 24: f (x) = sin x, a standard wave.



1. Explain each transformation of f (x) below.

(a) f (x + 1)

(b) f (x− 1)

(c) f (x) + 1

(d) f (x)− 1

(e) f (x− 1) + 1


(a) Shift f (x) one unit to the left.

(b) Shift f (x) one unit to the right.

(c) Shift f (x) one unit up.

(d) Shift f (x) one unit down.

(e) Shift f (x) one unit to the right and one unit up.

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2. Suppose you are given a function f (x) = x2. Explain each of the following transfor-mations in terms of f (x)’s graph.

(a) f (x + 2)

(b) f (x− 3)

(c) f (x)− 1

(d) f (x) + 4

(e) f (x + 1)− 2


(a) f (x + 2) is simply the graph of f (x) = x2 shifted two units to the left.

(b) f (x− 3) is simply the graph of f (x) = x2 shifted three units to the right.

(c) f (x)− 1 is simply the graph of f (x) = x2 shifted one unit down.

(d) f (x) + 4 is simply the graph of f (x) = x2 shifted four units upward.

(e) f (x + 1)− 2 is simply the graph of f (x) = x2 shifted one unit to the left andtwo units down.

3. Explain each transformation of f (x) below.

(a) f (3x)(b) 2 f (x)(c) f (x/3)

(d) f (x) /3

(e) f (−x)(f) − f (x)


(a) f (3x) is a horizontal compression by the factor of 3.

(b) 2 f (x) is a vertical expansion by the factor of 2.

(c) f (x/3) is a horizontal expansion by the factor of 3.

(d) f (x) /3 s a vertical compression by the factor of 3.

(e) f (−x) is a reflection in the y-axis.

(f) − f (x) is a reflection in the x-axis.

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4. Given the following graph15 (Figure 25, page 65).

-1 0 1 2 3

-1

1

Figure 25: f (x)

Graph each of the following.

(a) h (x) = − f (x)

(b) h (x) = f (−x)

(c) h (x) = − f (x) + 3

(d) h (x) = f (−x)− 2

(e) h (x) = f (2x) + 1

(f) h (x) = f (2x− 1)


There are five characteristic points of interest on the parent graph and you shouldalso list five characteristic points on each of the graphs below.

(a) h (x) = − f (x)The five key points are (−1, 0) ; (0, −1) ; (1, 0) ; (2, 1) ; (3, 0).

(b) h (x) = f (−x)The five key points are (1, 0) ; (0, 1) ; (−1, 0) ; (−2, −1) ; (−3, 0).

(c) h (x) = − f (x) + 3The five key points are (−1, 3) ; (0, 2) ; (1, 3) ; (2, 4) ; (3, 3).

15Key points on f are: (−1, 0) ; (0, 1) ; (1, 0) ; (2, −1) ; (3, 0)

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(d) h (x) = f (−x)− 2The five key points are (1, −2) ; (0, −1) ; (−1, −2) ; (−2, −3) ; (−3, −2).

(e) h (x) = f (2x) + 1The five key points are (−1/2, 1) ; (0, 2) ; (1/2, 1) ; (1, 0) ; (3/2, 1).

(f) h (x) = f (2x− 1)The five key points are (0, 0) ; (1/2, 1) ; (1, 0) ; (3/2, −1) ; (2, 0).

5. Given the following graph (Figure 26, page 66). Answer each of the following.

-5 -4 -3 -2 -1 0 1 2 3 4

-5

-4

-3

-2

-1

1

2

3

Figure 26: f (x)

(a) What are the characteristic points on f (x)?

(b) How are the characteristic points on on f (x) connected?

(c) Graph h (x) = −2 f (0.5x− 1) + 2

(d) Graph h (x) = 3 f (3− x) + 1

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(a) (−3, 1) ; (−1, 0) ; (0, −3) ; (2, −4)

(b) Line segments.

(c) Plot these points (−4, 0) ; (0, 2) ; (2, 8) ; (6, 10) and connect the dots in orderwith line segments.

(d) lot these points (6, 4) ; (4, 1) ; (3, −8) ; (1, −11) and connect the dots in orderwith line segments.




Again, you may need to review Chapter 11 of Schaum’s Precalculus if you are havingtrouble following what we are doing in class, or you have missed the lecture material. Itis your responsibility to master this material, and you will need to solve all problems inthis section. Get help if you can not do these problems.


1. Given the following graph16 (Figure 27, page 68).

Graph each of the following.

(a) h (x) = − f (x)

(b) h (x) = f (−x)

(c) h (x) = − f (x)− 1

(d) h (x) = f (−x) + 2

(e) h (x) = f (0.5x)

(f) h (x) = f (2x)

16Key points on f are: (−2, 2) ; (0, −4) ; (2, 2)

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-3 -2 -1 0 1 2 3

-4

-3

-2

-1

1

2

Figure 27: f (x)

Solution:

There are three characteristic points of interest on the parent graph and you shouldalso list three characteristic points on each of the graphs below.

(a) h (x) = − f (x)The three key points are: (−2, −2) ; (0, 4) ; (2, −2).

(b) h (x) = f (−x)The three key points are: (2, 2) ; (0, −4) ; (−2, 2).

(c) h (x) = − f (x)− 1The three key points are: (−2, −3) ; (0, 3) ; (2, −3).

(d) h (x) = f (−x) + 2The three key points are: (2, 4) ; (0, −2) ; (−2, 4).

(e) h (x) = f (0.5x)The three key points are: (−4, 2) ; (0, −4) ; (4, 2).

(f) h (x) = f (2x)The three key points are: (−1, 2) ; (0, −4) ; (1, 2).

(g) h (x) = − f (x)The three key points are: (−2, −2) ; (0, 4) ; (2, −2).

2. Given

f (x) =

√

4− x2 −2 ≤ x ≤ 02− x 0 < x ≤ 2

−x2 + 6x− 8 x > 2

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and its graph (Figure 28, page 69) of f (x).

-2 -1 0 1 2 3 4

1

2

Figure 28: f (x)

Answer each of the following.

(a) What are the characteristic points on f (x)?

(b) How are the characteristic points on on f (x) connected?

(c) Graph h (x) = −2 f (0.5x− 1) + 2

(d) Graph h (x) = 3 f (3− x) + 1

Solution:

(a) Five key points: (−2, 0) ; (0, 2) ; (2, 0) ; (3, 1) ; (4, 0)

(b) First segment is circular, second segment is linear, third segment is parabolic.

(c) Plot these points (−2, 2) ; (2, −2) ; (6, 2) ; (8, 0) ; (10, 2) and connect the dotsin order using the proper segments (circular, linear, parabolic.).

(d) Plot these points (5, 1) ; (3, 7) ; (1, 1) ; (0, 4) ; (−1, 1) and connect the dots inorder using the proper segments (circular, linear, parabolic.).

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12 Quadratic Functions

12.1 Polynomial Functions

A polynomial function of degree n ∈ Z+ has the form

f (x) = anxn + an−1xn−1 + · · ·+ a2x2 + a1x + a0,

where an 6= 0.This section covers the basics of just one type of polynomial function, the quadratic of

the form f (x) = Ax2 + Bx + C. Keep in mind that higher degree polynomials are moredifficult to graph and will require more time and effort, but we’ll get there. Furthermore, agraphics calculator is still helpful, but not always necessary when we are asked to graph.

12.1.1 Parabolas

-6 -5 -4 -3 -2 -1 0 1 2 3

-4

-3

-2

-1

1

2

Figure 29: Partial graph of f (x) = y = x2 + 2x− 3, with important features indicated.

You should be able to graph (Figure 29, page 70) a simple case (A = 1, B = 2, andC = −3) of the general form of a parabola,

y = Ax2 + Bx + C.

I strongly suggest you start by creating a table with simple points, at least six. I do expectthat you are capable to finding the following key features though:

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• x-intercepts by setting y = 0. They’re not always there, but in general they’re worthlooking for. In our example above we would do the following.

y = x2 + 2x− 3

0 = x2 + 2x− 30 = (x + 3) (x− 1)

So the x-intercepts are: (−3, 0) and (1, 0).

• y-intercept by setting x = 0. That’s just too easy! The y-intercept in the exampleabove is: (0, −3).

• The vertex, which is the point (at least in our examples) that will either be the highest(maximum) or lowest (minimum) point on our graph. This point is dead-center of thex values of the x-intercepts. If you use the quadratic formula and take the averageyou’ll get a nice formula for the vertex.(

− B2A

, f(− B

2A

))So in our example above we have (−1, −4) as the vertex.

• The axis-of-symmetry which is the dashed-line in our graph above. It should be notedthat this line is a folding line of symmetry. The equation of this line is x = −1.

The examples that follow are really not difficult and are representative of what you havealready covered in MTH 100.


1. Find the coordinates of the vertex and the equation of the axis of symmetry for theparabola given by the equation. Then graph the equation.

y = x2 − 4x− 1


Vertex: (2, −5); AOS: x = 2

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2. Find the coordinates of the vertex and the equation of the axis of symmetry for theparabola given by the equation. Then graph the equation.17

y = − (x + 1)2 − 5


Vertex: (−1, −5); AOS: x = −1

3. Find two quadratic functions, one that opens upward and one that opens downward,whose graphs have the given x-intercepts.

(−2, 0) , (3, 0)


There’s an infinite number of answers of the form y = k (x + 2) (x− 3), k ∈ R, k 6= 0,but here’s two simple ones:

y1 = (x + 2) (x− 3) = x2 − x− 6y2 = − (x + 2) (x− 3) = 6 + x− x2

4. Find the x and y-intercepts of the graph of the parabola given by the equation.

y = x2 − 5x + 6


x-intercepts: (2, 0), (3, 0); y-intercept: (0, 6)

17This form is referred to as standard f (x) = A (Bx + C)2 + D and can easily be seen to relate to the par-ent f (x) = x2. Yes, we can use what we learned in prior sections, particularly transformation/translation.

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5. Find the values of b (there’s two) such that the function has the given maximum valueof 52. Graph both cases to verify that the maximum does indeed occur where weexpect it to.

f (x) = −x2 + bx− 12


This parabola opens downward and its maximum occurs at the vertex (b/2, 52).You need to solve the following equation to find b.

f(

b2

)= −b2

4+

b2

2− 12 = 52

A little algebra will give you b = ±16.

6. Find the zeros (x coordinate of the x-intercepts) of f .

f (x) = 12 + 7x + x2


The zeros are: −4 and −3.

7. Find the two positive real numbers whose sum is 86 and whose product is a maxi-mum.


Let x > 0 and 86− x > 0 be the two positive numbers. We’re trying to make theirproduct a big as possible. That is we’re looking for the vertex of f (x) = x (86− x).

The two numbers are: 43 and 43.



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1. Find the coordinates of the vertex and the equation of the axis of symmetry for theparabola given by the equation. Then graph the equation.

y = 3x2 + 12x− 7

Solution: Vertex: (−2, −19); AOS: x = −2

2. Find the coordinates of the vertex and the equation of the axis of symmetry for theparabola given by the equation. Then graph the equation.18

y = −3 (x− 2)2 + 1

Solution: Vertex: (2, 1); AOS: x = 2

3. Find two quadratic functions, one that opens upward and one that opens downward,whose graphs have the given x-intercepts.

(−1/2, 0) , (2/3, 0)

Solution:

y1 = (2x + 1) (3x− 2) = 6x2 − x− 2y2 = − (2x + 1) (3x− 2) = 2 + x− 6x2

18This form is referred to as standard f (x) = A (Bx + C)2 + D and can easily be seen to relate to the par-ent f (x) = x2. Yes, we can use what we learned in prior sections, particularly transformation/translation.

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4. Find the x and y-intercepts of the graph of the parabola given by the equation.

y = x2 − 8x + 15

Solution:

x-intercepts: (3, 0), (5, 0); y-intercept: (0, 15)

5. A projectile is thrown up from the ground with an initial velocity of 144 feet persecond. Its height h (t) at time t is given by h (t) = 144t − 16t2. Find its maximumheight. At what time does the projectile hit the ground.

Solution:

The maximum height is 324 feet. It hits the ground at 9 seconds.

6. State the domain and range.

f (x) = 3 (x− 2)2 + 5

Solution:

Domain: (−∞, ∞); range: [5, ∞)

7. Find the two real numbers whose difference is 100 and whose product is a minimum.

Solution:

The two numbers are: 50 and −50.

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13 Algebra of Functions; Inverse Functions

13.1 Combinations and Compositions of Functions

Sums, Differences, Products, and Quotients of Functions: Let f and g be two functionswith overlapping domains. Then, for all x common to both domains, the sum, difference,product, and quotient of f and g are defined as follows.

Sum: The domain is the intersection of the domains f and g.

( f + g) (x) = f (x) + g (x)

Difference: The domain is the intersection of the domains f and g.

( f − g) (x) = f (x)− g (x)

Product: The domain is the intersection of the domains f and g.

( f · g) (x) = f (x) · g (x)

Quotient: The domain is the intersection of the domains f and g, but we must excludethe x’s in g’s domain where g (x) = 0.(

fg

)(x) =

f (x)g (x)

, g (x) 6= 0

The composition of functions is a bit more difficult, and here you’re going to have towork over some pretty vexing concepts. Don’t despair, you’ll get it if you try!

The Composition: of the function f with the function g is

( f g) (x) = f (g (x))

The domain of f g is the set of all x in the domain of g such that g (x), i.e. the range ofg, is in the domain of f . That sounds scary, but we’ll see in the examples to follow that itjust requires a little patience.

13.2 Inverse Functions

Definition: A function f is called one-to-one if it never takes on the same value twice;that is

f (x1) 6= f (x2)

whenever x1 6= x2. We can often visually verified that a function is one-to-one by usingthe horizontal line test, which states, “A function is one-to-one if an only if no horizontalline intersects its graph more than once.”

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For example, the following function (Figure 30, page 77), is not one-to-one.

-2 -1 0 1 2

1

2

3

Figure 30: f (x) = x2√

x3 + 5

However, this function (Figure 31, page 77) is one-to-one.

-2 -1 0 1 2

1

2

3

Figure 31: f (x) =√

x3 + 1

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Some however may be mislead by the horizontal line test, because it is hard to detectfrom the visual alone that f (x) =

√x3 + 1 is one-to-one. It should be clear though that

f (x) = x2√

x3 + 5 is not one-to-one from the graph alone. To show that f (x) =√

x3 + 1is one-to-one we proceed as follows. Since for all x1, x2 ≥ −1, and x1 6= x2, we have

x31 6= x3

2, which further implies that x31 + 1 6= x3

2 + 1, and finally that√

x31 + 1 6=

√x3

2 + 1.Definition: Let f be a one-to-one function with domain A and range B. Then its

inverse function f−1 has a domain B and range A and is defined by

f−1 (y) = x ⇔ f (x) = y.

To find the inverse of a one-to-one function, follow these steps.

1. In the equation for f (x), replace f (x) by y.

2. Interchange the roles of x and y, and solve for y.

3. Replace y by f−1 (x) in the new equation.

4. Although not necessary, you may want to verify that

f(

f−1 (x))= f−1 ( f (x)) = x.


1. Given f (x) = 2x− 3 and g (x) = 3x2 + 2x− 5, find each of the following.

(a) The domain of f (x).

(b) The domain of g (x).

(c) f (x) + g (x) and its domain.

(d) f (x)− g (x) and its domain.

(e) f (x) · g (x) and its domain.

(f) f (x) /g (x) and its domain.

(g) g (x) / f (x) and its domain.


(a) (−∞, ∞)

(b) (−∞, ∞).

(c) f (x) + g (x) = 3x2 + 4x− 8; (−∞, ∞)

(d) f (x)− g (x) = 2− 3x2; (−∞, ∞)

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(e) f (x) · g (x) = (2x− 3)(3x2 + 2x− 5

)= 6x3 − 5x2 − 16x + 15; (−∞, ∞)

(f) f (x) /g (x) =2x− 3

3x2 + 2x− 5; (−∞, −5/3) ∪ (−5/3, 1) ∪ (1, ∞)

(g) g (x) / f (x) =3x2 + 2x− 5

2x− 3; (−∞, 3/2) ∪ (3/2, ∞)

2. Given f (x) =√

4− x2 and g (x) =√

1− x, find each of the following.



(c) f (x) + g (x) and its domain.

(d) f (x)− g (x) and its domain.

(e) f (x) · g (x) and its domain.

(f) f (x) /g (x) and its domain.

(g) g (x) / f (x) and its domain.


(a) [−2, 2]

(b) (−∞, 1]

(c) f (x) + g (x) =√

4− x2 +√

1− x; [−2, 1]

(d) f (x)− g (x) =√

4− x2 −√

1− x; [−2, 1]

(e) f (x) · g (x) =√

4− x2 ·√

1− x; [−2, 1]

(f) f (x) /g (x) =

√4− x2√

1− x; [−2, 1)

(g) g (x) / f (x) =√

1− x√4− x2

; (−2, 1]

3. Given f (x) = x2 − 1 and g (x) =√

1− x2, find each of the following.



(c) The range of f (x).

(d) The range of g (x).

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(e) ( f g) (x) and its domain.

(f) (g f ) (x) and its domain.


(a) (−∞, ∞)

(b) [−1, 1]

(c) [−1, ∞)

(d) [0, 1]

(e) ( f g) (x) = −x2; [−1, 1]

(f) (g f ) (x) =√

2x2 − x4 = |x|√

2− x2;[−√

2,√

2]

4. Given f (x) =3

x2 − 1and g (x) = x + 1, find each of the following. A graph (Figure

32, page 80) of f is given to help determine f ’s range.

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

Figure 32: A partial graph of f .



(c) The range of f (x).

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(d) The range of g (x).(e) ( f g) (x) and its domain.(f) (g f ) (x) and its domain.


(a) (−∞, −1) ∪ (−1, 1) ∪ (1, ∞)

(b) (−∞, ∞)

(c) (−∞, −3] ∪ (0, ∞)

(d) (−∞, ∞)

(e) ( f g) (x) =3

x2 + 2x; (−∞, −2) ∪ (−2, 0) ∪ (0, ∞)

(f) (g f ) (x) =x2 + 2x2 − 1

; (−∞, −1) ∪ (−1, 1) ∪ (1, ∞)

5. Verify that f (x) = 2x− 1 is a one-to-one function.


Doing a graph here is sufficient to show that the function is one-to-one, that is, itpasses the horizontal line test.

6. Given f (x) = 2x− 1, find its domain, range, and inverse.

Solution: This work will also be discussed in class. Make sure you take notes, orget notes from a reliable classmate if you are not present during lecture.

First off we know f is one-to-one with domain x ∈ R and range f (x) ∈ R, so itsinverse, denoted f−1 (x), has a domain x ∈ R and range f−1 (x) ∈ R. Using simplealgebra (method described in this guide) to find f−1 (x).

f (x) = 2x− 1y = 2x− 1 write y = f (x)x = 2y− 1 interchange x and y

x + 12

= y solve for y

x + 12

= f−1 (x)

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Here’s a graph (Figure 33, page 82) that will be discussed in class.

