How to use these notes - math.utah.edukjohnson/math15sp15/Math_15.pdf · How to use these notes ......

Chapter 1

How to use these notes

These notes were prepared for the University of Utah’s Math 1050/1060 refresher course. They asssumethat the user has had the Math 1050/1060 courses College Algebra/Trigonometry (“Precalculus”) or itsequivalent (in high school or college). The notes work like this: in order to help you focus on the areaswhere you need the most improvement, you will be taking a series of diagnostic quizzes. After taking aquiz, you may check your answers against the answers found in the Answers section. You will then gradeyourself. If you have made only a few small mistakes, read the complete worked solutions to thoseproblems in the Solutions section. If you have made more than a few mistakes, the notes will refer you toonline lectures to brush up on the appropriate topics.On the websites http://www.math.utah.edu/lectures/math1050.html andhttp://www.math.utah.edu/lectures/math1060.html, you will find four links for each section:

1. Streaming video: to watch a video lecture right now

2. Downloadable video: to download a video lecture for watching offline

3. Pre-notes: a PowerPoint file outlining the notes, with gaps

4. Post-notes: the PowerPoint file marked up with all the work done during the lecture

The fastest, but not necessarily best, way to review the material would be to go straight to the post-notes. Amore thorough approach is to print a copy of the pre-notes, then watch the video, and fill them in. Whatyou choose to do will depend on how much review you need of a particular section.

1

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Chapter 2

Quizzes

2.1 Warmup

2.1.1 Some computations

1. 434−3

78

2. 434·37

8

3. 1.5 ·2.25

4. 2.25÷1.5

5. 6−3+2

6. 6÷3 ·2

7. (14−5)2

8.43−2 ·56−4÷2

9. 24

10. 2−4

11. 92

12. 912

13. 9−12

14. −42

15. (−4)2

2.1.2 Linear equations and inequalities

Solve for x:

1. 3x+4 = 6− x

2. −3x+4 = 6− x

3.3x=

914

4.x7=

35

5.3+ x

5=

710

6.3− x

x=

35

7.x−2

4=

x+112

Solve for x, writing your answer in interval notation:

1. 3x+4 < 6− x

2. −3x+4≥ 6− x

3.x7<

35

4.3+ x

5≤ 7

10

5.x−2

4≥ x+1

12

6. −3 < 2x+5≤ 7

3

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Now check your answers with the answers section. Give yourself one point for each correct answer. Thereare 28 questions in this section. If you got fewer than 23... Try the ones you missed again (carefully!)WITHOUT looking at the worked solutions.Still got fewer than 23? This material is not really 1050 material... Consider jumping over to the Math1010 refresher, or just enrolling in Math 1010. Your math may be a little rustier than you think.

2.2 Functions

Do the following equations, tables, or graphs represent y as a function of x?

1. y = 2x−3

2. 3x−2y2 =−3

3. x2 + y5 =−1

4. x =√

y+1

5.x y−2 4−1 10 01 12 4

6.x y−2 4−1 40 41 42 4

7.x y−1 4−1 10 01 11 4

8.

9.

Stop for a second, and check out 1.4 from the online notes if you’re getting confused.For the next problems,

f (x) =1

1− x,g(x) = x2 +1

1. Find the domain of f (x).

2. Find the domain of g(x).

3. Find and simplify f (x)/g(x), and give itsdomain.

4. Find and simplify g(x)/g(x), and give its

domain.

5. Find and simplify ( f ◦g)(x) and give itsdomain.

6. Find and simplify (g◦ f )(x) and give itsdomain.

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7. Evaluate f (−2).

8. Evaluate g(−2).

9. Evaluate f (1/x) and write the result as asimple fraction.

10. Find f−1(x)

11. Why does g not have an inverse?

12. Find the inverse of the function

h(x) =2x−3x+5

For more help, see the online 1.4 (for the “is this a function” problems) and 1.6 (for the second group).

2.3 Sequences and Series

Find the missing terms of the following sequences:

1. 1,4,7, ,

2. 3,6,12, ,

3. 1,−1/2,1/4,−1/8, ,

Find a formula for the nth term of the following sequences:

1. 1,4,7, . . .

2. 3,6,12, . . .

3. 1,−1/2,1/4,−1/8, . . .

Find the sum of the following series:

1. 1+4+7+ · · ·+31

2. 3+6+12+ · · ·+192

3. 1−1/2+1/4−1/8+ · · ·−1/512

Evaluate the factorial expressions:

1.8!6!

2.9!

5!4!

3.(n+2)!(n−1)!