-5 -4 -3 -2 -1 0 1 2 3 4 5

-4

-3

-2

-1

1

2

3

4

Figure 33: f (x), and f−1 (x) in blue.

You should observe the symmetry along the line y = x.

7. Here’s a more difficult example, however it is still algebraically manageable. Usingf (x) =

√x3 + 1, let’s find its inverse.


First off we know f is one-to-one with domain x ≥ −1 and range f (x) ≥ 0, so itsinverse, denoted f−1 (x), has a domain x ≥ 0 and range f−1 (x) ≥ −1. Using simplealgebra to find f−1 (x).

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f (x) =√

x3 + 1

y =√

x3 + 1 write y = f (x)

x =√

y3 + 1 interchange x and y

x2 = y3 + 1 solve for yx2 − 1 = y3

3√

x2 − 1 = y3√

x2 − 1 = f−1 (x) write y = f−1 (x)

Here’s a graph (Figure 34, page 83) that will be discussed in class.

-4 -3 -2 -1 0 1 2 3 4 5

-2

-1

1

2

3

4

5


You should observe the symmetry along the line y = x.

8. An even more difficult example, given,

f (x) =1−√

x1 +√

x.

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find f−1 and its domain and range. A graph (Figure 35, page 84) is provided as avisual aid.

-1 0 1 2 3

1

2

3



First off we know f is one-to-one with domain x ≥ 0 and range f (x) ∈ (−1, 1], soits inverse, denoted f−1 (x), has a domain x ∈ (−1, 1] and range f−1 (x) ≥ 0. Usingsimple algebra to find f−1 (x).

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f (x) =1−√

x1 +√

x

y =1−√

x1 +√

x

x =1−√y1 +√

yx + x

√y = 1−√y

√y + x

√y = 1− x

√y (1 + x) = 1− x

√y =

1− x1 + x

y =

(1− x1 + x

)2

We finally have,

f−1 (x) =(

1− x1 + x

)2

,

and its domain is (−1, 1] and its range is [0, ∞).






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1. Consider the following functions.

f (x) = 1− x2, x ≥ 0

g (x) =√

1− x, x ≤ 1

Show that f and g are inverse functions algebraically.

Solution: They’re inverses. You might want to show that f−1 (x) yields g (x). Graph-ing may help in determining the range and domain.

2. Given f (x) =2x− 3

4and g (x) =

2x + 13

, find the following.

(a) ( f g) (x) and its domain.

(b) (g f ) (x) and its domain.

Solution:

(a)

( f g) (x) =4x− 7

12, x ∈ (−∞, ∞)

(b)

(g f ) (x) =2x− 1

6, x ∈ (−∞, ∞)

3. Determine graphically whether the function is one-to-one.

f (x) =√

x− 4

Solution: This function is one-to-one. The graph clearly passes the horizontal linetest.

4. Find the inverse function of f algebraically.

f (x) =√

9− x2, x ∈ [−3, 0]

Use a graphing utility to graph both f and f−1 in the same viewing window. Describethe relationship between the graphs.

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Solution:

f−1 (x) = −√

9− x2, x ∈ [0, 3]

When you graph both f and its inverse you should note that the two functions aresymmetric with respect to the line y = x.

5. Find the inverse function of f algebraically.

f (x) =1√

x− 1

Use a graphing utility to graph both f and f−1 in the same viewing window. Describethe relationship between the graphs.

Solution:

f−1 (x) =1 + x2

x2 , x ∈ (0, ∞)

When you graph both f and its inverse you should note that the two functions aresymmetric with respect to the line y = x.

6. Given f (x) = x3 and g (x) = 2x− 1, find the following.

(a) ( f g) (x) and its domain.

(b) (g f ) (x) and its domain.

Solution:

(a) ( f g) (x) = (2x− 1)3 , x ∈ (−∞, ∞)

(b) (g f ) (x) = 2x3 − 1, x ∈ (−∞, ∞)

7. Use the functions f (x) = x/8− 4 and g (x) = x3 − 8 to find the indicated value.(g−1 f−1

)(−4)

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Solution:(g−1 f−1

)(−4) = 2

Many will find the inverses to do this problem, but it is actually not required.

8. Verify graphically that each of the following functions is one-to-one.

(a)f (x) = 3− 5x, x ∈ (−∞, ∞)

(b)

f (x) =x− 5

3x + 1, x ∈

(−∞, −1

3

)∪(−1

3, ∞)

(c)

f (x) =√

16− x2, x ∈ [0, 4]

(d)f (x) = 5−

√x− 2, x ∈ [2, ∞)

(e)f (x) = x2 − 1, x ∈ (−∞, 0]

Solution: Each graph clearly passes the horizontal line test. make sure you can useyour calculator to graph!

9. Given f (x) =√

1− x and g (x) =√

x2 − 4, find the following.

(a) ( f + g) (x) and its domain.

(b) (g− f ) (x) and its domain.

(c) (g · f ) (x) and its domain.

(d) ( f /g) (x) and its domain.

Solution:

(a) ( f + g) (x) =√

1− x +√

x2 − 4, x ∈ (−∞, −2]

(b) (g− f ) (x) =√

x2 − 4−√

1− x x ∈ (−∞, −2]

(c) (g · f ) (x) =√

x2 − 4 ·√

1− x, x ∈ (−∞, −2]

(d) ( f /g) (x) =√

1− x√x2 − 4

, x ∈ (−∞, −2)

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10. Find the inverse for each one-to-one function. Be sure to indicate the domain for eachinverse.

(a)f (x) = 3− 5x, x ∈ (−∞, ∞)

(b)

f (x) =x− 5

3x + 1, x ∈

(−∞, −1

3

)∪(−1

3, ∞)

(c)f (x) =

√16− x2, x ∈ [0, 4]

(d)f (x) = 5−

√x− 2, x ∈ [2, ∞)

(e)f (x) = x2 − 1, x ∈ (−∞, 0]

Solution:

(a)

f−1 (x) =3− x

5, x ∈ (−∞, ∞)

(b)

f−1 (x) =x + 5

1− 3x, x ∈

(−∞,

13

)∪(

13

, ∞)

(c)

f−1 (x) =√

16− x2, x ∈ [0, 4]

(d)

f−1 (x) = x2 − 10x + 27, (−∞, 5]

(e)

f−1 (x) = −√

x + 1, [−1, ∞)

11. Given f (x) =√

5− x and g (x) = x2 + 4x, find the following.

(a) ( f g) (x) and its domain.(b) (g f ) (x) and its domain.

Solution:

(a) ( f g) (x) =√

5− 4x− x2, x ∈ [−5, 1]

(b) (g f ) (x) = 5− x + 4√

5− x, x ∈ (−∞, 5]

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14 Polynomial Functions

In this sections we will use simple factoring (what you did in MTH 100) to find roots ofhigher degree polynomials. By example I hope to intuitively show that a polynomial ofdegree n has at most n real roots, and at most n− 1 relative extrema. We’ll also talk aboutrepeated19 roots, and how the leading coefficient (the constant multiple of the highestdegree term) affects the graph. Of particular importance will be how the even and odddegree polynomials differ. Again, simple factoring and reasoning skills will prove useful.We will also need to move on to more advanced methods of analyzing polynomials, andtheorems abound here, but much of what you’ll end of doing is quite simple.

The major tone of what we’re doing here is finding roots and getting a better idea howpolynomial functions behave graphically. Graphics calculators may help greatly as I willillustrate in the examples to follow.

14.1 Overview of Roots

Roots are just another name for the x values of the x-intercepts. You may recall from aprior section that we also used the term zero interchangeably with root. Roots come in avariety of forms: rational, irrational, and complex. We’ll see them all, but I plan to startsimple and move on. Some of what follows is review of MTH 100, and we will use muchof what you learned in prior courses, including long division, to move forward.

14.1.1 Rational Numbers

The rational numbers are indicated by Q and can be represented by a ratio of integers. Areally simple set of numbers and you’ve been dealing with rationals for a very long time.

14.1.2 Irrational Numbers

The irrational numbers are indicated by H and can not be represented by a ratio of inte-gers. This is really not so simple, but you’ve been working with these irrational numbersfor a long time, and they’ll certainly be necessary if you plan to move forward in mathe-matics.

14.1.3 Complex Numbers

The complex numbers are indicated by C and are written in the form a + bi, where irepresents the imaginary unit. Complex numbers are a bit troubling to some, but they’renumbers nonetheless.

Fact is, you’ve been told in MTH 100 that not all numbers are real and that some areimaginary. You may have further been told that they were invented and don’t really exists.That’s hogwash—they not only exists, but were a necessary consequence of solving a

19Referred to as multiplicity.

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certain class of problems—that is, they were discovered and not invented. I won’t dwellon this, and we’ll stick to some very basic concepts.

The unit imaginary is denoted by i, where

i =√−1.

A complex number is written as a linear combination of a real and imaginary part, and isusually denoted by the letter z, where

z = a + bi.

Both a and b are real numbers, and i is just the imaginary unit. You should note that ifi =√−1 then i2 = −1. That’s about it! However, since i is really a root you need to

follow the same conventions that were used in radical simplifications. And you shouldalways simplify your radicals first, that is, if you’re asked to multiple

√−8 by the

√−25,

you’d do this:√−25 ·

√−8 = 5i · 2i

√2

= 10i2√

2

= −10√

2

14.1.4 Operations on Complex Numbers

This is a review of what was covered in MTH 100. You may need to revisit MTH 100 ifyou’re having trouble remembering what you have already learned.

Addition/Subtraction: just as you did with any algebraic expression. For example ifyou’re asked to add 3 + 2i to 7− 3i you’ll get 10− i. And if you’re asked to subtract2− 7i from 5− 3i you’ll get 3 + 4i. Really simple!

Multiplication: again, just as you did with any algebraic expression. However, if you geti2 in your multiplications you need to rewrite this as −1. Here goes:

(2− 3i) (3 + 2i) = 6 + 4i− 9i− 6i2

= 6− 5i + 6= 12− 5i

Division: just as you did with radical divisions! For example:

13 + 2i

=1

3 + 2i· 3− 2i

3− 2i

=3− 2i9− 4i2

=3− 2i

13

=313− 2

13i

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I won’t bore you with a bunch of examples, and I think it is best to review this on yourown. Yes, you’ll have a simple homework set to complete and you will need to get helpif you do not understand this material.

14.2 Remainder Theorem

When a polynomial, f (x), is divided by x− a, the remainder is f (a). This is useful, andwe will do many examples to illustrate its use.

14.3 Fundamental Theorem of Algebra

Every polynomial equation having complex coefficients and degree at least one has atleast one complex root. This theorem was first proven by Gauss. It is equivalent to thestatement that a polynomial P (z) of degree n has n values zi (some of them possiblydegenerate) for which P (zi) = 0. Such values are called polynomial roots. An exampleof a polynomial with a single root of multiplicity greater than one is z2 − 2z + 1, whichhas as a root z = 1 of multiplicity 2.

The essential crux os this theorem is that a polynomial of degree n has n-roots. How-ever, it does not say how to get them. We’ll use all kinds of tricks to find the roots. Some-times they’ll be simple rational numbers, and other times they will be irrational, but theycan always be written as complex a + bi. One trick20 you’ll learn is that complex rootsoccur in conjugate pairs if the polynomials have real coefficients. For example if 6− 3i isa root of a polynomial, then so is 6 + 3i. This is useful, and you need to remember this.

14.4 Intermediate Value Theorem

Karl Weierstrass was a German mathematician who proved the Intermediate Value The-orem (IVT), which was not an easy task. Some have said that he proved the obvious, butit nonetheless had to be proved before it could become a theorem. In MTH-119 we willuse this theorem to show the existence of roots. Here’s what the theorem states:

Suppose f is continuous on the closed interval [a, b] and W is any numberbetween f (a) and f (b), where f (a) 6= f (b). Then, there is a number c ∈(a, b) for which f (c) = W.

14.5 Rational Root Theorem

It will be necessary to find key points on a polynomial graph, including the y-intercept,which can easily be found by setting x = 0. On the other hand the x-intercept(s) are typ-ically more difficult to find, but essentially involve setting f (x) = 0 and then factoring.

20In mathematics, the complex conjugate root theorem states that if P is a polynomial in one variable withreal coefficients, and a + bi is a root of P with a and b real numbers, then its complex conjugate a− bi is alsoa root of P.

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Factoring polynomials in general is very difficult, if not impossible, however the exam-ples I will give will move forward smoothly. And once again, you should also be awarethat x-intercepts are often referred to as roots or zeros of the polynomial.

The approach I will take in factoring polynomials will typically involve the RationalRoot Theorem, which states, that if a polynomial

f (x) = anxn + an−1xn−1 + · · ·+ a2x2 + a1x + a0

has integer coefficients, every rational zero of f has the form

± factors of a0

factors of an.

This is identical to the method that was used to factor quadratics and was often referredto as trial and error in lower-level courses.

14.6 An Overview

Fundamental Theorem of Algebra: If f is a polynomial function of degree n, for n >0, then f has precisely “n” zeros in the complex number system; this means thatthere are exactly “n” zeros, yet not all of which are real and some of which may berepeated.

Example: f (x) = x3 + 9x = x(x2 + 9

)has exactly 3 zeros (of which one is real and

two are complex) and exactly 3 linear factors; these zeros are x = 0, 3i, −3i.

The Linear Factorization Theorem: If f is a polynomial function of degree n, for n > 0,then f can be factored into precisely “n” linear factors, one for each complex zero;hence, if x = c is a zero, then (x− c) is a factor of f .

Example: f (x) = x3 + 9x = x(x2 + 9

)which has 3 zeros (x = 0, 3i, −3i) and can

be factored as f (x) = x (x− 3i) (x + 3i).

Real Coefficient Polynomials: If f is a polynomial function with real coefficients, then ifx = a + bi is a complex zero of f , then so is its conjugate x = a− bi.

Example: Since f (x) = x3 + 9x has x = 3i as a zero, then so is its conjugate x = −3i.

Intermediate Value Theorem Suppose f is continuous21 on the closed interval [a, b] andW is any number between f (a) and f (b), where f (a) 6= f (b). Then, there is anumber c ∈ (a, b) for which f (c) = W.

Factor Theorem: If a polynomial function f of degree n is divided by x − a by long di-vision and the remainder is 0, then x = a is a zero of f and (x− a) is a factor of f .Also, the quotient is the polynomial function (of degree n− 1) that is also a factor off and may be factorable to find other zeros of f .

21A continuous function on [a, b] is a function with no breaks in the graph.

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Example: When f (x) = 6x3 − 19x2 + 16x − 4 is divided by x − 2 by long divi-sion, the remainder is 0 so x = 2 is a zero of f and (x− 2) is a factor of f . Theremaining factor of f is the quotient 6x2 − 7x + 2 and since this factors further into(3x− 2) (2x− 1), this means that x = 1/2, x = 2/3 are also zeros of f .

Rational Root Theorem: If a polynomial function of the form anxn + an−1xn−1 + · · · +a1x + a0 has an as its leading coefficient and a0 as its constant term, then you canform a list of possible rational zeros of the function of the form ±p/q where p isa factor of a0 and q is a factor of an. These fractions can then be tested by longdivision to see if they are indeed zeros or not. A graphing calculator can help youto determine which fractions to test.

Example: Using the Rational Root Theorem on f (x) = 2x3 + 3x2 − 8x + 3, it isfound that all rational zeros of are of the form: ±p/q = ±1, 3

1, 2 , so the list of possiblerational zeros consists is ±1, ±3, ±1/2, ±3/2; of these, the zeros are found to bex = 1, x = −3, and x = 1/2.

Leading Coefficient Test: Given a polynomial function of degree n and a leading coeffi-cient a, then if n is even, and a is positive, the graph rises to the left and rises to theright; if n is even and a is negative, the graph falls to the left and falls to the right; ifn is odd, and a is positive, the graph falls to the left and rises to the right; and if n isodd, and a is negative, the graph rises to the left and falls to the right.

Zeros of a polynomial function f of degree n: A polynomial f has at most n real zerosand at most n− 1 relative maximum/minimum (or extrema) points.

Example: f (x) = x3 − x2 − 2x has at most 3 real zeros and 2 relative max/minpoints.

Multiplicities of real zeros of polynomial functions: If a polynomial function has a zeroat x = a of multiplicity k, then: if the multiplicity k is even, the graph of the functiononly touches the x-axis at x = a (and does not cross it) and if k is odd, then the graphof the function crosses the x-axis at x = a.

Example: f (x) = x5 − 6x4 + 9x3 = x3 (x− 3)2 has x = 0 as a zero of multiplicity 3and x = 3 as a zero of multiplicity 2, so the graph of crosses the x-axis at x = 0 andonly touches the x-axis at x = 3.

Remainder Theorem: If a polynomial function f is divided by x− a by using long divi-sion and the remainder is not 0, but some number k, then x = a is not a zero of f ,but f (a) = k; then, the point (a, k) is a point on the graph of f .

Example: f (x) = x4 − 10x2 − 2x + 4 when divided by x− 2 by using long divisionhas −24 as its remainder, so this means that f (2) = −24 or the point (2, −24) is onthe graph of f .

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1. Perform the operation and write the result in standard form.

4 + ii− 82

9− i


4 + ii− 82

9− i= −8− 5i

2. Solve the quadratic equation.

x2 + 2x + 9 = 0


x = −1± 2i√

2

3. Simplify the complex number 9i2 − 2i3 and write it in standard form.


−9 + 2i

4. Three of the zeros of a real coefficient fourth-degree polynomial function, f (x), are 7,6, and 2− 7i. What is the other zero of f ?

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2 + 7i

5. Find the zeros algebraically and use your calculator to graph the function.

f (x) = −5x5 + 20x3 − 20x


Five zeros: 0, −√

2, −√

2,√

2,√

2


f (x) = 48− 16x− 3x2 + x3


Three zeros: 3, ±4


f (x) = x4 + 5x2 + 4


Four zeros: ±i, ±2i

8. Use your calculator to find all roots and relative extrema of

f (x) = 4x5 − 3x2 − x + 1.

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Yes, you will need to use your graphic calculators to do this problem. Approximatereal roots: −0.637, 0.456, 0.895; relative minimum at x ≈ 0.718; relative maximumat x ≈ −0.164.

9. Find a polynomial of degree 4 that has roots 0, 1, 2, 3.


f (x) = x (x− 1) (x− 2) (x− 3) = x4 − 6x3 + 11x2 − 6x

10. Find a polynomial of any degree that has roots −7, 5, and 1±√

3.


f (x) = (x + 7) (x− 5)(

x− 1−√

3) (

x− 1 +√

3)= x4 − 41x2 + 66x + 70

11. Find a polynomial of degree 6 with leading coefficient −2 that has the given zeroswith indicated multiplicities: the root 1 occurs twice; the root−2 occurs once, and theroot −1 occurs three times.


f (x) = −2 (x− 1)2 (x + 2) (x + 1)3 = −2x6 − 6x5 + 12x3 + 6x2 − 6x− 4

12. Find and expand a fourth degree polynomial function f (x) with real coefficients thathas roots x = ±

√2 and x = 1± i.

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f (x) =(

x−√

2) (

x +√

2)(x− 1− i) (x− 1 + i) = x4 − 2x3 + 4x− 4

13. Given

f (x) = x4 − 13x3 + 17x2 − 3.

Use the Intermediate Value Theorem and a graphing utility to find graphically anyintervals of length 1 in which the polynomial function is guaranteed to have at leastone zero.


Some good choices are: [−1, 0], [0, 1]; [1, 2]; [11, 12]

14. Given

f (x) = 10x3 − 15x2 − 16x + 12

and its graph (Figure 36, page 99).

Use this information to answer the following questions.

(a) Use the Rational Root Theorem to list the candidate rational roots.

(b) Test these roots to see which ones are roots.

(c) Use long division to partially factor f (x).

(d) Factor the remaining quadratic factor using the quadratic formula.

(e) Factor f (x) completely.


(a) Writing the ratios out.

±1, 2, 3, 4, 6, 121, 2, 5, 10

Then you’ll need to figure out the set.

±

1, 2, 3, 4, 6, 12,12

,15

,1

10,

25

,32

,35

,3

10,

45

,65

,125

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-1 0 1 2

-15

-10

-5

5

10

15

Figure 36: Partial graph of f (x) = 10x3 − 15x2 − 16x + 12.

(b) You should be able to roughly determine the values of the candidate rationalroots by looking at the graph, and 2, 1/2 and −6/5 are good candidates ifthey’re rational. Testing them shows that only one candiadte is actually a root.

f (2) = 0

(c)

f (x) = (x− 2)(

10x2 + 5x− 6)

(d)

10x2 + 5x− 6 =1

40

(20x + 5 +

√265) (

20x + 5−√

265)

(e)

f (x) =1

40(x− 2)

(20x + 5 +

√265) (

20x + 5−√

265)

15. Given

f (x) = 36 + 18x− 28x2 − 13x3 + 5x4 + 2x5,



(a) What is the y-intercept?

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-3 -2 -1 0 1 2 3

-5

5

10

15

20

25

30

35

40

45

50

Figure 37: Partial graph of f (x) = 36 + 18x− 28x2 − 13x3 + 5x4 + 2x5.

(b) Use the Rational Root Theorem to list the candidate rational roots.

(c) Using the graph, and your answers from (b), what are reasonable candidates forrational roots. Test them to see if they actually are.

(d) Using (c), factor f (x) completely.

(e) List all roots, real and imaginary.


(a) Just set x = 0 and you’ll get (0, 36).

(b) Writing the ratios out.

±1, 2, 3, 4, 6, 9, 12, 18, 361, 2

Then you’ll need to figure out the set.±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36, ±1

2, ±3

2, ±9

2

(c) You should be able to roughly determine the values of the candidate rational

roots by looking at the graph, and −3, 2, −3/2 and 3/2 are good candidates ifthey’re rational. Testing them shows that only three of these are.

f (−3) = f (2) = f (−3/2) = 0

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(d) Now we know that x + 3, 2x + 3, x− 2 are factors of f . Just use long division tofind the missing quadratic factor.

f (x) = 2x5 + 5x4 − 13x3 − 28x2 + 18x + 36

= (x + 3) (2x + 3) (x− 2)(

x2 − 2)

= (x + 3) (2x + 3) (x− 2)(

x +√

2) (

x−√

2)

(e) 2, −3, −3/2,

√2, −√

2

Here’s a close-up (Figure 38, page 101) of the region between −2 and −1. You shouldnotice the two roots, what are they?22 You should also look back at the original graph.

-2 -1

Figure 38: Partial graph of f (x) = 36 + 18x− 28x2 − 13x3 + 5x4 + 2x5.

16. Here are some very manageable graphing problems that should be done without us-ing technology.

(a) f (x) = x3 + 4x2 + 4x

(b) f (x) = x3 − x2 − x + 1

22One is −√

2 and the other is −3/2. They’re very close together and actually appear as one in theoriginal graph. Don’t be deceived by appearances.

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(c) f (x) = −x3 + x2 + x− 1

(d) f (x) = x4 − 16

(e) f (x) = x4 − 5x2 + 4

(f) f (x) = 2x2 − 2x4

(g) f (x) = x5 − 2x4 − 4x + 8

(h) f (x) = −x3 + 7x− 6, hint: f (2) = 0

Solution:

(a) f (x) = x3 + 4x2 + 4x = x (x + 2)2. Plot the real roots and do sign analysis toget a rough graph. Your graphs should look similar to a computer generatedgraph (Figure 39, page 103).

(b) f (x) = x3 − x2 − x + 1 = (x− 1)2 (x + 1). Plot the real roots and do signanalysis to get a rough graph. Your graphs should look similar to a computergenerated graph (Figure 40, page 103).

(c) f (x) = −x3 + x2 + x − 1 = − (x− 1)2 (x + 1). Plot the real roots and do signanalysis to get a rough graph. Your graphs should look similar to a computergenerated graph (Figure 41, page 104).

(d) f (x) = x4 − 16 = (x− 2) (x + 2)(

x2 + 4)

. Plot the real roots and do signanalysis to get a rough graph. Your graphs should look similar to a computergenerated graph (Figure 42, page 104).

(e) f (x) = x4 − 5x2 + 4 = (x− 2) (x− 1) (x + 1) (x + 2). Plot the real roots anddo sign analysis to get a rough graph. Your graphs should look similar to acomputer generated graph (Figure 43, page 105).

(f) f (x) = 2x2 − 2x4 = −2x2 (x− 1) (x + 1). Plot the real roots and do sign analy-sis to get a rough graph. Your graphs should look similar to a computer gener-ated graph (Figure 44, page 105).

(g) f (x) = x5 − 2x4 − 4x + 8 = (x− 2)(

x2 − 2) (

x2 + 2)

. Plot the real roots anddo sign analysis to get a rough graph. Your graphs should look similar to acomputer generated graph (Figure 45, page 106).

(h) f (x) = −x3 + 7x− 6 = − (x− 2) (x− 1) (x + 3). Plot the real roots and do signanalysis to get a rough graph. Your graphs should look similar to a computergenerated graph (Figure 46, page 106).

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-4 -3 -2 -1 0 1

-2

-1

1

Figure 39: Partial graph of f (x) = x3 + 4x2 + 4x.

-2 -1 0 1 2 3

-1

1

Figure 40: Partial graph of f (x) = x3 − x2 − x + 1.

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-2 -1 0 1 2 3

-1

1

Figure 41: Partial graph of f (x) = −x3 + x2 + x− 1.

-3 -2 -1 0 1 2 3

-18-17-16-15-14-13-12-11-10-9-8-7-6-5-4-3-2-1

1234567891011

Figure 42: Partial graph of f (x) = x4 − 16.

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-3 -2 -1 0 1 2 3

-2

-1

1

2

3

4

Figure 43: Partial graph of f (x) = x4 − 5x2 + 4.

-1 0 1

-1

1

Figure 44: Partial graph of f (x) = 2x2 − 2x4.

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-2 -1 0 1 2

-3

-2

-1

1

2

3

4

5

6

7

8

9

10

Figure 45: Partial graph of f (x) = x5 − 2x4 − 4x + 8.

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

-16-15-14-13-12-11-10-9-8-7-6-5-4-3-2-1

12345678

Figure 46: Partial graph of f (x) = −x3 + 7x− 6.

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1. Verify that x = 8i is a root of f (x) = x3 + x2 + 64x + 64 and then factor this functioncompletely.

Solution:

f (8i) = 0f (x) = (x− 8i) (x + 8i) (x + 1)

2. Find all zeros of f (z) = 625z4 − 81.

Solution:

±35

, ±35

i


f (x) = x5 + 2x4 − 3x3

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Solution: Five zeros: 0, 0, 0, 1, −3

4. Simplify the complex number 3i60− 2i150− 5i82 + 6i109− 2i61 and write it in standardform.

Solution:

10 + 4i

5. Find and expand a third degree polynomial function f (x) with real coefficients thathas roots x = 2 and x = −3i.

Solution:

(x− 2) (x + 3i) (x− 3i) = x3 − 2x2 + 9x− 18


f (x) = x5 + 5x3 + 4x

Solution: Five zeros: 0, ±i, ±2i

7. Find a polynomial of degree 4 with leading coefficient 72 that has the given zeros withindicated multiplicities: the root 2/3 occurs twice; and the root −3/2 occurs twice.

Solution: f (x) = 2 (3x− 2)2 (2x + 3)2 = 72x4 + 120x3 − 94x2 − 120x + 72


f (x) = x4 − 4x2 + 1

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Solution: Four zeros: ±√

2±√

3

9. Given

f (x) = x2 − 2x + 10


-4 -3 -2 -1 0 1 2 3 4 5 6

5

10

15

20

25

30

35

Figure 47: Partial graph of f (x) = x2 − 2x + 10.



(b) Test these candidate roots to see which ones are roots.

(c) Use the quadratic formula to find the roots of f (x).

(d) Factor f (x).

Solution:


±1, 2, 5, 101

Then you’ll need to figure out the set.

±1, ±2, ±5, ±10

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(b) None of these candidate roots work.

(c) You need to use the quadratic formula.

x = 1± 3i

(d)

f (x) = (x− 1− 3i) (x− 1 + 3i)

10. Given

f (x) = 2x3 + 3x2 − 8x + 3


-4 -3 -2 -1 0 1 2

5

10

15

Figure 48: Partial graph of f (x) = 2x3 + 3x2 − 8x + 3.


(b) Test these candidate roots to see which ones are roots.

(c) Factor f (x) completely.



±1, 31, 2

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Then you’ll need to figure out the set.±1, ±3, ±1

2, ±3

2

(b) You should be able to roughly determine the values of the candidate rational

roots by looking at the graph, and −3, 1, and 1/2 are good candidates if they’rerational. Testing them shows that all three of these are.

f (−3) = f (1) = f (1/2) = 0

(c)

f (x) = (x + 3) (x− 1) (2x− 1)

11. Given that x = 1− i is a root of

f (x) = x4 − 2x3 − 3x2 + 10x− 10,

answer each of the following.

(a) Verify that x = 1− i and its conjugate are roots of f (x).

(b) Since x = 1− i and x = 1 + i are roots, we know that (x− 1 + i) and (x− 1 + i)are factors of f (x). Expand (x− 1 + i) (x− 1 + i)

(c) Long divide x4 − 2x3 − 3x2 + 10x− 10 by x2 − 2x + 2.

(d) Factor x2 − 5 into linear factors.

(e) Verify the factorization of f (x) is (x− 1 + i) (x− 1 + i)(

x−√

5) (

x +√

5)

(f) List all roots of f (x).

Solution:

(a) Verify that x = 1− i and its conjugate are roots of f (x).

f (1− i) = f (1 + i) = 0

(b)

(x− 1 + i) (x− 1 + i) = x2 − 2x + 2

(c)

x4 − 2x3 − 3x2 + 10x− 10x2 − 2x + 2

= x2 − 5

(d)

x2 − 5 =(

x−√

5) (

x +√

5)

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(e)

(x− 1 + i) (x− 1 + i)(

x−√

5) (

x +√

5)= x4 − 2x3 − 3x2 + 10x− 10

(f)

1 + i, 1− i,√

5, −√

5

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15 Rational Functions

15.1 Definition of a Rational Function

A rational function is any function which can be defined by a ratio of polynomials, i.e.an algebraic fraction such that both the numerator and the denominator are polynomials.Here we write the rational function as a ratio of two polynomials p (x) and q (x).

f (x) =p (x)q (x)

15.2 Vertical Asymptotes

The line x = a is a vertical asymptote for the graph of f (x) if, as x approaches a throughvalues from the left and or right of a, the values of f (x) grows without bound (±∞). Todo this we will initially be looking at the graph of f (x), but we must try to move beyondthis and start to look at and understand the following notation without first looking at agraph of f (x).

limx→a+

f (x) = ±∞

limx→a−

f (x) = ±∞

15.3 Horizontal Asymptotes

The line y = a is a horizontal asymptote for the graph of f (x) if, as x approaches ±∞,f (x) approaches a. Again, we need to look at limits, as we did above.

limx→∞

f (x) = a

limx→−∞

f (x) = a

15.4 Oblique Asymptotes

An oblique asymptote will be a line of the form y = mx + b. This will typically (notalways) happen if the degree of the numerator is one more than the degree of the denom-inator. We simply just need to long divide to get this form.

f (x) =p (x)q (x)

= mx + b +r (x)q (x)

Here r (x) is the remainder from long division.

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15.5 An Overview

Here we will be dealing with a ratio of polynomials and their graphs. We need to be espe-cially careful that we do not allow divisions by zero—but this is generally near where wewant to be, mainly because we love danger and there’s nothing in mathematics that’s moredangerous than dividing by zero. We’re also going to see behavior in these graphs thatappears linear, that is, somewhere along these curves it will look nearly, if not exactly, likea line. The idea that a curve may come arbitrarily close to a line without actually becom-ing the same is called asymptotic behavior. The details will be explained by examples,and I do encourage everyone to use their graphic calculators to help aid their desire tounderstand the difference between something that looks linear and something that is lin-ear. Perceptibly there may be no difference between them, but conceptually there is. Thatkind of differentiation takes considerable time to discern. Let’s start with an example . . .

15.5.1 First View

Given

f (x) =2x2 + 7x− 4

x2 + x− 2=

(2x− 1) (x + 4)(x− 1) (x + 2)

,

and its partial graph (Figure 49, page 114).

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

-15

-10

-5

5

10

15

Figure 49: Partial graph of f (x) with asymptotes indicated in red.

Use this information ( f (x) and its graph) to answer the following questions.

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• What is the y-intercept?

Solution: Work will be done in class. Make sure you take notes, or get notes froma reliable classmate if you are not present during lecture.

Just set x = 0 and you’ll get (0, 2).

• What are the x-intercepts?


Just set f (x) = y = 0 and you’ll get (−4, 0) and (1/2, 0).

• What is the equation of the horizontal asymptote?


By inspection, we get y = 2. However, you also need to be able to do these twolimits.

limx→∞

f (x) = 2

limx→−∞

f (x) = 2

• What are the equations of the two vertical asymptotes?


By inspection, we get x = −2 and x = 1. However, you also need to be able to dothese four limits.

limx→−2−

f (x) = −∞

limx→−2+

f (x) = ∞

limx→1−

f (x) = −∞

limx→1+

f (x) = ∞

• What is the domain of f (x)?

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By inspection, we get all real numbers except −2 and 1.

• What is the range of f (x)?


By inspection, we get all real numbers, this is often denoted R.

15.5.2 Second View

Here we’ll be introducing a third type of linear asymptote called a slant or oblique asymp-tote. They occur when the degree of the numerator is one more than the degree of thedenominator. Let’s start with a visual example.

Given

f (x) =x2 − 4x− 5

x− 3=

(x + 1) (x− 5)x− 3

= x− 1− 8x− 3

,


-15 -10 -5 0 5 10 15

-10

-5

5

10



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• What is the y-intercept?


Just set x = 0 and you’ll get (0, 5/3).

• What are the x-intercepts?


Just set y = 0 and you’ll get (−1, 0) and (5, 0).

• What is the equation of the vertical asymptote?


By inspection, we get x = 3. However, you also need to be able to do these fourlimits.

limx→3−

f (x) = ∞

limx→3+

f (x) = −∞

• What is the equation of the slant (oblique) asymptote?


By inspection, we get y = x− 1.

• What is the domain of f (x)?


By inspection, we get R, x 6= 3.

• What is the range of f (x)?

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By inspection, we get R.


Answering questions about rational functions is often aided by having a good graph.However, you should not rely solely on graphs and you should try to develop a skillset that does not heavily rely on using a graphic calculator. Even though graphs may beprovided you should still take the time graph each problem using your graphic calculator.

1. Given

f (x) = y =x2 − 3x− 4

2x2 + 4x=

(x + 1) (x− 4)2x (x + 2)

,


-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

-15

-10

-5

5

10

15



(a) What is/are the y-intercept(s)?

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Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.Just set x = 0 and you’ll see the dreaded division by zero, so there is none.

(b) What is/are the x-intercept(s)?

Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.Just set y = 0 and you’ll get (−1, 0) and (4, 0).

(c) What is the domain of f (x)?

Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.All real numbers except 0 and −2.

(d) What is/are the equation(s) of the horizontal asymptote(s)?

Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.We will need to look at two limits to determine this.

limx→∞

x2 − 3x− 42x2 + 4x

= limx→∞

1− 3/x− 4/x2

2 + 4/x= 1/2,

and

limx→−∞

x2 − 3x− 42x2 + 4x

= limx→−∞

1− 3/x− 4/x2

2 + 4/x= 1/2.

So the horizontal asymptote is y = 1/2.

(e) What is/are the equation(s) of the vertical asymptote(s)?

Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.You’ll need to look for divisions by zero, and then look at limits to see whathappens. We will discuss this in class.

limx→0−

x2 − 3x− 42x2 + 4x

= ∞

limx→0+

x2 − 3x− 42x2 + 4x

= −∞

limx→−2−

x2 − 3x− 42x2 + 4x

= ∞

limx→−2+

x2 − 3x− 42x2 + 4x

= −∞

So the vertical asymptotes occur at x = 0 and x = −2.

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(f) What is the range of f (x)?

Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.You’ll need the graph, and then it should be clear that the range is R.

2. Find the limit, if it exists.

limx→1−

x2 + 3x + 5x2 − 1

Use values close to x = 1, from the left.23 You should also consider graphing to see ifyou get the same result.


Using a numerical sequences. First from the left of x = 1.

y (x) =x2 + 3x + 5

x2 − 1, x 6= 1

y (0.9) ≈ −44.7895y (0.99) ≈ −449.754

y (0.999) ≈ −4499.75

Looks like this sequence is going towards −∞. So I write

limx→1−

x2 + 3x + 5x2 − 1

= −∞

3. Find the limit, if it exists.24

limx→∞

√x2 + 4− 22x− 3

You should also consider graphing to see if you get the same result. Although this isnot rational function it is being used here to demonstrate a way to take a limit.

23Use: 0.9, 0.99, 0.99924Use: 10000, 100000, 1000000

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Using a numerical sequence.

y (x) =

√x2 + 4− 22x− 3

y (10000) ≈ 0.499975y (100000) ≈ 0.499998

y (1000000) ≈ 0.500000

Looks like this sequence is going towards 1/2. So I write

limx→∞

√x2 + 4− 22x− 3

=12

4. Given

f (x) =x2 − 9xx2 − 9

=x (x− 9)

(x + 3) (x− 3),


-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-5

-4

-3

-2

-1

1

2

3

4

5

Figure 52: Partial graph of f (x) =x2 − 9xx2 − 9

with asymptotes indicated in red.



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Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.Just set x = 0 and you’ll get (0, 0)


Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.Just set y = 0 and you’ll (0, 0) and (9, 0).


Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.All real numbers except ±3.


Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.We will need to look at two limits to determine this.

limx→∞

x2 − 9xx2 − 9

= 1

and

limx→−∞

x2 − 9xx2 − 9

= 1

So the horizontal asymptote is y = 1.


Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.You’ll need to look for divisions by zero, and then look at limits to see whathappens. We will discuss this in class.

limx→−3−

x2 − 9xx2 − 9

= ∞

limx→−3+

x2 − 9xx2 − 9

= −∞

limx→3−

x2 − 9xx2 − 9

= ∞

limx→3+

x2 − 9xx2 − 9

= −∞

So the vertical asymptotes occur at x = ±3.

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Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.You’ll need the graph, and then it should be clear that the range is (−∞, ∞).

5. Given

f (x) =x3 − 8x2 + 7

,


-21 -20 -19 -18 -17 -16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

9

10

Figure 53: Partial graph of f (x) =x3 − 8x2 + 7



(a) What is the equation of the slant asymptote?

Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.You’ll need to do a long division to find

f (x) =x3 − 8x2 + 7

= x− 7x + 8x2 + 7

.

The equation of the slant asymptote is y = x

(b) What is/are the y-intercept(s)?

Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.Just set x = 0 and you’ll get (0, −8/7)

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(c) What is/are the x-intercept(s)?

Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.Just set y = 0 and you’ll get (2, 0).

(d) What is the domain of f (x)?

Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.All real numbers.

(e) What is the range of f (x)?

Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.You’ll need the graph, and then it should be clear that the range is all real num-bers.

6. Find the real zeros of the rational function.

f (x) = 1 +13

x2 + 1

A partial graph (Figure 54, page 124) is provided as a guide.

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-1

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

Figure 54: Partial graph of f (x) = 1 +13

x2 + 1with asymptotes indicated in red.


There are no real zeros.

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7. Use long division to rewrite.

f (x) =x3 − 2x2 + 3x− 5

x2 + x− 1


y = x− 3 +7x− 8

x2 + x− 1

8. Given

y =x2 − 9x + 3

,


-11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-7

-6

-5

-4

-3

-2

-1

1

2

Figure 55: Partial graph of y =x2 − 9x + 3

.



Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.Just set x = 0 and you’ll get (0, −3)

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Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.Just set y = 0 and you’ll see that this equation has only one solution. Don’tbe deceived by the equation, there’s a hole at the point (−3, −6), and the onlyx-intercept is (3, 0).


Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.All real numbers except −3.

(d) Using limits explain what happens at x = −3

Solution:

limx→−3−

x2 − 9x + 3

= −6

and

limx→−3+

x2 − 9x + 3

= −6

There’s a hole at the point (−3, −6), just like in my graph. You may want to dothis on your graphic calculator to see if you can see the hole.

(e) What is the range of f (x)?

Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.It should be clear that the range is all real number except −6.

9. Here are some very manageable graphing problems that should be done without us-ing technology.

(a) f (x) =2x

x− 4.

(b) f (x) =x + 2

x2 − 25.

(c) f (x) =x + 8x2 − 4

.

(d) f (x) =x2 + x− 42x2 − x− 30

=(x− 6)(x + 7)(x− 6)(x + 5)

.25

25Note that when x = 6 your graph should indicate a hole.

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Solution:

(a) Plot intercepts and asymptotes and do sign analysis to get a rough graph. Yourgraphs should look similar to a computer generated graph (Figure 56, page 127).

(b) Plot intercepts and asymptotes and do sign analysis to get a rough graph. Yourgraphs should look similar to a computer generated graph (Figure 57, page 128).

(c) Plot intercepts and asymptotes and do sign analysis to get a rough graph. Yourgraphs should look similar to a computer generated graph (Figure 58, page 128);and closer view near x = 8 (Figure 59, page 129).

(d) Plot intercepts and asymptotes and do sign analysis to get a rough graph. Yourgraphs should look similar to a computer generated graph (Figure 60, page 129).

-3 -2 -1 0 1 2 3 4 5 6 7 8 9

-12-11-10-9-8-7-6-5-4-3-2-1

123456789101112

Figure 56: Partial graph of f (x) =2x

x− 4.



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-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

-3

-2

-1

1

2

3

Figure 57: Partial graph of f (x) =x + 2

x2 − 25.

-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

-3

-2

-1

1

2

3

Figure 58: Partial graph of f (x) =x + 8x2 − 4

.

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-20 -19 -18 -17 -16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5

-0.05

0.05

0.1

0.15

Figure 59: Closer partial view of f (x) =x + 8x2 − 4

near x = −8.

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

-3

-2

-1

1

2

3

Figure 60: Partial graph of f (x) =x2 + x− 42x2 − x− 30

.

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1. Use long division to find the slant asymptote.

f (x) =x3 − 2x2 + 3x− 5

x2 + x− 1

Solution:

y = x− 3

2. Given

f (x) =2

(x− 1)3 ,









Solution:

(a) Just set x = 0 and you’ll get (0, −2)

(b) Just set y = 0 and you’ll get an equation that is never true. No x-intercepts.

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-3 -2 -1 0 1 2 3

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

9

Figure 61: Partial graph of f (x) =2

(x− 1)3 with asymptotes indicated in red.

(c) All real numbers except 1.

(d) You will need to look at two limits to determine this.

limx→∞

2

(x− 1)3 = 0

and

limx→−∞

2

(x− 1)3 = 0

So the horizontal asymptote is y = 0.

(e) You’ll need to look for divisions by zero, and then look at limits to see whathappens.

limx→1−

2

(x− 1)3 = −∞

limx→1+

2

(x− 1)3 = ∞

So the vertical asymptotes occur at x = 1.

(f) You’ll need the graph, and then it should be clear that the range is

(−∞, 0) ∪ (0, ∞) .


limx→1+

x2 + 3x + 5x2 − 1

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Use values close to x = 1, from the right.26 You should also consider graphing to seeif you get the same result.

Solution: Using a numerical sequences. First from the right of x = 1.

y (x) =x2 + 3x + 5

x2 − 1, x 6= 1

y (1.1) ≈ 45.2857y (1.01) ≈ 450.254

y (1.001) ≈ 4500.25

Looks like this sequence is going towards ∞. So you should write

limx→1+

x2 + 3x + 5x2 − 1

= ∞.

4. Given

f (x) =x2 − 2x− 1

− 2


-11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-7

-6

-5

-4

-3

-2

-1

1

2

Figure 62: Partial graph of f (x) =x2 − 2x− 1

− 2.


26Use: 1.1, 1.01, 1.001

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(a) Rewrite f (x) as a single fraction and then perform a long division.

(b) What is/are the y-intercept(s)?

(c) What is/are the x-intercept(s)?

(d) What is the equation of the vertical asymptote?

(e) What is the equation of the slant (oblique) asymptote?

(f) What is the domain of f (x)?

(g) What is the range?

Solution:

(a) Long divsion can be done before or after getting the single fraction.

f (x) =x2 − 2x− 1

− 2 =x2 − 2xx− 1

= x− 1− 1x− 1

(b) Just set x = 0 and you’ll get (0, 0)

(c) Just set y = 0 and you’ll see that this equation has two solutions, the x-interceptsare (0, 0) and (2, 0).

(d) You will need to take a look at limits.

limx→1−

x2 − 2xx− 1

= ∞

limx→1+

x2 − 2xx− 1

= −∞

So the vertical asymptote is x = 1.

(e) When x → ±∞ the function is asymptotic with y = x− 1.

(f) R, x 6= 1.

(g) By inspection, we clearly see that it is all real numbers.


limx→0

√x2 + 4− 2

x2

Use values close27 to x = 0, from both left and right. Although this is not a ratio ofpolynomials it illustrates the use of numerical tables to get a limit, you should alsoconsider graphing to see if you get the same result.

27Use: ±1,±0.5,±0.1,±0.05,±0.01.

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Solution: Using a numerical sequences. First from the right of x = 0.

y (x) =

√x2 + 4− 2

x2 , x 6= 0

y (1) ≈ 0.236068y (0.5) ≈ 0.246211y (0.1) ≈ 0.249844

y (0.05) ≈ 0.249961y (0.01) ≈ 0.249998

Again, it looks like this sequence is going towards 0.25 = 1/4. So I write

limx→0+

√x2 + 4− 2

x2 =14

.

Now from the left of x = 0.

y (x) =

√x2 + 4− 2

x2 , x 6= 0

y (−1) ≈ 0.236068y (−0.5) ≈ 0.246211y (−0.1) ≈ 0.249844

y (−0.05) ≈ 0.249961y (−0.01) ≈ 0.249998

Again, it looks like this sequence is going towards 0.25 = 1/4. So I write

limx→0−

√x2 + 4− 2

x2 =14

.

6. Given

f (x) =x2 + 2xx− 1



(a) First rewrite the function as has been done in prior examples.

(b) What is the y-intercept?

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-18 -17 -16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

9

10

11

Figure 63: Partial graph of f (x) =x2 + 2xx− 1

.

(c) What are the x-intercepts?

(d) What is the equation of the vertical asymptote?

(e) What is the equation of the slant (oblique) asymptote?

(f) What is the domain of f (x)?

(g) What is the range28 of f (x)?

Solution:

(a) You will need to do both a long division and a factorization.

f (x) = y =x2 + 2xx− 1

=x (x + 2)

x− 1= x + 3 +

3x− 1

(b) Just set x = 0 and you’ll get (0, 0).

(c) Just set y = 0 and you’ll get (−2, 0) and (0, 0).

(d) You will need to take a look a limits.

limx→1−

x2 + 2xx− 1

= −∞

limx→1+

x2 + 2xx− 1

= ∞

So the vertical asymptote is x = 1.

(e) When x → ±∞ the function is asymptotic with y = x + 3.

(f) R, x 6= 1.

28A little calculus will show that it’s(−∞, 4− 2

√3]∪[4 + 2

√3, ∞

)

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(g) By inspection, we clearly see that it is not all real numbers, however we can-not determine the range exactly with the math we know now. You shouldbe able to get a rough estimate using your graphic calculator, most calcula-tors will give (−∞, 0.5359] ∪ [7.4641, ∞). Keep in mind that if you continueto study and learn a little calculus, you will be able to show that it’s actually(−∞, 4− 2

√3]∪[4 + 2

√3, ∞

).

7. Find the limit, if it exists.29

limx→−∞

√x2 + 4− 22x− 3

You should also consider graphing to see if you get the same result.

Solution: Using a numerical sequences.

y (x) =

√x2 + 4− 22x− 3

y (−10000) ≈ −0.499825y (−100000) ≈ −0.499983

y (−1000000) ≈ −0.499998

Looks like this sequence is going towards −1/2. So I write

limx→−∞

√x2 + 4− 22x− 3

= −12

8. Given

f (x) =x2 + 2x + 13x2 − x− 4

=(x + 1) (x + 1)(3x− 4) (x + 1)

,






29Use: −10000, −100000, −1000000

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-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-5

-4

-3

-2

-1

1

2

3

4

5

Figure 64: Partial graph of f (x) =x2 + 2x + 13x2 − x− 4





Solution:

(a) Just set x = 0 and you’ll get (0, −1/4)

(b) Just set y = 0 and you’ll see that this equation has no solution. Don’t bedeceived by the graph, there’s a hole at the point (−1, 0), but it is not an x-intercept.

(c) All real numbers except −1 and 4/3.

(d) We will need to look at two limits to determine this.

limx→∞

x2 + 2x + 13x2 − x− 4

= 1/3

and

limx→−∞

x2 + 2x + 13x2 − x− 4

= 1/3

So the horizontal asymptote is y = 1/3.

(e) You’ll need to look for divisions by zero, and then look at limits to see what

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happens. We will discuss this in class.

limx→−1−

x2 + 2x + 13x2 − x− 4

= 0

limx→−1+

x2 + 2x + 13x2 − x− 4

= 0

limx→4/3−

x2 + 2x + 13x2 − x− 4

= −∞

limx→4/3+

x2 + 2x + 13x2 − x− 4

= ∞

So the vertical asymptotes occur at x = 4/3.

(f) You’ll need the graph, and then it should be clear that the range is all real num-ber except 1/3 and 0.

9. Graph (Figure 65, page 138).

f (x) =3x4 − 5x2 + 3

x4 + 2

and indicate what you have learned so far about the graphs of rational functions.

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

1

2

3

Figure 65: Partial graph of f (x) =3x4 − 5x2 + 3

x4 + 2.

Solution: Here you should be able to answer the following questions.

(a) You should look at the limit as x → ∞ and be able to see that you get a horizon-tal asymptote, y = 3.

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(b) You should look at the limit as x → −∞ and be able to see that you get ahorizontal asymptote, y = 3.

(c) You should be able to determine the domain, (−∞, ∞)

(d) You should be able to detremine the y intercept, (0, 3/2).

(e) You should be able to determine that there are no x-intercepts.

(f) You should be able to show, using an appropriate test, that this is an even func-tion.

(g) You should be able to find the approximate range using a calculator,[(9−√

59)

/4, 3)≈ [0.33, 3) .

10. Using a graphic calculator, graph (Figure 66, page 139).

y =x√

x2 + 1.

You should try to make sense out of this graph by looking at key points such as x andy-intercepts, sign-analysis, limits, and domain. From this graph try to determine therange.

-2 -1 0 1 2

-1

1

Figure 66: Partial graph of y =x√

x2 + 1.

Solution: Here you should be able to answer the following questions.

(a) You should look at the limit as x → ∞ and be able to see that you get a horizon-tal asymptote, y = 1.

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(b) You should look at the limit as x → −∞ and be able to see that you get ahorizontal asymptote, y = −1.

(c) You should be able to determine the domain, (−∞, ∞)

(d) You should be able to detremine the y intercept, (0, 0).

(e) You should be able to determine the x-intercepts, (0, 0).

(f) You should be able to show, using an appropriate test, that this is an odd func-tion.

(g) You should be able to find the range, (−1, 1).

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16 Algebraic Functions; Variation

This section of the textbook is not currently being covered. However, students may decideon their own to review Chapter 16 of Schaum’s Precalculus.

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17 Exponential Functions and Their Graphs

17.1 Exponentials

The function f (x) = ax, where x ∈ R, a > 0 and a 6= 1, is called the exponential function,with base a.

17.1.1 Graphs

Since a is a positive real number we have a variety of possible graphs. You should be ableto graph (Figure 67, page 139) a variety of exponential functions, and I can only suggestthat creating a table of values can often lead to a pattern that is easy to predict. Be sure toselect easy values.

-2 -1 0 1 2 3

1

2

3

4

5

6

7

8

9

Figure 67: y = 2x [red], y = 5x [blue], y = 20x [green]

17.1.2 Properties of Exponents

The exponent rules were extensively covered in MTH 100.If m, n, and p are real numbers, then:

1. x0 = 1, x 6= 0;

2. xm · xn = xm+n;

3. (xm)n = xmn;

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4. (xmyn)p = xmpynp;

5.xm

xn = xm−n, x 6= 0;

6.(

xm

yn

)p=

xmp

ynp , y 6= 0;

7. x−n =1xn , x 6= 0;

8. xn =1

x−n , x 6= 0;

9.(

xy

)−n=(y

x

)n, y 6= 0, x 6= 0;

17.1.3 Euler’s Number

The number e is Euler’s number and is often called a natural base—it is an extremelyimportant30 number in mathematics. It is defined as a limit.

limx→∞

(1 +

1x

)x

Euler’s number is irrational31 and can not be written as a ratio of integers. This numbercan be approximated on your calculator and you need to be able to use this number incomputation. Everyone should be able to locate Euler’s number (symbol is e) on yourcalculator. Again, it is often referred to as a natural base and we’ll see this quite often inmathematics.

17.1.4 Discrete Compound Interest

If a principal of P dollars is invested at an annual interest rate of r, and the interest iscompounded n times per year, then the amount of money A (t) generated at time t, inyears, is given by:

A (t) = P(

1 +rn

)nt.

30There are five numbers that are of upmost importance in mathematics and they are surprisingly relatedby a simple equation:

eiπ + 1 = 0.

31An approximate value for e is 2.7182818284590452353602874713526624978.

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17.1.5 Continuous Compound Interest

If a principal of P dollars is invested at an annual interest rate of r, and the interest iscompounded continuously, then the amount of money A (t) generated at time t, in years,is given by:

A (t) = Pert.

17.1.6 Application Problems

Everyone should be familiar with the growth of money (compound and continuous inter-est) and you should also try to become familiar with problems related to decay (radioac-tive) and growth (population)—what’s important to remember is that these problems areexponential in nature. Exponentials are important and may be used to model growth anddecay.

17.2 Examples

1. You build an annuity by investing P dollars every month at interest rate r, com-pounded monthly. Find the amount A accrued after n months using the formula

A = P[(1 + r/12)n − 1

r/12

]where r is in decimal form. Compute A given that P = $119, r = 0.09, n = 52 months.


A = 119

[(1 + 0.09/12)52 − 1

0.09/12

]≈ 7534.02

2. Use the discrete compound interest formula to compute the amount of money after20 years, if the initial deposit is 30000 dollars, compounded monthly, and the interestrate is 12%.


A (20) = 30000(

1 +0.1212

)20·12

≈ 326776.61

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3. Given

f (x) = −21−x,

create a table of values using easy (finger numbers) values for x, then plot the pointsand try to connect-the-dots. Your graph (Figure 68, page 145) should look similar tomine.

-2 -1 0 1 2

-7

-6

-5

-4

-3

-2

-1

1

2

Figure 68: f (x) = −21−x


As we do more graphs you’ll see that some increase as x increases, some decrease asx increases, and some may do both.

4. Use the continuous compound interest formula to compute the amount of moneyafter 40 years, if the initial deposit is 15000 dollars, and the interest rate is 9%.


A (40) = 15000e40·0.09 ≈ 548973.52

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5. What type of transformation of the graph of f (x) = 3x is the graph of f (x− 1)?


The graph of f (x− 1) is a horizontal shift, one unit to the right, of the graph off (x) = 3x.

6. Use a calculator to evaluate the function at the indicated value of x. Round your resultto four decimal places.

f (x) = 2.3x, x =23


f(

23

)= 2.32/3 ≈ 1.7424

7. What type of transformation of the graph of f (x) = 2x is the graph of − f (x)?


The graph of − f (x) is a rotation about the x-axis of the graph of f (x) = 2x.

8. Graph (Figure 69, page 147) the exponential function by hand. Identify any asymp-totes and intercepts and determine whether the graph of the function is increasing ordecreasing.

f (x) =(

23

)x


This is a decreasing function. The horizontal asymptote is y = 0. The y-intercept is1, no x-intecepts.

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-4 -3 -2 -1 0 1 2 3 4 5

-1

1

2

3

Figure 69: f (x) =(

23

)x


f (x) =(

32

)x−1


This function is increasing on (−∞, ∞), the horizontal asymptote is y = 0, the y-intercept is 2/3, no x-intecepts.

10. Choose the correct symbol (< or>) between the pair of numbers.

83/4 ?(

34

)8

Solution: We’ll do this is class.

83/4 >

(34

)8

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-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4

-1

1

2

3

4

Figure 70: f (x) =(3

2

)x−1






1. What type of transformation of the graph of f (x) = 5x is the graph of f (x + 1)?

Solution: The graph of f (x + 1) is a horizontal shift, one unit to the left, of the graphof f (x) = 5x.

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2. Complete the balance A for $1,100 invested at rate r = 3% for t = 41 years andcompounded n = 1, 2, 12, 365 times per year.