Solve the following counting problems:

1. How many flags are possible to make from red, white, and blue vertical, equal-width stripes.

2. How many ways are there to make four horizontal stripes using four colors chosen at random from abox of eight crayons?

3. I have six pairs of pants and five shirts. How many combinations of these can I make?

4. How many ways are there to order a three-topping pizza from a menu with ten possible toppings?

There are sixteen problems in this part. I hope you got more than 12 right. If not, check out:

1. 6.1, 6.2, 6.3 for the first three parts.

2. 6.5 for the last two parts.

2.4 Translation and reflection of graphs

Sketch the graphs of the following:

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1. y =−(x+2)2−3

2. y = (x−3)2 +5

3. y = |x−1|+4

4. y = (x−2)3−3

5. y =−(x+2)3

6. y =−1

x+3

In the online lectures, this is about halfway through 1.5.

2.5 Polynomials

2.5.1 Multiplying

Multiply out and write in standard form:

1. (x−1)2(x+2)

2. (x−2)3

3. 3(x−2)(x+3)

4. (5x+3)(x2−3x−4)

5. (x− y)5 (binomial theorem)

6. (x+2a)4 (binomial theorem)

2.5.2 Factoring

Factor as completely as possible:

1. x2 +7xy−144y2

2. 10x2 +6xy−25xy−15y2

3. 3x3y2−75xy2

4. 36x3 +12x2−48x

5. x3 +3x2− x−3

6. x4− y4

This material is technically 1010. Hope you did OK! Check out sections 5.4-5.5 from the notes athttp://www.math.utah.edu/lectures/math1010.html if you need help.

2.5.3 Polynomial equations

Solve for x:

1. x2−5x = 6

2. x2−2x = 0

3. x2− 14= 0

4. x2 +16 = 0

5. 3x2 +6x = 24

6. 2x2−3x−2 = 0

7. 6x2 = 7x+3

8. 12x2−13x+3 = 0

9. x4−2x2 +1 = 0

10. x−8√

x+7 = 0

11. x3−5x2 = 6x

12. x4−4x2 = 0

13. (x−2)2 = 7

14. x2−2x−5 = 0

15. x2 +7x+3 = 0

16. x2 +2x+3 = 0

You need to be doing these by factoring or completing the square. For more help with completing thesquare, check Extras: How to complete the square.

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Also check out sections 8A-D from the notes athttp://www.math.utah.edu/lectures/math1010.html.

2.5.4 Finding roots of an equation with a given root

In these problems, one root of a polynomial is given. Use it to find all the other roots.

1. 2x3−5x2 + x+2 = 0, x = 2 is a root.

2. 2x3−3x2−11x+6 = 0, x =−2 is a root.

3. 12x3 +16x2−5x−3 = 0, x =−32

is a root.

4. 3x3 +7x2−22x−8 = 0, x =−13

is a root.

The technique of synthetic division may be useful. For a quick refresher, look at the explanations in thesenotes. For further explanation, see 2.3 in the online lectures.

2.5.5 Long division

Divide these polynomials, writing the remainders.

1. (6x3 +7x2 +12x−5)÷ (3x−1)

2.3x2−2x+5

x−3

3.4x4−4x2 +6x

x−4

2.5.6 Graphs of polynomial functions

1. Write y = 2(x−1)2 +4 in the formy = ax2 +bx+ c.

2. Write y = 2x2 +4x+5 in the formy = a(x−h)2 + k.

3. Find the coordinates of the vertex of theparabola y = 2(x−1)2 +4

4. Find the coordinates of the vertex of theparabola y = 2x2 +4x+5.

5. Find the equation of a parabola with vertex(−1,3) with y-intercept (0,−2).

6. Find the equation of a parabola with roots −2and 3.

7. Graph y = 2(x−1)2 +4.

8. Graph y = 2x2 +4x+5.

Check out 2.1 in the notes for more about graphing.

2.5.7 Graphs of higher-degree polynomial functions

Match the equations with their graphs.

1. y =−x4 + x2

2. y = x3−4x2

3. y = (x−3)2

4. y =−x3− x2 +5x−3

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1.

2.

3.

4.

2.6 Rational expressions

Simplify the following rational expressions as completely as possible:

1.2x2 +5x−3x2 +2x−3

2.−4x3y2

2x2y3 ·−4xy3

10xy2

3.3

x−2+

3x2x+3

4.4x−48x2−144

5.x2−92x+2

· x2 +2x+1(x−3)(x+1)

6.x2 + x−23x2−9x

÷ x2 +4x−56x−18

· x2 +2x−15x2− x−6

If you need help with these, go to the notes athttp://www.math.utah.edu/lectures/math1010.html. The sections are 6.2-6.3.

2.6.1 Complex fractions

It’s not considered good mathematical writing to have fractions living inside of other fractions. Suchexpressions are called complex fractions. Have no fear: a fraction bar signifies division, so one fractionover another can be rewritten as a division of fractions problem.Write these as simple fractions:

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1.

x+53x2

x2−256x3

2.4+

12

13+

16

3.

5x−5

+3

x+35

x+3+

3x−5

If you need help with these, go to the notes athttp://www.math.utah.edu/lectures/math1010.html. The section is 6.4.

2.6.2 Graphs of rational functions

Sketch graphs of the following rational functions:

1. y =x+2x−3

2. y =x2−4x+3x2 + x−2

3. y =x−1

x2− x−2

4. y =x2 +3x+2

x−1

The section 2.6 of the video lectures has a detailed example and outlines the general procedure.

2.7 Exponential and Logarithmic functions

Evaluate:

1. 82/3

2. 8−2/3

3. (−8)2/3

4. (−8)−2/3

5. −82/3

6.(

49

)3/2

7. log2(1/2)

8. log3 27

9. log9 3

Look at the notes in sections 3.1-3.2 for help.Solve:

1. logx =−3

2. log2 x =−12

3. e2x−2 = 0

4. 8−3 ·20.5x =−40

5. log3 9x = 3

6. e2x−3ex = 28

7. log(x+1)− log(x−1) = log3

8. ln(x2−4)− ln(x+2) = ln(3− x)

Look at the notes in sections 3.4-3.5 for help.Combine the logarithms into a single logarithm:

1.12

log2 x+ log2y 2. 3 logx−2logy− logz

Expand the single logarithm as fully as possible:

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1. log(

x√

yz

)2. log4 4x2y3

Look at the notes in section 3.3 for help.

2.8 Matrices and Linear Algebra

NOTE: If you’re coming in to this section and you haven’t heard of matrices before, you should startdirectly with the video lectures in Chapter 5. This might be the case, for example, if you took the 1050equivalent in high school, it may not have been covered. If the material is new to you, it is a lot to learn in acouple of hours. Don’t worry too much: matrices are a useful tool for doing something you probablyalready know how to do another way. You may not even really see them again until Calc III.Multiply the matrices, or indicate why the product cannot be formed.

1. 1 −3 0 60 2 1 02 0 4 −4

9 −30 −20 52 0

2. [

3 −2 30 1 −1

][1 0 −76 11 −2

]

3. [1 34 2

][−2 31 −2

]

4. [−2 31 −2

][1 34 2

]

Solve the following linear systems (is there a way to save time on the first four?):

1.

2x1 +3x2 = 1

x1 +4x2 = 2

2.

2x1 +3x2 = 3

x1 +4x2 = 4

3.

2x1 +3x2 = 5

x1 +4x2 = 6

4.

2x1 +3x2 = 7

x1 +4x2 = 8

5.

x+ z = 1

2x+2y+4z = 2

2x−2y+8z = 6

6.

3x1 + x2 +4x3 = 0

9x1 +5x2 +16x3 = 0

21x1 +11x2 +36x3 = 0

Find the inverses of the following matrices, if possible:

1. [4 −3−3 2

]2. [

4 −2−2 1

]

3. 1 2 32 3 50 1 2

4. 1 0 2

3 3 −46 3 2

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Compute the following determinants:

1. [0 12 4

]

2. [−1 −5−3 −12

]

3. 1 2 32 3 50 1 2

4. 1 0 2

3 3 −46 3 2

2.9 Measuring and describing angles

1. Convert to radian measure: a) 30◦; b) 150◦.

2. Convert to degree measure: a)3π

2; b) −7π

6.

3. Find a positive and a negative angle coterminal to a) 390◦; b)17π

6.

4. Find a) the complement of 72◦; b) the supplement of 72◦; c) the complement ofπ

12; d) the

supplement of5π

6.

Check out section 1.1 from the trigonometry lectures onhttp://www.math.utah.edu/lectures/1060.html. This will be where all the lectures come from forthe trigonometry sections.

2.10 Trigonometric functions

Evaluate:

1. sin(45◦)

2. cos(60◦)

3. sec(135◦)

4. tan(300◦)

5. cot(0◦)

6. sinπ/6

7. cos(−5π/3)

8. tan(3π/4)

9. csc(4π/3)

10. secπ

1. θ is acute and sinθ = 0.6. Find a) cosθ , b) tanθ .