A = P(

1 +rn

)nt

Solution:

A = 1100(

1 +0.03

1

)1·41

≈ $3695.89

A = 1100(

1 +0.03

2

)2·41

≈ $3729.14

A = 1100(

1 +0.0312

)12·41

≈ $3757.58

A = 1100(

1 +0.03365

)365·41

≈ $3763.16


f (x) = 31−x2

Solution:

This function is increasing on (−∞, 0), decreasing (0, ∞), the horizontal asymptoteis y = 0, the y-intercept is 3, no x-intecepts.

4. What type of transformation of the graph of f (x) = 7x is the graph of f (−x)?

Solution: The graph of f (−x) is a rotation about the y-axis of the graph of f (x) =7x. It might be a good idea to see if you can graph both of these by hand and on agraphics calculator.

5. Use a graphing utility to graph (Figure 72, page 150) the function. Using limits findthe equation of the horizontal asymptote and analyze the behavior of f as x ap-proaches zero using limits.

f (x) =3

1 + e−0.5/x

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-3 -2 -1 0 1 2 3

1

2

3

Figure 71: f (x) = 31−x2

-2 -1 0 1 2

1

2

3

Figure 72: f (x) =3

1 + e−0.5/x

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Solution:

limx→0−

f (x) = 0

limx→0+

f (x) = 3

limx→−∞

f (x) =32

limx→∞

f (x) =32

The horizontal asymptote is y = 3/2.

6. You build an annuity by investing P dollars every month at interest rate r, com-pounded monthly. Find the amount A accrued after n months using the formula

A = P[(1 + r/12)n − 1

r/12

]where r is in decimal form. P = $115.26, r = 0.085, n = 123 months.

Solution:

A = 115.26

[(1 + 0.085/12)123 − 1

0.085/12

]≈ 22497.14


f (x) = 2x − 3

Solution: This function is increasing on (−∞, ∞), the horizontal asymptote is y =−3, the y-intercept is −2, x-intercepts is approximately 1.585.

8. Show that the value of f (x) approaches the value of g (x) as x increases withoutbound (x → ∞) graphically (Figure 74, page 152) and numerically.

f (x) =(

1 +3x

)x, g (x) = e3

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-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5

-4

-3

-2

-1

1

2

Figure 73: f (x) = 2x − 3

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150

-5

5

10

15

20

Figure 74: f (x) and g (x)

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Solution: You need to graph both f and g on the same coordinate axis and verify asx increases the two graphs appear to merge. You also need to create a table of valuesfor both f and g for large values of x and verify the difference between f and g tendto zero as x increases.

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18 Logarithmic Functions

18.0.1 Inverses of Exponentials

Here’s the graph (Figure 75, page 154) of y = 2x and its inverse.32

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

4

5

Figure 75: y = 2x and x = 2y

The function, x = 2y, is the inverse if y = 2x. However, since f (x) = 2x, and we knowit has an inverse, we need to know what f−1 (x) is in terms of the variable x. This inversefunction is written as:

f−1 (x) = log2 x.

So we have this basic relationship to consider.

y = loga x ⇒ ay = x

As before, a > 0 and a 6= 1, and y ∈ R, so x > 0. You really need to understand thisrelationship if you plan to fluidly move between logarithms and exponentials.

18.0.2 Special Bases

Logarithms, base 10, are usually called common, and were once used extensively to sim-plify computation. In fact the discovery of logarithms was a substantial event in speed-ing up computation, similar to the invention of the digital computer. In fact, once people

32You may want to review inverses at this point.

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learned to use logarithms they were able to rapidly compute, but now we use computersto rapidly compute. However, logarithms are still used, just not for computation.

The other special base is Euler’s number, represented by the letter e. Logarithms withthis base are usually called natural. Certainly e is a strange enough number and can bewritten as a limit.

e = limx→∞

(1 +

1x

)x≈ 2.718281828

Here’s the graph (Figure 76, page 155).

0 5 10 15 20

1

2

Figure 76: y =(

1 + 1x

)xand y = e

In most textbooks at this level, and on your calculators, the common logarithms arewritten as,

log x,

and the natural are written as,

ln x.

This, I believe, is due to the French (Mercator, although German, may have coined theterm in France), although European logarithms originated in Switzerland and Scotland.

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-6 -5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

4

5

Figure 77: Partial graph of y = 2x and x = 2y.

18.0.3 Graphing Logarithmic Functions

In the beginning of this section we graphed (Figure 77, page 156) both y = 2x and itsinverse on the same axis.

And we now know (or should be familiar with) that x = 2y can be rewritten asy = log2 x. You should also recall that when dealing with logarithmic functions, a ba-sic relationship should always be considered. That basic relationship is:

y = loga x ⇔ ay = x

As before, a > 0 and a 6= 1, and y ∈ R, so x > 0.Okay, let’s graph a simple logarithmic function.

y = f (x) = log3 (x− 2) + 1

First, I strongly suggest that you rewrite the logarithmic function as an exponential, orat least until you become more experienced with logarithmic functions. This is typicallybest done by solving for x in simple steps, as follows.

y = log3 (x− 2) + 1y− 1 = log3 (x− 2)

3y−1 = x− 23y−1 + 2 = x

Now, pick simple values for y and evaluate to find x. For example, if y = 1 we have:

31−1 + 2 = 30 + 2 = 1 + 2 = 3 = x.

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Continue to pick easy points and you’ll eventually see a pattern. And the pattern shouldlook like this graph (Figure 78, page 157).

0 1 2 3 4

-1

1

2

Figure 78: Partial graph of y = log3 (x− 2) + 1.

18.1 Properties of Logarithms

18.1.1 Basic Rules

Product Rule: For any positive numbers M and N, and any logarithmic base a,

loga (M · N) = loga M + loga N.

Proof: Let

loga M = x and loga N = y.

Now rewrite these as exponentials.

M = ax and N = ay.

We also know that

M · N = ax · ay = ax+y

Converting this back to logarithm

M · N = ax+y ⇒ loga (M · N) = x + y

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So we have, after looking back,

loga (M · N) = loga M + loga N.

Q.E.D.This product rule can easily be extended to include powers. The Power Rule states,

that for any positive numbers M, and any logarithmic base a, and any real number p,

loga Mp = p · loga M.

Now using the power and product rules, we can include subtraction. The SubtractionRule states, that for any positive numbers M and N, and any logarithmic base a,

logaMN

= loga M− loga N.

18.1.2 Changing Bases

You will have to evaluate logs for a variety of bases, and on occasion you will need to useyour calculator to carry out the computation. However your calculator most likely onlyhas natural (base e) and common logs (base 10), so what should you do if you’re givenanother base? Here goes . . .

loga x = yx = ay

ln x = ln ay

ln x = y ln a

y =ln xln a

This leads us to a simple change of base formula.

loga x =ln xln a

=log xlog a

18.2 Examples

1. Use the definition of logarithmic function to evaluate log10 100.


log10 100 = 2

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2. Evaluate each of the following.

(a) ln e

Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.

ln e = 1

(b) ln e2


ln e2 = 2

(c) ln e3


ln e3 = 3

(d) ln en


ln en = n

(e) log 10


log 10 = 1

(f) log 102


log 102 = 2

(g) log 103

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log 103 = 3

(h) log 10n


log 10n = n

(i) eln 2


eln 2 = 2

3. What exponential equation is equivalent to the logarithmic equation loga b = c?


ac = b



log9 3 =12

5. Graph

y = f (x) = 1− 12

log2 (1− x) ,

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being sure to indicate key features including asymptote(s).


Your graph (Figure 79, page 161) should look like mine.

-3 -2 -1 0 1

1

2

3

Figure 79: Partial graph of y = 1− 12 log2 (1− x).

6. Use a graphing utility to find the domain, range, x and y-intercept, any asymptotesof the logarithmic function and sketch its graph.

y =12 ln

(x2 + 1

)x− 2


Domain is all reals x 6= 2; x and y-intercept are both 0; vertical asymptote is x = 2;horizontal asymptote is y = 0. Your graph (Figure 80, page 162) should look likemine.

of 254




-35 -30 -25 -20 -15 -10 -5 0 5 10 15 20 25 30 35 40

-25

-20

-15

-10

-5

5

10

15

20

25

Figure 80: Partial graph of y =12 ln

(x2 + 1

)x− 2

.

7. Evaluate the logarithm using the change-of-base formula. Round your result to threedecimal places.

log11 4


log11 4 =ln 4

ln 11≈ 0.578



log2 8 = 3

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9. Use the change-of-base formula and a graphing utility to graph the function.

f (x) = log4 (x− 1)


f (x) = log4 (x− 1) =ln (x− 1)

ln 4=

log (x− 1)log 4


-1 0 1 2 3 4 5 6 7 8

-3

-2

-1

1

2

Figure 81: Partial graph of f (x) = log4 (x− 1).

10. Expand.

log7x2y3

z4

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log7x2y3

z4 = 2 log7 x + 3 log7 y− 4 log7 z

11. Condense.

5 log x− 3 log y +12

log z


5 log x− 3 log y +12

log z = logx5√z

y3

12. Consider the following.

y1 = 2[ln 5 + ln

(x2 + 1

)], y2 = ln

[25(

x2 + 1)2]

(a) Use a graphing utility to graph the two equations in the same viewing window.

Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.Graphs (Figure 82, page 165) are identical.

(b) Are these two graphs identical? Explain in English.

Solution: Work will be done in class. Make sure you take notes, or get notesfrom a reliable classmate if you are not present during lecture.The statements are equivalent. The process of simplification has no restrictions.This is an identity for all real numbers.

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-20 -19 -18 -17 -16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

-2-1

1234567891011121314151617181920212223

Figure 82: Partial graph of y1 and y2.








log6 1 = 0

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2. The model

t = 16.625 · ln xx− 750

, x > 750

approximates the length of a home mortgage of $150, 000 at 6% in terms of the monthlypayment. In the model, t is the length of the mortgage in years and x is the monthlypayment in dollars.

(a) Use the model to approximate the lengths of a $150, 000 mortgage at 6% whenthe monthly payment is $897.72 and when the monthly payment is $1, 659.21.(Round your answers to the nearest year.)


t1 = 16.625 · ln 897.72897.72− 750

≈ 30

t2 = 16.625 · ln 1659.211659.21− 750

≈ 10

(b) Approximate the total amounts paid over the term of the mortgage with a monthlypayment of $897.72 and with a monthly payment of $1,659.21. (Round your an-swers to nearest dollar.)

Solution:

T1 = 16.625 · ln 897.72897.72− 750

· 12 · 897.72 = 323184.28

T2 = 16.625 · ln 1659.211659.21− 750

· 12 · 1659.21 = 199110.83

(c) What amount of the total is interest costs for each payment? (Round your an-swers to nearest cent.)

Solution:I1 = 323184.28− 150000 = 173184.28I2 = 199110.83− 150000 = 49110.83

(d) Use trial and error (and your head) to figure out the monthly payment that willpay the loan off in fifteen years.

Solution: A good place to start would be between 897.72 and 1659.21, and afterseveral tries you should arrive at 1261.89.

3. Use the definition of logarithmic function to evaluate the function at the indicatedvalue of x without using a calculator.

f (x) = log36 x, x =16

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Solution:

f(

16

)= log36

16= −1

2

4. Use the properties of logarithms to rewrite and simplify the logarithmic expression.

log9 27

Solution:

log9 27 = 3 log9 3 = 3 log9

√9 =

32

5. Rewrite the expression in terms of ln 4 and ln 5. Verify the equivalence using yourcalculator.

ln 1600

Solution:

ln 1600 = ln(

5243)= 2 ln 5 + 3 ln 4

6. Condense.

12

log2 x + 4 log2 y− 3 log2 z

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Solution:

12

log2 x + 4 log2 y− 3 log2 z = log2y4√x

z3

7. Evaluate the logarithm using the change-of-base formula. Round your result to threedecimal places.

log637

Solution:

log637=

ln 3/7ln 6

≈ −0.473

8. Consider the following.

y1 = 4 (ln 2 + ln x) , y2 = ln 16x4

(a) Use a graphing utility to graph the two equations in the same viewing window.

Solution: These graphs (Figure 83, page 169) are not identical. The curve is y1is only to the right of the y-axis and y2 lies on top of this, as well as the reflectionof y1 about the y axis.

(b) Are these two graphs identical? Explain in English.

Solution: The domain for y1 is x > 0 and the domain for y2 is all real numbersx 6= 0. The equivalence only works for x > 0 and that’s why the graphs onlyagree there.

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-20 -19 -18 -17 -16 -15 -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

-2-1

1234567891011121314151617181920212223

Figure 83: Partial graph of y1 and y2.

9. Expand.

log113

√x2z3

y5

Solution:

log113

√x2z3

y5 =13[2 log11 x + 3 log11 z− 5 log11 y] =

23

log11 x + log11 z− 53

log11 y

10. Find the domain, x-intercept, and vertical asymptote of the logarithmic function andsketch its graph.

f (x) = ln (x + 3)

Solution: Domain is x > −3; x-intercept is −2; vertical asymptote is x = −3. Yourgraph (Figure 84, page 170) should look like mine.

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-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

Figure 84: Partial graph of f (x) = ln (x + 3).

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19 Exponential and Logarithmic Equations

19.1 Equations

19.1.1 Exact and Approximate

There’s a big difference between exact and approximate answers. For example the exactsolution of the equation

x2 = 2

is

x = −√

2 or x =√

2.

However, if I were to ask for an approximate solution to this equation I would first haveto tell you how precise I want the approximate to be. For example, if I want the answer tothe nearest tenth, then the solution would be written as:

x ≈ −1.4 or x ≈ 1.4

but if I instead ask for six decimal places, then the solution would be written as:

x ≈ −1.414214 or x ≈ 1.414214

You should pay careful attention to this matter.

19.1.2 Solving Exponential Equations

For any a > 0, and a 6= 1,

ax = ay ⇔ x = y.

This should be somewhat obvious. However if a 6= b will will need to use logs. For anya, b > 0, and a, b 6= 1,

ax = b ⇔ x =ln bln a

.

This for most students is not obvious and I will discuss this in detail in class.

19.1.3 Solving Logarithmic Equations

For any a > 0, and a 6= 1, (recall that x and y must be positive)

loga x = loga y ⇔ x = y.

This should be somewhat obvious. Furthermore, you should not forget how to rewrite alogarithmic equation as an exponential.

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As before, you should be especially mindful about the difference between exact andapproximate solutions. And please don’t forget to check your solutions once obtained—some may be extraneous. For example, here’s a problem from MTH 100 that has an extra-neous solution.

√x + 2 = x square both sidesx + 2 = x2 solve for zero

0 = x2 − x− 2 factor0 = (x + 1) (x− 2)

The last line above has the solution x = −1 or x = 2; however, if you check these solutionsin the original problem you should quickly note that x = −1 does not work! Hence wesay that x = −1 is an extraneous solution. Finally the solution to the equation

√x + 2 = x

is x = 2 only.

19.2 Examples

1. Solve for x.

2x = 16


x = 4

2. Solve for x.

log5 (2x− 1) = log5 (3x− 4)


x = 3

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3. Solve for x, both exact and approximate (4 decimal places).

ln 2x = 3


x =e3

2≈ 10.0428

4. Solve for x.

ln (x + 1)2 = 0


x = 0, −2

5. Solve for x.

log4 x− log4 (x− 1) =12


x = 2

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6. Solve for x.

27 = 35x9x2


x =12

, −3

7. Use a graphing utility to graph f (x) = 8 log4 x and g (x) = 3 in the same viewingwindow. Approximate the point of intersection of the graphs of f and g. Then solvethe equation of f (x) = g (x) algebraically. (Round your answer to three decimalplaces.)


x = 43/8 ≈ 1.682


8. Solve for x. Both exact and and approximate (4 decimal places).

e2x − 5ex + 6 = 0


Let u = ex.

e2x − 5ex + 6 = 0u2 − 5u + 6 = 0

(u− 2) (u− 3) = 0

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Figure 85: Partial graph of f and g with point of intersection indicated.

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Using the zero-product rule we get.

u = 2 = ex

u = 3 = ex

Solving for x.

x = ln 2 ≈ 0.6931, ln 3 ≈ 1.0986

9. Solve for x. Both exact and and approximate (4 decimal places).

5251 + e−x = 325


5251 + e−x = 325

525325

= 1 + e−x

2113

= 1 + e−x

813

= e−x

138

= ex

ln138

= x

x = ln138≈ 0.4855

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10. Solve for x.

−x2e−x + 3xe−x = 0


−x2e−x + 3xe−x = 03xe−x − x2e−x = 0

xe−x (3− x) = 0

Using the zero-product rule we get.

x = 0, 3






1. Solve for x. Both exact and an approximate (4 decimal places).

2x = 5

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Solution:

x =ln 5ln 2

=log 5log 2

≈ 2.3219

2. Solve for x.

3 + 2 log10 x = 5

Solution:

x = 10

3. Solve for x.

log3 x + log3 (x− 8) = 2

Solution:

x = 9

Note that x = −1 does not work; it is extraneous.

4. Solve for x.

43x−5 = 16

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Solution:

x =73

5. Solve for x.

ln x = −2

Solution:

x = e−2

6. Solve for x.

ln (x + 1) + ln (x− 2) = ln x

Solution:

x = 1 +√

3

Note that x = 1−√

3 does not work; it is extraneous.

7. Solve for x.

3x2+4x =1

27

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Solution:

x = −1, −3

8. This one’s a bit tricky—try it though. Graphing (Figure 86, page 180) may help!

5x − 5−x

5x + 5−x =23

Solution:

x =12

-2 -1 0 1 2

-1

1

Figure 86: Partial graph of y = 2/3 and y = (5x − 5−x) (5x + 5−x)

9. This one’s a bit tricky too—try it though. Both exact and an approximate (4 decimalplaces).

ex − 6e−x = 1

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Solution:

x = ln 3 ≈ 1.0986

10. Find the time t required for a $1050 investment to double (that is, become $2100) atinterest rate r = 5%, compounded continuously. Round your results to two decimalplaces.

Solution:

2 · 1050 = 1050 · e0.05t, t =ln 20.05

≈ 13.86

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30 Systems of Linear Equations

As you may recall from MTH 100, you solved linear systems with two variables—usuallyx and y, but other variables may have been used—of the following form.

2x + 3y = 55x − 2y = −16

The methods that you used were: elimination, substitution, and graphical (Figure 87,page 182). We will review these three methods in class, and you should be very familiarwith these three methods as they were extensively covered in MTH 100.

Here’s a graph of this linear system to start us off.

-4 -3 -2 -1 0 1 2 3 4 5

-2

-1

1

2

3

4

5

6

Figure 87: Linear system, 2x + 3y = 5 and 5x− 2y = −16.

You should be able to read the solution directly from the graph; solve by substitution;and solve by addition elimination. These methods will be briefly reviewed.

30.1 Terminology

Systems of linear of linear equations fall into three categories.

Consistent and Independent: A system with exactly one solution.

Inconsistent: A system with no solution.

Dependent: A system with an infinite number of solutions.

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30.2 Examples

1. Solve by graphing. (If equations are independent, give answer as a point; If the equa-tions are dependent, enter a for the value of x and y in terms of a; if the system isinconsistent, just state inconsistent.)

x + y = −3x − y = 5


Answer: (1, −4). See graph (Figure 88, page 183).