2. tanθ =−15

8and sinθ < 0. Find a) sinθ ; b) cosθ .

3. sinθ =45

and θ lies in the second quadrant. Find a) cosθ , b) tanθ .

Check the notes from 1.2, 1.4 for help.

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2.11 Inverse trigonometric functions

Evaluate (give your answer in radians):

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1. sin−1 1

2. tan−1 1

3. cos−1(−√

3/2)

4. sin−1(−1/2)

5. tan−1(−√

3)

6. cos−1(−√

2/2)

The notes for this material are section 1.7.

2.12 Trigonometric Equations

Solve the following trigonometric equations, finding solutions between 0 and 2π .

1. sinx = cosx

2. 3sinx−2 = 5sinx−1

3. 2cos2 x+ cosx−1 = 0

4. 2cos2 x+3sinx = 0

2.13 Graphs of trigonometric functions

1. What is the amplitude of the functiony = 3cosx?

2. What is the period of the functiony = sin(πx)?

3. What is the period of the functiony = tan(x/2)?

4. What is the phase shift of the functiony = cos(2x−π/2)?

Sketch the graphs of the following trigonometric functions:

1. y = 2sin(x−π/2)

2. y = cos(2x+π/2)

3. y = tan(x−π/4)

4. y = 3cos(πx−π)+2

Check out 1.5-1.6 for some tips on graphing trig functions.

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Chapter 3

Answers

3.1 Warmup

3.1.1 Some operations

1.78

2. 181332

3. 3.375

4. 1.5

5. 5

6. 4

7. 81

8. 13.5 (= 1312=

544

)

9. 16

10.116

11. 81

12. 3

13.13

14. -16

15. 16


1.12

2. −1

3.143

4.215

5.12

6.158

7.72

1. x ∈ (−∞, 12)

2. x ∈ [−1,∞)

3. x ∈ (−∞, 215 )

4. x ∈ (−∞, 12 ]

5. x ∈ [72 ,∞)

6. x ∈ (−4,1]

3.2 Functions

15

Name 16

1. Yes.

2. No.

3. Yes.

4. Yes.

5. Yes.

6. Yes.

7. No.

8. No.

9. Yes.

For the next problems,

f (x) =1

1− x,g(x) = x2 +1

1. x 6= 1

2. all real numbers

3.1

1− x+ x2− x3 =1

(1− x)(x2 +1), x 6= 1

4. (x2 +1)(1− x),x 6= 1

5. − 1x2 ,x 6= 0

6.1

(1− x)2 ,x 6= 1

7.13

8. 5

9.x

x−1

10.x−1

x

11. For example, g(1) = g(−1) = 2, so there canbe no inverse at 2.

12.5x+32− x


1. 10, 13.

2. 24, 48.

3. 1/16, −1/32.

1. an = 1+3(n−1).

2. an = 3 ·2n−1.

3. an =(−1

2

)n−1.

1. 176

2. 381

3. 343512

Evaluate the factorial expressions:

1. 56

2. 126

3. (n+2)(n+1)n

Solve the following counting problems:

1. 6

2. 1680

3. 30

4. 120

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Name 18

1.

2.

3.

4.

5.

6.

3.5 Polynomials

3.5.1 Multiplying

1. 5x2−12x+4

2. x3−3x+2

3. x3−6x2−4x−8

4. 3x2 +3x−18

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5. 5x3−12x2−29x−12

6. 45−108i

7. x5−5x4y+10x3y2−10x2y3 +5xy4− y5

8. x4 +8x3a+24x2a2 +32xa3 +16a4

3.5.2 Factoring

1. (x+16y)(x−9y)

2. (2x−5y)(5x+3y)

3. 3xy2(x+5)(x−5)

4. 12x(x−1)(3x+4)

5. (x+3)(x+1)(x−1)

6. (x2 + y2)(x+ y)(x− y)


1. -1,6

2. 0,2

3. ±12

4. no real solution

5. -4,2

6. −12

, 2

7. −13

,32

8.13

,34

9. -1,1

10. 1,49

11. -1,0,6

12. -2,0,2

13. 2±√

7

14. 1±√

6

15.−7±

√37

2

16. no real solution


1. −12

and 1.

2.12

and 3.

3. −13

and13

4. −4 and 2

3.5.5 Long division

1. 2x2 +3x+5

2. 3x+7+26

x−3

3. 4x3 +16x2 +60x+246+984

x−4


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1. 2x2−4x+6

2. 2(x+1)2 +3

3. (1,4)

4. (−1,3)

5. y =−5(x+1)+3

6. y = (x+2)(x−3) = x2− x−6. Any nonzeromultiple of this is also OK.

7.

8.


1. Graph 2

2. Graph 3

3. Graph 1

4. Graph 4


1.2x−1x−1

,x 6=−3

2.4x5,x,y, 6= 0

3.3(x2 +3)

(2x+3)(x−2)

4.4

x+12,x 6= 12

5.(x+3)

2,x 6=−1,1,3

6.2x,x 6=−5,−2,1,3


1.2x

x−5,x 6=−5,0 2. 9

3.x

x−2,x 6=−3,5


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1.

2.

3.

4.


1. 4

2.14

3. 4

4.14

5. -4

6.827

7. -1

8. 3

9.12

1. 0.001

2. 1√2

3.12

ln2 = ln√

2

4. 8

5. 3

6. ln7

7. 2

8.52

1. log2 y√

x 2. logx3

y2z

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1. logx+12

logy− logz 2. 1+2log4 x+3log4 y


1. 21 30 1

10 14

2. not possible

3. [1 −3−6 8

]4. [

10 0−7 −1

]

1. x1 =−2/5,x2 = 1

2. x1 = 0,x2 = 11/5

3. x1 = 2/5,x2 = 17/5

4. x1 = 4/5,x2 = 23/5

5. x = 5,y =−2,z =−1

6. x =−2t,y =−6t,z = 3t for any real number t

1. [−2 −3−3 −4

]

2. No inverse.

3. −1 1 −14 −2 −1−2 1 1

4. No inverse

1. 2

2. -3

3. -1

4. 0


1. a) π/6; b) 5π/6.

2. a) 270◦; b) −210◦.

3. a) 30◦, −330◦; b) 5π/6, −7π/6

4. a) 18◦; b) 108◦; c) 5π/12; d) π/6.


1.√

2/2

2.12

3. −√

2

4. −√

3

5. not defined

6.12

7.12

8. −1

9. 2√

33

10. −1

1. a) 0.8; b) 34 .

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2. a) −1517 ; b) 8

17 .

3. a) −35 ; b) −4

3 .

3.10.1 Inverse trigonometric functions

1. π/2

2. π/4

3. 5π/6

4. −π/6

5. −π/3

6. 3π/4


1. π/4,5π/4

2. π/6,5π/6

3. π/3,5π/3,π

4. 7π/6,11π/6


1. 3

2. 2

3. 2π

4. π/4

1.

2.

3.

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4.

Chapter 4

Explanations

4.1 Warmup

4.1.1 Some operations

1. This is how I would do it. First you need a common denominator

434−3

78= 4

68−3

78

Now exchange a “1” from the 4 for 8 eighths:

= 3148−3

78

Now the 3s go, and

=148− 7

8=

78.

Here, changing to “improper fractions” also works fine, but is more work.

2. If you got 122132

, be careful: you can’t just multiply the whole number and fraction parts separately.

Don’t believe me? Try that on 112·11

2and see if your answer seems reasonable. Here, we better use

improper fractions:

434·37

8=

194· 31

8

=19 ·314 ·8

=58932

3. Remember the “rule”? Multiply, ignoring the decimal points, count the number of places total afterthe decimal in the question, and place the decimal point that many digits from the right. It works,because

1.5 ·2.25 =1510· 225

100

=15 ·22510 ·100

=33751000

= 3.375

25

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4. This is equivalent to 22.5÷15. General comment: Many people learn the order of operations by themnemonic PEMDAS (“please excuse my dear Aunt Sally”) – parentheses, exponentiation,multiplication, division, addition, subtraction. If you do this, remember that “MD” are really donetogether, left to right, when they occur. You don’t do “M” before “D”, just before “A” and “S”. Samefor “AS”.

5. If you got 1, read what I wrote above.

6. If you got 1, read what I wrote above.

7. Parens tell you what to do before squaring.

8. Evaluate the whole numerator and whole denominator (with proper order of operations) first. Thenreduce the resulting fraction. No cancelling until you’ve done all the addition and subtraction!

9. No comment.

10. Negative exponents don’t make the result negative. They mean to take the reciprocal before applyingthe corresponding positive power. This is done to make the rules of exponents consistent (ask me!).So

2−4 =124 =

116

11. No comment.

12. Fractional exponents correspond to roots: 912 =√

9 = 3.

13. Combination of exponent rules.

14. Exponents have higher precedence than the minus sign! So we raise 4 to the second power first, thentake the opposite of that.