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5

-5

-4

-3

-2

-1

1

Figure 88: Partial graph of system indicating the point of intersection.

2. Solve by substitution. (If equations are independent, give answer as a point; If theequations are dependent, enter a for the value of x and y in terms of a; if the systemis inconsistent, just state inconsistent.)

2x − 4y = 0y = x + 1

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Answer: (−2, −1). See graph (Figure 89, page 184).

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5

-5

-4

-3

-2

-1

1


3. Solve by the addition method. (If equations are independent, give answer as a point;If the equations are dependent, enter a for the value of x and y in terms of a; if thesystem is inconsistent, just state inconsistent.)

2x + 3y = 45x + 2y = −1


Answer: (−1, 2). See graph (Figure 90, page 185).



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-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

-3

-2

-1

1

2

3





1. Solve by graphing. (If equations are independent, give answer as a point; If the equa-tions are dependent, enter x for the value of x and y in terms of x; if the system isinconsistent, just state inconsistent.)

2x + 3y = 6−3x − y = 5

Solution:

Answer: (−3, 4)

2. Solve by substitution. (If equations are independent, give answer as a point; If theequations are dependent, enter x for the value of x and y in terms of x; if the systemis inconsistent, just state inconsistent.)

2x + 3y = 6−3x − y = 5

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Solution:

Answer: (−3, 4)

3. Solve by the addition method. (If equations are independent, give answer as a point;If the equations are dependent, enter x for the value of x and y in terms of x; if thesystem is inconsistent, just state inconsistent.)

2x + 3y = 6−3x − y = 5

Solution:

Answer: (−3, 4)

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31 Gaussian and Gauss-Jordan Elimination

31.0.1 Introduction to Linear Algebra: Linear 2× 2 System

Here we will have two linear equations, with two unknowns. Graphically we know thatthe lines may intersect once, always intersect (same line), or possibly never intersect (dif-ferent parallel lines). Here’s some typical terms (introduced in MTH 100) that apply tolinear systems:

Consistent: At least one solution. The term Independent is often used here and indicatesthat the system is not dependent.

Inconsistent: No solution.

Dependent: Infinite number of solutions.

The basics of Gaussian elimination was extensively covered in MTH 100, and the methodhere is no different. It essentially relies on the following.

1. Interchange any two equations.

2. Multiply both sides of one equation by a non-zero constant.

3. Add a nonzero multiple of one equation to another equation.

Although the method I’ll present today is no different, the essential structure will need tobe adapted to allow for extending the system to more variables and equations. I assureyou however, that the method is not different from what you learned in MTH 100. Forexample, you might be asked to solve.

2x − 3y = 7x + 4y = −2

But now to speed the process we will write this system using matrix notation, as[2 −3 71 4 −2

]This matrix is called an augmented matrix. It’s dimension is 2× 3, that is we report thedimension of the augmented matrix to be the number of rows× the number of columns.For example, the coefficient matrix of this example is 2× 2 (this is referred to as a squarematrix33)and looks like[

2 −31 4

].

The constant matrix of this example is a single column and looks like[7−2

],

33Whenever the number of rows equal the number of columns.

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and its dimension is 2× 1.You should clearly make the connection between augmented matrix and the original

system of equations. It is also important to note that these two forms are identical ininformation, as long as we know the first column corresponds to x; the second columncorresponds to y; the vertical bar corresponds to an equal sign; and finally the last columncorresponds to the constants. Now, instead of operating on equations, we’re operating onrows. The allowable operations (mention above) are now called row-equivalent opera-tions, and that’s all we’re going to use to rewrite our systems into more readable formats.

1. Interchange any two rows.

2. Multiply each entry of a row by a non-zero constant.

3. Add a nonzero multiple of one row to another row.

Let’s proceed using these operations. (This will be done in class.)[2 −3 71 4 −2

]∼[

2 −3 70 1 −1

]∼[

1 0 20 1 −1

]The last matrix is no different than that of the others, but it is very readable. The lastmatrix simple states that x = 2 and y = −1.

A matrix that has undergone this process of Gaussian elimination is said to be inreduced row echelon form. Such a matrix has the following characteristics:

1. All zero rows are at the bottom of the matrix.

2. The leading entry of each nonzero row after the first occurs to the right of the leadingentry of the previous row.

3. The leading entry in any nonzero row is 1.

4. All entries in the column above and below a leading 1 are zero.

Another common definition is row echelon form and that only requires zeros below theleading ones, while the above definition also requires them above the leading ones. Bythe way, your calculators are quite capable of getting a matrix into either reduced rowechelon form (rref) or row echelon form (ref).34

31.1 Linear 3× 3 System

Now that we’ve just done a 2 × 2 system, you should be ready to follow the methodforward.35 Be aware that you will have solve this problem—and others to follow—by

34I strongly suggest that you look at your calculator’s manual to figure this out.35Don’t worry, I’ll be reasonable.

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using an augmented matrix and elementary row operations. Here’s a 3 × 3 system tosolve. Again we’ll do this in class, and I will slowly show the details.

2x + 3y − z = −73x − 3y + z = 122x + 4y + z = −3

Augmented matrix form of the system: 2 3 −1 −73 −3 1 122 4 1 −3

Elementary row operations, in order given:

R1 + R2 → R2

R1 + R3 → R3

Produces: 2 3 −1 −73 −3 1 122 4 1 −3

∼ 2 3 −1 −7

5 0 0 54 7 0 −10

The second row gives x = 1; using x = 1 in row three, gives y = −2; finally, using x = 1and y = −2 in row one, gives z = 3. Now that we know the solution we can actually goone step further and write the matrix in a more readable format. 2 3 −1 −7

3 −3 1 122 4 1 −3

∼ 2 3 −1 −7

5 0 0 54 7 0 −10

∼ 1 0 0 1

0 1 0 −20 0 1 3

This final form is called reduced row echelon form, and your calculator can actually doit. However, you may be required to do this by hand, and you must show work to get fullcredit. At first it looks impossible, but with practice you will be able to fly through theseproblems.

31.2 Examples

1. Identify the elementary row operation performed to obtain the new row-equivalentmatrix.[

3 −1 −4−4 7 9

]∼[

3 −1 −48 3 −7

]

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4R1 + R2 → R2

2. Use the matrix capabilities of a graphing utility to write the matrix in reduced rowechelon form. 1 3 2

20 60 362 6 10


1 3 00 0 10 0 0

3. Use the indicated row operations (in order given) to write the matrix in reduced rowechelon form. 1 3 2

20 60 362 6 10

−20R1 + R2 → R2

−2R1 + R3 → R3

−14

R2 → R2

16

R3 → R3

−2R2 + R1 → R1

−R2 + R3 → R3

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1 3 00 0 10 0 0

4. Write the system of linear equations represented by the augmented matrix. Then useback-substitution to solve. (Use the variables x, y, and z, if applicable.)

1 2 −2 30 1 1 10 0 1 3


x + 2y− 2z = 3y + z = 1

z = 3

Using back-substitution, z = 3, y = −2, and x = 13.

5. Solve the following system using a augmented matrix and row-equivalent operations.Your final augmented matrix should be in reduced row echelon form.

x + 2y − 3z = −82x − y + 2z = 53x − y − 4z = −7


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x = 0, y = −1, and z = 2. 1 0 0 00 1 0 −10 0 1 2

6. Solve the following system.x + y − z = 5−x + 2y + 3z = 1

3y + 2z = 6


x =9 + 5a

3

y =6− 2a

3z = a

7. Use a system of equations to find the quadratic function f (x) = Ax2 + Bx + C thatsatisfies the equations. Solve the system using matrices.

f (−2) = −15, f (1) = −6, f (2) = −23


f (x) = −5x2 − 2x + 1



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1. Write the system of linear equations represented by the augmented matrix. Then useback-substitution to solve. (Use the variables x, y, and z, if applicable.)

1 2 2 −30 1 1 −110 0 1 4

Solution:

x + 2y + 2z = −3y + z = −11

z = 4

Using back-substitution, z = 4, y = −15, and x = 19.

2. Solve the following system using a augmented matrix and row-equivalent operations.Your final augmented matrix should be in reduced row echelon form.

2x − y + 4z = 11x − 2y − 10z = −17

3x + 4z = 11

Solution:

x = 1, y = −1, and z = 2. 1 0 0 10 1 0 −10 0 1 2

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3. Solve the following system.2x − 3y − z = −73x − 4y + z = −2−x + 2y − 5z = −12

Solution:

x = 1y = 2z = 3

4. Solve the following system.x + y + 2z = 3

y − z = 2x + 2y + z = 5

Solution:

x = 1− 3ay = 2 + az = a

5. A corporation borrowed $1.49 million to expand its line of shoes. Some of the moneywas borrowed at 3%, some at 4%, and some at 6%. Use a system of equations todetermine how much was borrowed at each rate given that the annual interest was$72, 200 and the amount borrowed at 6% was four times the amount borrowed at 3%.Solve the system using matrices.

Solution: $180, 000 at 3%; $590, 000 at 4%; $720, 000 at 6%. I strongly suggest youlearn how to set-up problems using matrices, and to use your calculators to put yourmatrix into reduced row echelon form. It’s the reading that causes the most studentstrouble—reading is fundamental to your eventual success.

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32 Partial Fraction Decomposition

32.0.1 The Pre-Calculus of Partial Fraction Decomposition

Given a rational function in lowest terms of the formP (x)Q (x)

.

To do a partial fraction decomposition you’ll need to follow the steps below.

1. If the degree of the numerator is greater than or equal to the denominator, longdivide. You may be left with a rational function. For example:

6x3 + 5x2 − 73x2 − 2x− 1

= 2x + 3 +8x− 4

3x2 − 2x− 1.

2. If the degree of the numerator is less than the degree of the denominator, then factorthe denominator into linear factors or irreducible36 quadratic factors. An irreduciblequadratic factor is a quadratic that cannot not be factored into linear factors with realcoefficients. Example continued from above.

6x3 + 5x2 − 73x2 − 2x− 1

= 2x + 3 +8x− 4

(3x + 1) (x− 1).

3. Assign each linear factor (px + q)n the sum of n partial fractions:

A1

(px + q)1 +A2

(px + q)2 +A3

(px + q)3 + · · ·+ An

(px + q)n

Continuing with the example above.

6x3 + 5x2 − 73x2 − 2x− 1

= 2x + 3 +8x− 4

(3x + 1) (x− 1)= 2x + 3 +

k1

(3x + 1)+

k2

(x− 1)

Of course we will need to determine the values of the constants k1 and k2.

4. Assign each irreducible quadratic factor(ax2 + bx + c

)m the sum of m partial frac-tions:

B1x + C1

(ax2 + bx + c)1 +B2x + C2

(ax2 + bx + c)2 +B3x + C3

(ax2 + bx + c)3 + · · ·+ Bmx + Cm

(ax2 + bx + c)m .

Again you’ll need to determine these constants.

5. This all looks nightmarish, I know, but the algebra—at least for the problems thatwe’ll be doing—is really quite easy.

36A quadratic polynomial ax2 + bx + c is called irreducible if it cannot be factored into linear factorswithout using complex numbers. e.g. x2 + 1 = (x− i) (x + i)

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32.0.2 Examples

1. Decompose into partial fractions.

4x− 132x2 + x− 6


The first step is to break the fraction into two parts.

4x− 132x2 + x− 6

=A

x + 2+

B2x− 3

Now add the fractions on the right hand side together.

4x− 132x2 + x− 6

=A (2x− 3) + B (x + 2)

2x2 + x− 6

Now equate the numerators, we really no longer need to worry about the denomina-tor, after-all they’re the same. Basically we just need to find the constants that makethese numerators equal. You should, of course, pick values for x that make solvingfor A and B easy. Let’s see what values for x make our job easier.

4x− 13 = A (2x− 3) + B (x + 2)

I say we choose x = −2 to see what happens.

4x− 13 = A (2x− 3) + B (x + 2)−8− 13 = A (−4− 3) + B (−2 + 2)−21 = −7A

3 = A

Okay, now that we know that A = 3, let’s find B by choosing an easy value for x,let’s try x = 0, but any value for x will do just fine.

4x− 13 = 3 (2x− 3) + B (x + 2)−13 = 3 (−3) + B (2)−13 = −9 + 2B−4 = 2B−2 = B

Now we finally have it.

4x− 132x2 + x− 6

=3

x + 2− 2

2x− 3

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6x3 + 5x2 − 73x2 − 2x− 1


Since the degree of the numerator is bigger than the denominator we will first dolong division.

6x3 + 5x2 − 73x2 − 2x− 1

= 2x + 3 +8x− 4

3x2 − 2x− 1.

Now break the fraction into parts.

6x3 + 5x2 − 73x2 − 2x− 1

= 2x + 3 +8x− 4

(3x + 1) (x− 1)= 2x + 3 +

A(3x + 1)

+B

(x− 1).

Now let’s determine the constants A and B.

8x− 4 = A (x− 1) + B (3x + 1)

Let x = 1 and you’ll get:

4 = 4B, ⇒ B = 1.

Now let x = 0 and you’ll get:

−4 = −A + 1, ⇒ A = 5.

The we have:

6x3 + 5x2 − 73x2 − 2x− 1

= 2x + 3 +8x− 4

3x2 − 2x− 1= 2x + 3 +

53x + 1

+1

x− 1


11− x2

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11− x2 =

A1− x

+B

1 + x

=1/2

1− x+

1/21 + x

We’ll do this in class.

32.0.3 Greek To Me!

Let’s be thankful that we have native Greek speakers (Yes, they’re here at ECC) to help usout when we get stuck. Don’t worry, I won’t be testing you on this.37 Oh, by the way, Iget confused by the use of Greek letter too! Greek to me so-to-speak.

Lower case greek letter used in mathematics. alpha = α, beta = β, gamma = γ, delta =δ, epsilon = ε, var epsilon = ε, zeta = ζ, eta = η =, theta = θ, var theta = ϑ, iota = ι, kappa =κ, lambda = λ, mu = µ, nu = ν, xi = ξ, pi = π, var pi = v, rho = ρ, var rho = $, sigma = σ,var sigma = ς, tau = τ, upsilon = υ, phi = φ, var phi = ϕ, chi = χ, psi = ψ, omega = ω.

Upper case Greek letters used in mathematics: Gamma = Γ, Delta = ∆, Theta = Θ,Lambda = Λ, Xi = Ξ, Pi = Π, Sigma = Σ, Upsilon = Υ, Phi = Φ, Psi = Ψ, Omega = Ω.






37No need to memoirize.

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1. Determine the values of A, and B.

4x2 − 1

=A

x− 1+

Bx + 1

Solution:

A = 2, B = −2

2. Determine the values of A, B, and C.

x2 + 7x− 2x3 − x

=Ax+

Bx− 1

+C

x + 1

Solution:

A = 2, B = 3, C = −4


x2 + 2(x− 1) (x− 4) (x + 2)

=A

x− 4+

Bx + 2

+C

x− 1

Solution:

A = 1, B =13

, C = −13

4. Long divide.

x3 + 1x2 − 4

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Solution:

x3 + 1x2 − 4

= x +4x + 1x2 − 4

5. Determine the values of A, and B.

x3 + 1x2 − 4

= x +4x + 1x2 − 4

= x +A

x + 2+

Bx− 2

Solution:

A =74

, B =94


3x− 9

(x− 1) (x + 2)2 =A

x + 2+

Bx− 1

+C

(x + 2)2

Solution:

A =23

, B = −23

, C = 5


18(x + 3) (x2 + 9)

=A

x + 3+

Bx + Cx2 + 9

Solution:

A = 1, B = −1, C = 3


13x− 126x2 − 13x + 6

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Solution:

13x− 126x2 − 13x + 6

=2

3x− 2+

32x− 3


x− 9(x2 + 4) (x + 1)

Solution:

x− 9(x2 + 4) (x + 1)

=2x− 1x2 + 4

− 2x + 1


4x− 1x3 − 1

Solution:

4x− 1x3 − 1

=1

x− 1− x− 2

x2 + x + 1

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33 Nonlinear Systems of Equations

33.1 Introduction to Non-linear Systems

The method of elimination works best for linear systems. However, a visual (Figure 91,page 202) along with substitution generally works best for non-linear systems. Here, itis strongly advised that you graph the system first. The use of a graphics calculator isencouraged, but not entirely necessary for many systems.

-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

Figure 91: Where to these two graphs (red and green curves) intersect?

Looking at the visual (Figure 91, page 202) above you should note that two curvesare intersecting at four distinct points. Not saying these points are easy to determine,but you should have some idea that we can read an approximate from a good graph. Yes,your calculators are quite capable of finding points-of-intersection, at least approximately,however, we are very interested in the exact values if they can be found. Simple exampleswill follow.

33.2 Examples

1. Solve the non-linear system for x and y.x + y = 0

x3 − 5x = y

A graph (Figure 92, page 203) of this non-linear system is provided as a guide. Unlikea linear system, having a visual is very important. Here, we at least know that thereare three points of intersection.

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-3 -2 -1 0 1 2 3

-5

-4

-3

-2

-1

1

2

3

4

5

Figure 92: Linear system, x + y = 0 and x3 − 5x = y.


You should be able to visually determine the points of intersection and then verifythem. They appear to be (−2, 2), (0, 0), and (2, −2), and when these values aresubstituted into the original system they all work. Now let’s do the algebra to seewhat we get. Solving the first equation for y we get y = −x, now we will justsubstitute this into the second equation and solve for x.

y = −xx3 − 5x = −xx3 − 4x = 0

x(

x2 − 4)

= 0

Yes, the solutions are x = −2, x = 0, x = 2 which gives the expected points (−2, 2),(0, 0), and (2, −2).

2. Solve the non-linear system for x and y.2 − 2x2 = y

2x4 − 4x2 = y − 2

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Again, here’s a graph (Figure 93, page 204) of this non-linear system. Unlike linearsystem, having a visual is very important. Here, we at least know that there are threepoints of intersection.

-1 0 1

1

2

Figure 93: Linear system, 2− 2x2 = y and 2x4 − 4x2 = y− 2.


You should be able to visually determine the points of intersection and then verifythem. They appear to be (−1, 0), (0, 2), and (1, 0), and when these values aresubstituted into the original system they all work. Now let’s do the algebra to seewhat we get. The first equation is already solved for y = 2− 2x2, now we will justsubstitute this into the second equation and solve for x.

y = 2− 2x2

2x4 − 4x2 = 2− 2x2 − 22x4 − 2x2 = 0

Yes, the solutions are x = −1, x = 0, x = 1 which gives the expected points (−1, 0),(0, 2), and (1, 0).

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33.4 Homework: MTH119 HW Sec 33



1. Solve for x and y. Use both a graph and algebra.

y = x− 20 = x2 + y2 − 4x + 6y + 4

Solution: You’ve got a circle and a line that intersect at: (2, 0) and (−1, −3). Yourgraph (Figure 94, page 205) should look like mine.