15. Here the parentheses tell us to take the opposite of four and raise that to the second power.


General tips: If two things are equal, and we do the same thing to both of them, they stay equal. Thisjustifies “adding the same thing to both sides”, and so on.A good strategy, when trying to get all the expressions with x on the same side is to move so that thecoefficient of x is positive. Fewer arithmetic errors are made when dividing by positive numbers.

1.

3x+4 = 6− x ⇒ 4x = 2

⇒ x =12

2.

−3x+4 = 6− x ⇒ −2 = 2x

⇒ x =−1

3.

3x=

914

⇒ 9x = 3 ·14

⇒ x =3 ·14

9=

143

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4.x7=

35⇒ 5x = 3 ·7

⇒ x =215

5.3+ x

5=

710

⇒ 10(3+ x) = 5 ·7

⇒ 30+10x = 35

⇒ 10x = 5

⇒ x =12

6.3− x

x=

35⇒ 5(3− x) = 3x

⇒ 15−5x = 3x

⇒ 15 = 8x

⇒ x =815

7.x−2

4=

x+112

⇒ 12(x−2) = 4(x+1)

⇒ 12x−24 = 4x+4

⇒ 8x = 28

⇒ x =288

=72

1.

3x+4 < 6− x ⇒ 4x < 2

⇒ x <12.x ∈ (−∞,

12)

2.

−3x+4≥ 6− x ⇒ −2≥ 2x

⇒ −1≥ x.x ∈ [−1,∞)

3.x7<

35⇒ 5x < 21

⇒ x <215.x ∈ (−∞,

215)

4.3+ x

5≤ 7

10⇒ 10(3+ x)≤ 5 ·7

⇒ 30+10x≤ 35

⇒ 10x≤ 5

⇒ x≤ 12.x ∈ (−∞,

12]

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5.

x−24≥ x+1

12⇒ 12(x−2)≥ 4(x+1)

⇒ 12x−24≥ 4x+4

⇒ 8x≥ 28

⇒ x≥ 288.x ∈ [

72,∞)

6.

−3 < 2x+5≤ 7 ⇒ −8 < 2x≤ 2

⇒ −4 < x≤ 1.x ∈ (−4,1]

4.2 Functions

1. Yes. This is already solved for y, so a value of x uniquely determines y.

2. No. If we try to solve for y, we end up taking a square root, which could have positive or negativevalues. So y is not uniquely determined.

3. Yes. Odd degree roots are unique.

4. Yes. Here we can solve for y. We square both sides at some point, but that does not cause a problem.

5. Yes. Each x value appears no more than once.

6. Yes. Each x value appears no more than once.

7. No. x values appear multiple times.

8. No. Fails the vertical line test.

9. Yes. Passes the vertical line test.


1. 1,4,7, , . The pattern is that each term is three greater than the previous term. A sequence that hasa common difference between subsequent terms is called an arithmetic sequence. The next two termsof this sequence are thus 10 and 13.

2. 3,6,12, , . The pattern is that each term is twice the previous term. A sequence that has a commonratio between subsequent terms is called a geometric sequence. The next two terms of this sequenceare thus 24 and 48.

3. 1,−1/2,1/4,−1/8, , . This sequence is also geometric. Each term is −1/2 of the previous term.Thus the next two terms are 1/16 and −1/32.

1. This sequence starts from 1, and each term is three greater than the last. So by time we get to the nthterm, we have n−1 times added three. So an = 1+3(n−1).

2. This sequence starts from 3, and each term is double the last. So by time we get to the nth term, wehave doubled n−1 times. So an = 3 ·2n−1.

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3. Same idea: we start from 1, and multiply by −1/2 to get to subsequent terms. an =(−1

2

)n−1. Notethat this even works for n = 1, since raising a number to the zeroth power yields 1.

1. 1+4+7+ · · ·+31. We need to know how many terms there are here: the first and last terms differby 30, so there are 11 terms here (the first term and 30/3 = 10 more). If we write:

1+4+7+ · · · +31

31+28+25+ · · · +1

we have two copies of the series. Note that the numbers in columns always add up to 32. So we haveeleven copies of 32 when we take the series twice. So the sum is 11 ·32/2 = 176.

2. The reasoning for the sum of geometric series is somewhat complicated. It results in the formula:

a+ar+ar2 + · · ·+arn−1 = arn−1r−1

Here a is the first term, r is the common ratio, and n is the number of terms. For our sequence, weget to 192, starting from 3 and doubling. 192/3 = 64, and 64 = 26, so our sequence has 6+1 = 7terms (remember we didn’t multiply to get the first term). So the sum is:

327−12−1

= 31271

= 381

3. 1−1/2+1/4−1/8+ · · ·−1/512 Here the common ratio is −1/2, and since 512 = 29, there are 10terms. The sum is:

1(−1/2)10−1(−1/2)−1

=(1/1024)−1−3/2

=1023 ·23 ·1024

=343512

1. The factors 1-6 cancel out:8!6!

=8 ·7 ·6!

6!= 8 ·7 = 56

2.9!

5!4!=

9 ·8 ·7 ·6 ·5!5!4 ·3 ·2 ·1

=9 ·2 ·7=

126

3.(n+2)!(n−1)!

=(n+2)(n+1)n(n−1)!

(n−1)!= (n+2)(n+1)n

1. Order matters, three choices for the first, two for the second, one for the third. 6.

2. 8 ·7 ·6 ·5 = 1680

3. 6 ·5 = 30

4. Ten choices for the first, nine choices for the second, eight choices for the third. But order doesn’tmatter, so you need to divide by the number of ways of reordering three toppings:

10 ·9 ·83 ·2 ·1

= 120

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Since this is a refresher, only the final results will be presented here.If we start with the basic graph of a function y = f (x), it may be modified:

1. y = f (x)+a is the same graph translated up by a units (the net translation is “down” when a isnegative). Adding a after applying the function adds a to y.

2. y =− f (x) is the same graph reflected over the x-axis. Applying the negative sign after the functionchanges all function values (y-values) to their opposites.

3. y = f (x+a) is the graph translated left by a units (the net translation is “right” when a is negative).The rough idea is that to solve for x, we would undo f , and then subtract a.

4. y = f (−x) is the graph reflected over the y-axis. It is the x-values whose opposites are being taken.

In addition to these basic rules, it is never a bad idea to check your work by plugging a point or two fromyour graph into the original equation.

4.5 Polynomials

Certain expressions we meet a lot. Since we meet them a lot, there is a lot of terminology for them.A polynomial is an expression of the form:

anxn +an−1xn−1 + · · ·+a1x+a0

where the an are real numbers, and n is a whole number.Some terminology:

1. n is called the degree of the polynomial (note that n is the highest power occuring on an x if the termshappen not to be written in descending order, as here).

2. The numbers ai are called coefficients.

3. The number an is called the leading coefficient.

4. The number a0 is called the constant term.

When

• n = 0, the polynomial is called constant.

• n = 1, the polynomial is called linear.

• there is a total of one term, the polynomial is called a monomial.

• there is a total of two terms, the polynomial is called a binomial.

• there is a total of three terms, the polynomials is called a trinomial.

Once we enter the world of polynomial expressions, there is one more thing we can do with expressions:

• factor: write an expression as the product of two other expressions.

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Factoring out the greatest monomial factor: here we try to find the factors that are common to every term inan expression, both those made of variables, and not.Examples:

1. 2x2 +6x−4 = 2(x2 +3x−2)

2. x3y2− x2y+ x4y3 = x2y(xy−1+ x2y2)

3. 4x(x+2)−3(x+2) = (x+2)(4x−3)

This is the first thing to try when factoring anything.

4.5.1 Multiplying

1.(x−2)(5x−2) = 5x2−2x−10x+4 = 5x2−12x+4)

2.

(x−1)2(x+2) = (x2−2x+1)(x+2)

= x3 +2x2−2x2−4x+ x+2

= x3−3x+2

3.

(x−2)3 = (x2−4x+4)(x−2)

= x3−2x2−4x2−8x+4x−8

= x3−6x2−4x−8

4.3(x−2)(x+3) = 3(x2 +3x−2x−6) = 3(x2 + x−6) = 3x2 +3x−18

5.(5x+3)(x2−3x−4) = 5x3 +3x2−15x2−20x−9x−12 = 5x3−12x2−29x−12

6.

(9−6i)2 = (9−6i)(9−6i)

= 81−54i−54i+36i2

= 81−108i−36

= 45−108i

7. The binomial theorem tells us that the coefficients of the terms in the expansion of a binomial raisedto the nth power are the numbers in the nth row of Pascal’s triangle. So

(x− y)5 = 1x5(−y)0 +5x4(−y)1 +10x3(−y)2 +10x2(−y)3 +5x1(−y)4 +1x0(−y)5

= x5−5x4y+10x3y2−10x2y3 +5xy4− y5

8.