-4 -3 -2 -1 0 1 2 3 4 5 6 7 8

-6

-5

-4

-3

-2

-1

1

Figure 94: Partial graph of y = x− 2 and 0 = x2 + y2 − 4x + 6y + 4.

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y = x2 − 1y = x + 1

Solution: You’ve got a parabola and a line that intersect at: (−1, 0) and (2, 3). Yourgraph (Figure 95, page 206) should look like mine.

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

-2

-1

1

2

3

4

Figure 95: Partial graph of y = x2 − 1 and y = x + 1.


y =√

xy = 2− x

Solution: You’ve got a half parabola and a line that intersect at: (1, 1). Your graph(Figure 96, page 207) should look like mine.


y = x2 − 8x + 16y = 6− x

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-1 0 1 2 3 4 5 6

1

2

3

Figure 96: Partial graph of y =√

x and y = 2− x.

Solution: You’ve got a parabola and a line that intersect at: (2, 4) and (5, 1). Yourgraph (Figure 97, page 207) should look like mine.

-3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11

1

2

3

4

5

6

Figure 97: Partial graph of y = x2 − 8x + 16 and y = 6− x.


xy− 2 = 03x− 2y = −4

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Solution: You’ve got a line and a rational that intersect at: (2/3, 3) and (−2, −1).Your graph (Figure 98, page 208) should look like mine.

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

-3

-2

-1

1

2

3

4

Figure 98: Partial graph of xy− 2 = 0 and 3x− 2y = −4.

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y = 2x− 1

y =√

x + 1

Solution: You’ve got a line and a square root function that intersect at: (5/4, 3/2).Your graph (Figure 99, page 209) should look like mine.

-2 -1 0 1 2 3 4

-1

1

2

Figure 99: Partial graph of y = 2x− 1 and y =√

x + 1.

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34 Introduction to Matrix Alegebra

34.1 Equal Matrices

Two matrices are equal if they are of the same dimension and corresponding entries areequal. For example 2 3 −1

3 −3 12 4 1

=

2 3 −13 −3 12 4 1

,

are equal. If you’re given 2x 3 −13 y− 2 12 4 −6

=

x− 1 3 −13 3− y 12 4 x− 2y

,

then x = −1 and y = 5/2 is a necessary condition for equality.

34.2 Matrix Arithmetic

You’ll be asked to add matrices, and to multiply matrices by scalars38. We’ll keep thearithmetic very simple.

34.2.1 Matrix Addition

If we have two matrices of the same dimensions, we can add them together by simplyadding their corresponding entries. For example. −1 −3 2

5 4 −5−3 2 −1

+

2 3 50 −5 71 −1 −1

=

1 0 75 −1 2−2 1 −2

34.2.2 Scalar Multiplication

Again, a scalar is a real number, and when we multiply a matrix by a scalar we are simplymultiplying each element of the matrix by the scalar. For example.

5 ·

−1 −3 25 4 −5−3 2 −1

=

−5 −15 1025 20 −25−15 10 −5

38In mathematics, real numbers are called scalar. They’re used here to simply scale matrices through scalar

multiplication.

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34.2.3 Simple Properties

Given that A, B, and C are matrices of the same dimensions, and c and d are scalars. Wehave the following:

1. Associative Property of Scalar Multiplication

(cd)A = c (dA)

2. Commutative Property of Matrix Addition

A + B = B + A

3. Scalar Identity

1A = A

4. Distributive Property

c (A + B) = cA + cB

5. Associative Property of Matrix Addition

A + (B + C) = (A + B) + C

34.3 Examples

1. Find x and y. x + 2 −4 −3−2 2y 2x−8 −1 y + 2

=

2x + 6 −4 −3−2 8 −8−8 −1 6


x = −4, y = 4

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2. Find, if possible, A + B, A − B, 3A, and 3A − 2B. Use the matrix capabilities of agraphing utility to verify your results.

A =

[−1 2 −3

5 0 4

], B =

[2 3 −50 1 5

]


A + B =

[1 5 −85 1 9

]A− B =

[−3 −1 2

5 −1 −1

]3A =

[−3 6 −915 0 12

]3A− 2B =

[−7 0 115 −2 2

]






1. Find x and y. x2 − 1 3 y2 + 2y x xy2 −x −y

=

0 3 3−x −y xy2

2 1 −1

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Solution:

x = −1, y = 1

2. Find, if possible, A + B, B−A, −2A, and −2A + 3B. Use the matrix capabilities of agraphing utility to verify your results.

A =

−1 −2 34 −3 23 −4 5

, B =

2 −3 −50 1 −51 −1 2

Solution:

A + B =

1 −5 −24 −2 −34 −5 7

B−A =

3 −1 −8−4 4 −7−2 3 −3

−2A =

2 4 −6−8 6 −4−6 8 −10

−2A + 3B =

8 −5 −21−8 9 −19−3 5 −4

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35 Matrix Multiplication and Inverses

35.0.1 Something Old . . .

Suppose you’re asked to solve this system2x − 3y = 7

x + 4y = −2,

but not as we did in MTH 100. Instead used matrix notation, along with elementary rowoperations to rewrite the augmented matrix into reduced row echelon form.[

2 −3 71 4 −2

]∼[

2 −3 70 1 −1

]∼[

1 0 20 1 −1

]Again, the last matrix is no different than the others, but it is very readable. The lastmatrix simple states that x = 2 and y = −1. This final matrix is the reduced row echelonform.

35.0.2 Something New . . .

As before, we are now going to rewrite our original system in terms of matrices alone,but we’re actually going to use three separate matrices. The three matrices are as follows.

Coefficient Matrix Here we are going to just list the coefficients of the variables.

A =

[2 −31 4

]

Variable Matrix Here we are going to just list the variables.

X =

[xy

]

Constant Matrix Here we are going to just list the constants.

B =

[7−2

]

Our equation is now of the form:

A · X = B,

or, if we use the matrices we have:[2 −31 4

]·[

xy

]=

[7−2

].

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In class I will show that this matrix multiplication39 is equivalent to the original problem.Our matrix multiplications will be no different than this, and we will do other examples.What’s most interesting about this form is that we can now easily solve for X, by doingthe following.

A · X = BX = A−1 · B

This new matrix A−1 is called the inverse of A. Computing this inverse is not difficult,but it can become quite involved for larger matrices. For now, I will tell you what theinverse is.

A−1 =

[2 −31 4

]−1

=

[4/11 3/11−1/11 2/11

]Now try this multiplication (we’ll do it in class).[

2 −31 4

]·[

xy

]=

[7−2

][

2 −31 4

]−1

·[

2 −31 4

]·[

xy

]=

[2 −31 4

]−1

·[

7−2

][

4/11 3/11−1/11 2/11

]·[

2 −31 4

]·[

xy

]=

[4/11 3/11−1/11 2/11

]·[

7−2

][

1 00 1

]·[

xy

]=

[2−1

][

xy

]=

[2−1

]

Now we will verify that

A · X = B.

That is, verify[2 −31 4

]·[

2−1

]=

[7−2

]We’ll certainly discuss the basics of matrix multiplication in class. You’ll need to be ableto multiply matrices by hand and when necessary, using a calculator.

Now let’s take an even bigger example.2x + 3y − z = −73x − 3y + z = 122x + 4y + z = −3

39We’ll be doing this by example and I also suggest that your learn how to do this on your calculators.

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Here goes . . . 2 3 −1 −73 −3 1 122 4 1 −3

∼ 2 3 −1 −7

5 0 0 54 7 0 −10

∼ 1 0 0 1

0 1 0 −20 0 1 3

Again x = 1; y = −2; and z = 3.

Now however, I want to rewrite it using matrix multiplication. As before, we are nowgoing to rewrite our original system in terms of matrices alone, but we’re actually goingto use three separate matrices. The three matrices are as follows.

Coefficient Matrix Here we are going to just list the coefficients of the variables.

A =

2 3 −13 −3 12 4 1

Variable Matrix Here we are going to just list the variables.

X =

xyz

Constant Matrix Here we are going to just list the constants.

B =

−712−3

Our equation is now of the form:

A · X = B,

or, if we use the matrices we have: 2 3 −13 −3 12 4 1

· x

yz

=

−712−3

.

In class I will show that this matrix multiplication is equivalent to the original problem.Our matrix multiplications will be no different than this, and we will do other examples.What’s most interesting about this form is that we can now easily solve for X, by doingthe following.

A · X = BX = A−1 · B

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This new matrix A−1 is called the inverse of A. Computing this inverse is not difficult,but it can become quite involved for larger matrices. For now, I will tell you what theinverse is.

A−1 =

2 3 −13 −3 12 4 1

−1

=

1/5 1/5 01/35 −4/35 1/7

−18/35 2/35 3/7

Now try this multiplication (we’ll do it in class). 2 3 −1

3 −3 12 4 1

· x

yz

=

−712−3

x

yz

=

2 3 −13 −3 12 4 1

−1

·

−712−3

Yes, I know this looks difficult, but you can do this. x

yz

=

2 3 −13 −3 12 4 1

−1

·

−712−3

x

yz

=

1/5 1/5 01/35 −4/35 1/7

−18/35 2/35 3/7

· −7

12−3

x

yz

= ·

1−2

3

And once again you’ll need to verify that

A · X = B.

35.0.3 Inverse of a Square Matrix

If there exists an n× n matrix A−1 such that AA−1 = In = A−1A, then A−1 is called theinverse or inverse matrix of A. In is an n× n matrix whose entries are: ai=j = 1 (diagonal)and ai 6=j = 0, and is called the identity matrix. In has the property that InAn×n = An×nInIf a matrix A has an inverse, it is called invertible or nonsingular; if it does not have aninverse, it is called singular. Not all n× n matrices are invertible.

In this section we will cover how to verify if two n × n matrices are inverses of oneanother. And we will also develop a simple technique to find inverses of nonsingular ma-trices. However, in general, finding inverses can be quite painful and I strongly suggestyou learn how to find inverses using a calculator.

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I know that finding inverses may prove difficult for many students, but I will outlinea procedure in the examples that follow that will hopefully become second nature. How-ever, I do want to encourage the use of a graphics calculator to compute inverses, just asI have encouraged their use to compute reduced-row-echelon-form. Again, computationis doable by hand, but calculators sure do make computation painless.

35.1 Examples

1. Given

A =

−2 3 −54 −1 1−7 6 0

,

verify AI3 = I3A = A.


2. What is the dimension of AB given A is a 3× 2 matrix and B is a 2× 5 matrix?


3× 5

3. Find the product.

[3 1 04 −2 5

]·

0 1 2−1 −5 −1

0 1 −1


[3 1 04 −2 5

]·

0 1 2−1 −5 −1

0 1 −1

=

[−1 −2 5

2 19 5

]

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4. Show that B is the inverse of A. To show that B is the inverse of A, we need to showthat AB = I = BA.

A =

[3 −5−1 2

], B =

[2 51 3

]


AB = BA =

[1 00 1

]

5. Show that B is the inverse of A. To show that B is the inverse of A, we need to showthat AB = I = BA.

A =

1 0 −1−1 2 0

1 5 0

, B =17

0 −5 20 1 1−7 −5 2


AB = BA =

1 0 00 1 00 0 1

6. Solve using matrix multiplication and inverses.40

2x + 3y + z = −13x + 3y + z = 12x + 4y + z = −2

40Inverse will be given in class.

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x = 2, y = −1, z = −2

7. Solve using matrix multiplication and inverses.415x + 4y − 3z = 82x − y = 2

x + 4y + 7z = −6


x = 1, y = 0, z = −1

8. Find the inverse, if possible.[1 34 13

]


This method along with using your calculator will be throughly discussed in class. Iknow it’s difficult, and I will try to go easy on you with regards to finding inverseson exams. Nonetheless, a calculator certainly makes this process much easier.

[1 3 1 04 13 0 1

]∼[

1 0 13 −30 1 −4 1

]So the inverse is[

13 −3−4 1

].

41Inverse will be given in class.

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9. Find the inverse, if possible. 1 1 13 5 43 6 5


This method along with using your calculator will be throughly discussed in class. Iknow it’s difficult, and I will try to go easy on you with regards to finding inverseson exams. Nonetheless, a calculator certainly makes this process much easier.

1 1 1 1 0 03 5 4 0 1 03 6 5 0 0 1

∼ 1 0 0 1 1 −1

0 1 0 −3 2 −10 0 1 3 −3 2

So the inverse is 1 1 −1

−3 2 −13 −3 2

.






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[1 −3 2−5 1 −2

]·

1 2 3−3 −2 −1

4 −5 6

Solution:

[18 −2 18−16 −2 −28

]

2. In general, matrix multiplication is not commutative. Given

A =

1 −2 3−4 3 2

1 2 −3

, B =

0 1 −2−3 4 3

2 −1 0

,

verify

A · B 6= B ·A.

Solution: Make sure you know what to do here.

3. The formula for finding the inverse of a 2× 2 matrix

A =

[a bc d

]is

A−1 =1

ad− bc

[d −b−c a

],

provided ad− bc 6= 0. Use this formula to find the inverse of

A =

[7 6−5 −12

].

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Solution:

A−1 =

[2/9 1/9

−5/54 −7/54

].

4. If possible, find the value of the constant k such that B = A−1.

A =

[−4 6

3 5

], B =

[−5/38 3/19

k 2/19

]

Solution:

k =3

38

5. Find the inverse. 1 1 −11 −1 11 1 1

Solution:

1/2 1/2 00 −1/2 1/2

−1/2 0 1/2

6. Verify. 1/2 1/2 00 −1/2 1/2

−1/2 0 1/2

· 1 1 −1

1 −1 11 1 1

= I3 1 1 −11 −1 11 1 1

· 1/2 1/2 0

0 −1/2 1/2−1/2 0 1/2

= I3

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Solution: Yes, it does verify.

7. Solve using matrix multiplication and inverses.4x − 2y + 3z = 62x + 2y + 5z = 68x − 5y − 2z = 52

Solution:

x = 10, y = 8, z = −6

8. Solve using matrix multiplication and inverses.x + y − z = 0x − y + z = 2x + y + z = 6

Solution:

x = 1, y = 2, z = 3

9. Consider a company that specializes in potting soil. Each bag of potting soil forseedlings requires 2 units of sand, 1 unit of loam, and 1 unit of peat moss. Eachbag of potting soil for general potting requires 1 unit of sand, 2 units of loam, and 1unit of peat moss. Each bag of potting soil for hardwood plants requires 2 units ofsand, 2 units of loam, and 2 units of peat moss. Find the numbers of bags of the threetypes of potting soil that the company can produce with the given amounts of rawmaterials: 750 units of sand; 850 units of loam; 600 units of peat moss.

Solution: You need to be able to do problems like this using matrices! Setting upthe system is probably the toughest step for most students. Solving the system canbe done using a variety of methods, including matrix algebra involving inverses, orusing an augmented matrix to find the reduced row echelon form. These problemsare writen to have very nice integer solutions.

150 bags of seedling soil; 250 bags of general soil; and 100 bags for hardwood plants.

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36 Determinants and Cramer’s Rule

36.1 Determinants of a Square Matrix

You may recall that the formula for finding the inverse of a 2× 2 matrix

A =

[a bc d

]is

A−1 =1

ad− bc

[d −b−c a

],

provided ad− bc 6= 0.Here the matrix A is invertible if and only if ad− bc 6= 0. We called this number the

determinant of A. It certainly would be nice to have a similar result for bigger squarematrices. This determinant has many uses, but for now we’re just interested in using thedeterminant to determine if a square matrix is invertible or not. You may also recall thata square matrix that is not invertible is called singular (degenerate). A square matrix issingular if and only if its determinant is 0.42 First let us use the following notation for thedeterminant.

det[

a bc d

]=

∣∣∣∣ a bc d

∣∣∣∣ = ad− bc

To find the determinant of a bigger square matrix you will have to resort to variety oftechniques, many are outlined in your textbook and on the web, but I generally just wantyou to learn how to do so on your calculators.

36.2 Application of Matrices and Determinants

36.2.1 Invertibility

If the determinant of a square matrix is zero, this indicates that matrix does not have aninverse.

36.2.2 Area

Given the vertices of a triangle in the coordinate plane you can find the area using thethree vertices entered into a 3× 3 matrix (see below) and its corresponding determinant.This technique also works for for a parallelogram because the parallelogram is composedto two congruent triangles.

42A very good intuitive reason is that det (AB) = det (A)det (B). If AB = I then det (A)det (B) = 1 not0 so neither det (A) nor det (B) can be 0.

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For the there points (xa, ya), (xb, yb), (xc, yc), you can just put the x and y values inas the first two rows of the matrix, then fill the third row with ones: xa xb xc

ya yb yc1 1 1

.

The absolute value of the determinant is the area of the parallelogram, and 1/2 that valueis the area of the triangle.

36.2.3 Cramer’s Rule

Turing a simple method into a mechanical rule is often used to automate a repetitive task.Certainly, the methods we’ve covered so far have been pretty simple, however, at timeswe prefer to automate the simple procedure into a simple rule, and this is in fact whatCramer’s Rule is all about. So, say we have the two equation, with two unknowns (x andy are the variables, and the other letters are constants),

ax + by = cdx + ey = f

To solve for the variable y we would do the following: multiple the first equation by dand the second equation by −a.

adx + bdy = cd−adx − aey = −a f

Now add the two equations together and you’ll get,

(bd− ae) y = cd− a f

y =cd− a fbd− ae

=a f − cdae− bd

=

∣∣∣∣ a cd f

∣∣∣∣∣∣∣∣ a bd e

∣∣∣∣ =Dy

D

You should note the top and bottom sure do look like determinants (they’re being indi-cated as well). Now, let’s see what happens when we solve for x. Here we’ll multiply thefirst equation by e and the second equation by −b.

aex + bey = ce−bdx − bey = −b f

Now add the two equations together and you’ll get,

(ae− bd) x = ce− b f

x =ce− b fae− bd

=

∣∣∣∣ c bf e

∣∣∣∣∣∣∣∣ a bd e

∣∣∣∣ =Dx

D

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Once again you should note the top and bottom sure do look like determinants. In eachcase the bottom determinant is just the determinant of the coefficient matrix, whereasthe the top determinant is just the determinant if the column of variable’s coefficients arereplaced by the constants. You should note that the variable coefficients being used forDx don’t include x variables coefficients.

Remarkably this method can be extended to larger systems, and in fact would bea fairly simple procedure to program if you know a procedural language such as C orPython. However, it should be noted that Computer Algebra Systems (CAS) are alreadyadept at solving systems of equations. The main point here is to illustrate a careful anal-ysis of what is being done in general to develop a simple rule that can be applied asneeded.

36.3 Examples

1. Solve for x

det([

x 4−1 x

])= 20


x = ±4

2. Find the determinate of 1 1 −1−3 2 −1

3 −3 2

.


1

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3. Use a determinant to find the area of the triangle with the given vertices.

(−2, 3) , (2, 3) , (2, −6)


A = ±12

∣∣∣∣∣∣−2 2 2

3 3 −61 1 1

∣∣∣∣∣∣ = ±12[−36] = ±18

The area is 18 square units. This one is also easy to do using simple geometry andI encougage everyone to do so—that is, you’ll need to draw the triangle and thencompute its area using a simple grade school formula.

4. Find y such that the triangle has an area of 13 square units.

(−4, 2) , (−3, 5) , (−2, y)


A = ±12

∣∣∣∣∣∣−4 −3 −2

2 5 y1 1 1

∣∣∣∣∣∣ = ±12(y− 8) = 13

This results in two equations: y− 8 = 26 and y− 8 = −26.

y = −18, 34

5. Solve using Cramer’s Rule.2x + 3y = 6−3x − y = 5

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Solution:

y =Dy

D=

∣∣∣∣ 2 6−3 5

∣∣∣∣∣∣∣∣ 2 3−3 −1

∣∣∣∣ =10 + 18−2 + 9

= 4

x =Dx

D=

∣∣∣∣ 6 35 −1

∣∣∣∣∣∣∣∣ 2 3−3 −1

∣∣∣∣ =−6− 15−2 + 9

= −3

Answer: (−3, 4)

6. Solve using Cramer’s Rule.4x − 2y + 3z = 62x + 2y + 5z = 68x − 5y − 2z = 52

Solution:

z =Dz

D=

∣∣∣∣∣∣4 −2 62 2 68 −5 52

∣∣∣∣∣∣∣∣∣∣∣∣4 −2 32 2 58 −5 −2

∣∣∣∣∣∣=

492−82

= −6

Again, this may not be the best way to solve a system like this, but it sure does looklike it can easily be repeated by a machine.

y =Dy

D=

∣∣∣∣∣∣4 6 32 6 58 52 −2

∣∣∣∣∣∣∣∣∣∣∣∣4 −2 32 2 58 −5 −2

∣∣∣∣∣∣=−656−82

= 8

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x =Dx

D=

∣∣∣∣∣∣6 −2 36 2 5

52 −5 −2

∣∣∣∣∣∣∣∣∣∣∣∣4 −2 32 2 58 −5 −2

∣∣∣∣∣∣=−820−82

= 10

x = 10, y = 8, z = −6






1. Find the determinate of 3 1 −22 4 13 6 5

.

Solution:

35

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2. Solve for x

det([

2x 1−x x

])= 1

Solution:

x = −1,12

3. Use a determinant to find the area of the triangle with the given vertices.

(−7, −1) , (0, 3) , (4, 8)

Solution:

A = ±12

∣∣∣∣∣∣−7 0 4−1 3 8

1 1 1

∣∣∣∣∣∣ = ±12[19] = ±19

2

The area is192

square units.

4. Solve using Cramer’s Rule.x − 3y = 4

3x + 2y = 1

Solution: Answer: (1, −1)

5. Solve using Cramer’s Rule.2x + 3y = 6−3x − y = 5

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Solution: Answer: (−3, 4)

6. Solve using Cramer’s Rule.5x + 4y − 3z = 82x − y = 2

x + 4y + 7z = −6

Solution:

x = 1, y = 0, z = −1

7. Show for a 2× 2 that det (AB) = det (A)det (B).

Solution: You’ll need to show this.

det (AB) = det (A)det (B)

det([

a1 a2a3 a4

] [b1 b2b3 b4

])= det

([a1 a2a3 a4

])det

([b1 b2b3 b4

])

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37 Final Exam Review

These questions are being provided as review for the final exam. They will not be coveredin class, and if you need help with these problems I suggest you get help during myscheduled office hours.


y = x3 − x2

y = x− 1

Solution: You’ve got a cubic and a line that intersect at: (1, 0) and (−1, −2) .Your graph (Figure 100, page 233) should look like mine.

-3 -2 -1 0 1 2 3

-2

-1

1

Figure 100: Partial graph of y = x3 − x2 and y = x− 1.

2. Find the domain of

f (x) =x2 − 49√x2 + 9− 5

.

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Solution: x2 + 9 is always positive so we don’t have to worry about the square root.However, to find the domain we need to solve

√x2 + 9− 5 = 0.√

x2 + 9− 5 = 0√x2 + 9 = 5x2 + 9 = 25

x2 = 16x = ±4

So, the domain is R, x 6= ±4 .

3. Solve the inequality.

|14− x| − 3 < 17

Solution:

|14− x| − 3 < 17|14− x| < 20

We know that for a > 0, that |x| < a is equivalent to −a < x < a, so

|14− x| < 20−20 < 14− x < 20−34 < −x < 6

34 > x > −6−6 < x < 34

Here the proper interval is: (−6, 34) .

4. Solve by using an augmented matrix and elementary row operations.2x + 3y − z = −73x − 3y + z = 122x + 4y + z = −3

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Solution: Augmented matrix form of the system: 2 3 −1 −73 −3 1 122 4 1 −3

Elementary row operations, in order given:

R1 + R2 → R2

R1 + R3 → R3

Produces: 2 3 −1 −73 −3 1 122 4 1 −3

∼ 2 3 −1 −7

5 0 0 54 7 0 −10

The second row gives x = 1 ; using x = 1 in row three, gives y = −2 ; finally, using

x = 1 and y = −2 in row one, gives z = 3 .

5. Use long division to find the quotient and remainder when x4 − 4x2 + 2x + 5 is di-vided by x− 2.

Solution:

A remainder of 9 and a quotient of x3 + 2x2 + 2 .

6. Is x = −1 a root of the polynomial function f (x) = 2x3 − 5x2 − 4x + 3?

Solution: Yes , because f (−1) = 0.

7. Factor43 f (x) = 2x3 − 5x2 − 4x + 3.

Solution: Using the fact that x = −1 is a root, we can divide f (x) by x + 1, getting

2x3 − 5x2 − 4x + 3x + 1

= 2x2 − 7x + 3 = (2x− 1) (x− 3) ,

43Previous problem may be helpful.

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so the complete factorization of f (x) is

(x + 1) (2x− 1) (x− 3) .

8. Find the area of the triangle with vertices (−1, 0), (2, 1) and (1,−2).

Solution:

A = ±12

∣∣∣∣∣∣−1 2 1

0 1 −21 1 1

∣∣∣∣∣∣ = ±4

The area is 4 square units.

9. Solve each inequality, and use interval notation to express the solution.

(a)1x≥ 1

x3

Solution:

1x≥ 1

x3

1x− 1

x3 ≥ 0

x2 − 1x3 ≥ 0

(x− 1) (x + 1)x3 ≥ 0

Using a number line, you’ll get [−1, 0) ∪ [1, ∞) .

(b) (1− x)2 ≤ 10− 2x

Solution:

(1− x)2 ≤ 10− 2x1− 2x + x2 ≤ 10− 2x

x2 − 9 ≤ 0(x− 3) (x + 3) ≤ 0

Using a number line, you’ll get [−3, 3] .

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10. Solve for x.

(a) log5 x = 2

Solution:log5 x = 2 ⇔ x = 52 ⇒ x = 25

(b) logx 64 = 3

Solution:logx 64 = 3 ⇔ x3 = 64 ⇒ x = 4

(c) log319= x

Solution:

log319= x ⇔ 3x =

19

⇒ x = −2

11. Factor

f (x) =(

x2 + x + 1) (

6x2 + 5x− 6)

into linear factors.

Solution: You’ll need to use the quadratic formula on x2 + x + 1 = 0, which gives:

x =−1±

√−3

2=−1±

√3i

2,

so x2 + x + 1 factors into two linear factors, as follows

x2 + x + 1 =

(x− −1 +

√3i

2

)(x− −1−

√3i

2

)

=14

(2x + 1−

√3i) (

2x + 1 +√

3i)

.

You can also use the quadratic formula to factor 6x2 + 5x− 6, but I think it is easierto use intelligent trial and error, as follows

6x2 + 5x− 6 = (2x + 3) (3x− 2) .

Finally, as requested, the complete factorization is

f (x) =14(2x + 3) (3x− 2)

(2x + 1−

√3i) (

2x + 1 +√

3i)

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12. A total of $17,500 is invested at an annual rate of 9.3%, compounded weekly. Find thebalance after 35 years.

Solution:

17500 ·(

1 +0.093

52

)(52·35)

≈ 452276.46

So, the balance is $452, 276.46 .


3− 2xx− 1

+ 2 ≥ 0

Solution:

3− 2xx− 1

+ 2 ≥ 0

3− 2x + 2 (x− 1)x− 1

≥ 0

3− 2x + 2x− 2x− 1

≥ 0

1x− 1

≥ 0

Using simple sign analysis, the proper interval is: (1, ∞) .

14. Solve for x.

log3 (2x + 1) + log3 (2x− 1) = 1

Solution:

log3 (2x + 1) + log3 (2x− 1) = 1log3 (2x + 1) (2x− 1) = 1

log3

(4x2 − 1

)= 1

4x2 − 1 = 34x2 − 4 = 0

x2 − 1 = 0x = ±1

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The solution (If you’re limited to using real numbers only, whereas complex num-bers can look mighty strange, such as the famous—or infamous—identity eiπ + 1 =

0.) is x = 1 and you should check this.


[3 1 04 −2 5

]·

0 1 2−1 −5 −1

0 1 −1

Solution:

[3 1 04 −2 5

]·

0 1 2−1 −5 −1

0 1 −1

=

[−1 −2 5

2 19 5

]

16. Given

f (x) =2x− 12− 3x

,

find and simplify the difference quotient

f (x + h)− f (x)h

, h 6= 0.

Solution: Do I need to say it? Yes, the following is true if and only if h 6= 0.

f (x + h)− f (x)h

=

2x + 2h− 12− 3x− 3h

+2x− 12− 3x

h

=(2x + 2h− 1) (2− 3x)− (2x− 1) (2− 3x− 3h)

h (2− 3x− 3h) (2− 3x)

=4x + 4h− 2− 6x2 − 6xh + 3x− 4x + 2 + 6x2 − 3x + 6xh− 3h

h (2− 3x− 3h) (2− 3x)

=h

h (2− 3x− 3h) (2− 3x)

=1

(2− 3x− 3h) (2− 3x)

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17. Find the inverse, if possible. 1 1 13 5 43 6 5

Solution: 1 1 1 1 0 0

3 5 4 0 1 03 6 5 0 0 1

∼ 1 0 0 1 1 −1

0 1 0 −3 2 −10 0 1 3 −3 2

So the inverse is 1 1 −1

−3 2 −13 −3 2

.

18. Solve the system, any method is fine.x + y − z = 4−3x + 2y − z = −63x − 3y + 2z = 5

Solution: You may have noticed that you found the inverse of the coefficient matrixin the prior problem. So I’ll solve using the inverse. 1 1 −1

−3 2 −13 −3 2

· x

yz

=

4−6

5

x

yz

=

1 1 13 5 43 6 5

· 4−6

5

=

321

I get x = 3 , y = 2 and z = 1 which can be easily checked.

19. Identify the elementary row operation performed on the original matrix (left) to ob-tained the new row-equivalent matrix (right).[

−2 5 13 −1 −8

]∼[

13 0 −393 −1 −8

]

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Solution: By inspection:

5R2 + R1 → R1 .

20. Find the domain of

f (x) =x + 2√

2x2 − 18.

Solution: Finding the domain requires solving 2x2 − 18 > 0 for x.

2x2 − 18 > 0x2 − 9 > 0

(x− 3) (x + 3) > 0

Simple sign analysis gives (−∞, −3) ∪ (3, ∞) .

21. Given

f (x) = (x + 3)2 , x ≥ −3,

find f−1 (x). Graphing may be helpful, but is not required.

Solution: The domain of f (x) is [−3, ∞) and the range is [0, ∞), so the domain off−1 (x) is [0, ∞) and the range is [−3, ∞).

f (x) = (x + 3)2

y = (x + 3)2

x = (y + 3)2

±√

x = y + 3±√

x− 3 = y

However, since y ≥ −3 we have

f−1 (x) =√

x− 3, x ≥ 0 .

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22. Given

f (x) =√

x + 6 and g (x) = x2 − 5,

answer each of the following questions.

(a) Find ( f g) (x)

Solution:

( f g) (x) = f (g (x)) = f(

x2 − 5)=√

x2 − 5 + 6 =√

x2 + 1

(b) The domain of ( f g) (x)

Solution: R .

(c) Find (g f ) (x)

Solution:

(g f ) (x) = g ( f (x)) = g(√

x + 6)=(√

x + 6)2− 5 = x + 6− 5 = x + 1

(d) The domain of (g f ) (x)

Solution: x ≥ −6 .

23. Given

f (x) = 2x4 + 7x3 − 4x2 − 27x− 18, and f (2) = f (−3) = 0

answer the following questions.

(a) Use the given roots, and long division, to completely factor f (x).

Solution: Since we are given two roots, we know two factors of f (x) are (x− 2)and (x + 3). Dividing f (x) by the product of these two factors gives

2x2 + 5x + 3 = (2x + 3) (x + 1) .So the complete factorization of f (x) is

(2x + 3) (x + 1) (x− 2) (x + 3)

(b) Graph f (x) using the roots, y-intercept and sign-analyses.

Solution: Your graph (Figure 101, page 243) does not need to have such precisedetail as mine, but it should still reflect the key pre-calculus analysis.

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-4 -3 -2 -1 0 1 2 3 4

-40

-30

-20

-10

10

Figure 101: Graph of f (x) = 2x4 + 7x3 − 4x2 − 27x− 18.

24. Given

f (x) =x3 + 2x2

x2 + 1,

answer the following questions.

(a) x-intercept(s) in point form.

Solution: Set f (x) = 0 and solve for x.

0 =x3 + 2x2

x2 + 1

0 =x2 (x + 2)

x2 + 1Clearly, (−2, 0) and (0, 0) .

(b) y-intercept in point form.

Solution: Set x = 0 and evaluate. Clearly, (0, 0) .

(c) All linear asymptotes in equation form.

Solution: Since the degree of the numerator is one more than the denominator,we have the possibility of getting a slant asymptote. The long division gives

f (x) =x3 + 2x2

x2 + 1= x + 2− x + 2

x2 + 1,

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so the slant asymptote is y = x + 2 .

(d) Graph f (x) using the information above and sign-analyses.

Solution: Your graph (Figure 102, page 244) does not need to have such precisedetail as mine, but it should still reflect the key pre-calculus analysis.

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

-5

-4

-3

-2

-1

1

2

3

4

5

Figure 102: Partial graph of f (x) =x3 + 2x2

x2 + 1= x + 2− x + 2

x2 + 1.

25. Given

f (x) = x2 − x + 1,

find and simplify the difference quotient

f (x + h)− f (x)h

, h 6= 0.

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Solution: Do I need to say it? Yes, the following is true if and only if h 6= 0.

f (x + h)− f (x)h

=(x + h)2 − (x + h) + 1−

(x2 − x + 1

)h

=x2 + 2xh + h2 − x− h + 1− x2 + x− 1

h

=2xh + h2 − h

h= 2x + h− 1


|14− 3x| − 5 > −2

Solution:

|14− 3x| − 5 > −2|14− 3x| > 3

We know that |x| > a is equivalent to x > a or x < −a, so

|14− 3x| > 3 ⇒ 14− 3x > 3 or 14− 3x < −3x < 11/3 or 17/3 < x

Here the proper intervals are: (−∞, 11/3) ∪ (17/3, ∞) .

27. Write the system of linear equations represented by the augmented matrix. Then useback-substitution to find the solution. (Use the variables x, y, and z.) 1 −1 2 4

0 1 −1 20 0 1 −2

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Solution:x − y + 2z = 4

y − z = 2z = −2

The last line gives z = −2 , then using the value for z in line two, we get y = 0 ,

finally using these two values in line one we get x = 8 .

28. Condense the expression to the logarithm of a single quantity.

13

[2 ln (x + 3) + ln x− ln

(x2 − 1

)]Solution:

13

[2 ln (x + 3) + ln x− ln

(x2 − 1

)]=

13

[ln (x + 3)2 + ln x− ln

(x2 − 1

)]=

13

[ln

x (x + 3)2

x2 − 1

]

= ln3

√x (x + 3)2

x2 − 1

29. Given the following functional graph (Figure 103, page 247) . Answer the followingquestions.

(a) Is f (x) invertible? Explain your answer.

Solution: No . It fails the horizontal line test.

(b) Graph

2 · f(

1− x2

)− 1

Solution: The points translated, in order are(10, 3) , (6, 5) , (2, −3) , (−6, −5) .

You graph (Figure 104, page 248) should look like mine.

30. Evaluate the expression, if possible. −1 6−4 5

0 3

· [ 2 30 9

]

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-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

Figure 103: Graph of f (x).

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-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

Figure 104: Graph of 2 · f(

1− x2

)− 1.

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Solution: −1 6−4 5

0 3

· [ 2 30 9

]=

−2 51−8 33

0 27

31. Sketch the graph of the function.

f (x) = 3x+2

Solution: Your graph (Figure 105, page 249) should look similar to mine.

-15 -12.5 -10 -7.5 -5 -2.5 0 2.5

-2.5

2.5

5

7.5

10

12.5


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32. Solve for x.(23

)x=

8116

Solution:(23

)x=

8116(

23

)x=

(32

)4

(23

)x=

(23

)−4

So, x = −4 .


x + 6x + 1

− 2 < 0

Solution:

x + 6x + 1

− 2 < 0

x + 6x + 1

− 2 (x + 1)x + 1

< 0

4− xx + 1

< 0

Using simple sign analysis, the proper intervals are: (−∞, −1) ∪ (4, ∞) .

34. Solve algebraically (exact answer) and then approximate to three decimal places.

−2 + 2 ln 3x = 17

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Solution:

−2 + 2 ln 3x = 172 ln 3x = 19

ln 3x =192

3x = e19/2

x =e19/2

3≈ 4, 453.242

35. Given

f (x) =x2 − 3x + 2

2x2 + 5x + 3=

(x− 1) (x− 2)(2x + 3) (x + 1)

,

answer the following questions:

(a) x-intercepts in point form.

Solution: (1, 0) ; (2, 0)

(b) y-intercept in point form.

Solution: (0, 2/3)

(c) Equation of the horizontal asymptote.

Solution: y = 1/2

(d) Equation of the vertical asymptotes.

Solution: x = −1; x = −3/2

(e) Graph f (x) using the above information and sign analysis.

Solution: Your graph (Figure 106, page 252) should look similar to mine.


y = x2 − x− 2y = 2x + 2

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-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

-70

-60

-50

-40

-30

-20

-10

10


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Solution: You’ve got a parabola and a line that intersect at: (−1, 0) and (4, 10) .Your graph (Figure 107, page 253) should look like mine.

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

-2

-1

1

2

3

4

5

6

7

8

9

10

11

Figure 107: Partial graph of y = x2 − x− 2 and y = 2x + 2.

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38 Technology

A great place to learn about your particular calculator is by visiting the manufacturer’swebsite and viewing their video tutorials. The key factor here, is that you need to learnhow to use your calculator, and by doing so, you will certainly become more proficientat using technology in general. Ideally, you will also need to focus on the mathematicsyou are learning, but the calculator may in fact add greatly to your understanding of themathematics, and will almost certainly allow you to solve problems that would otherwisebe impossible. Not all problems require a calculator, but you will be served well if youtake the time to learn how to best use your calculator.

Your teacher may provide guidance, but it is your responsibility to learn how to useyour calculator.

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