(x+2a)4 = 1x4(2a)0 +4x3(2a)1 +6x2(2a)2 +4x1(2a)3 +1x0(2a)4

= x4 +8x3a+24x2a2 +32xa3 +16a4

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4.5.2 Factoring

1. 144 = 16 ·9. Factors are probably like (x+?y)...

x2 +7xy−144y2 = (x+16y)(x−9y)

2.

10x2 +6xy−25xy−15y2 = 2x(5x+3y)−5y(5x+3y)

= (5x+3y)(2x−5y)

3.3x3y2−75xy2 = 3xy2(x2−25) = 3xy2(x+5)(x−5)

4.

36x3 +12x2−48x = 12x(3x2 + x−4)

= 12x(3x2−3x+4x−4)

= 12x[3x(x−1)+4(x−1)]

= 12x(x−1)(3x+4)

5.

x3 +3x2− x−3 = x2(x+3)−1(x+3)

= (x+3)(x2−1)

= (x+3)(x+1)(x−1)

6.

x4− y4 = (x2)2− (y2)2

= (x2 + y2)(x2− y2)

= (x2 + y2)(x+ y)(x− y)


When the same variable occurs in the same equation to different powers, there are basically no validalgebraic operations that will allow you to isolate the variable. As an alternative, in this case, we move allthe terms on to on side of the equation, leaving the other side zero. If the resulting expression is factorable,we can use the zero property of multiplication: if a number of factors are multiplied and the result is zero,then one of those factors had to be zero itself.Some polynomial equations do not require this. Using some clever manipulations (one of which iscompleting the square), these equations can be solved by taking roots.

1.x2−5x = 6⇒ x−5x−6 = 0⇒ (x−6)(x+1) = 0⇒ x =−1,6

2.x2−2x = 0⇒ x(x−2) = 0⇒ x = 0,2

3.

x2− 14= 0⇒ x2 =

14⇒ x =±

√14⇒ x =±1

2

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4.x2 +16 = 0⇒ x2 =−16

No real number squares to a negative. This equation has complex solutions x =±4i.

5.

3x2 +6x = 24 ⇒ 3x2 +6x−24 = 0

⇒ 3(x2 +2x−8) = 0

⇒ 3(x+4)(x−2) = 0

⇒ x =−4,2

6.

2x2−3x−2 = 0 ⇒ 2x2−4x+ x−2 = 0

⇒ 2x(x−2)+1(x−2) = 0

⇒ (x−2)(2x+1) = 0

⇒ x =−12,2

7.

6x2 = 7x+3 ⇒ 6x2−7x−3 = 0

⇒ 6x2−9x+2x−3 = 0

⇒ 3x(2x−3)+1(2x−3) = 0

⇒ (2x−3)(3x+1) = 0

⇒ x =32,−1

3

8.

12x2−13x+3 = 0 ⇒ 12x2−9x−4x+3 = 0

⇒ 3x(4x−3)−1(4x−3) = 0

⇒ (4x−3)(3x−1) = 0

⇒ x =34,13

9.

x4−2x2 +1 = 0 ⇒ (x2)2−2(x2)+1 = 0

⇒ (x2−1)2 = 0

⇒ x2−1 = 0

⇒ x =±1

10.

x−8√

x+7 = 0 ⇒ (√

x)2−8(√

x)+7 = 0

⇒ (√

x−1)(√

x−7) = 0

⇒√

x = 1 or√

x = 7

⇒ x = 1,49

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11.

x3−5x2 = 6x ⇒ x3−5x2−6x = 0

⇒ x(x2−5x−6) = 0

⇒ x(x−6)(x+1) = 0

⇒ x =−1,0,6

12.x4−4x2 = 0⇒ x2(x2−4) = 0⇒ x =−2,0,2

13.(x−2)2 = 7⇒ x−2 =±

√7⇒ x = 2±

√7

14. Think: x2−2x+1 = (x−1)2

x2−2x−5 = 0 ⇒ (x−1)2−6 = 0

⇒ (x−1)2 = 6

⇒ x−1 =±√

6

⇒ x = 1±√

6

15. Quadratic formula

x2 +7x+3 = 0 ⇒ x =−7±

√72−4 ·1 ·32 ·1

⇒ x =−7±

√37

2

16. For x2 +2x+3 = 0, the discriminant 22−4 ·1 ·3 =−8 < 0, so the quadratic equation gives no realsolutions.


1.2 −5 1 2

2 4 −2 −2

2 −1 −1 0

This means:2x3−5x2 + x+2 = (x−2)(2x2− x−1)

Now:

2x2− x−1 = 2x2−2x+ x−1

= 2x(x−1)+1(x−1)

= (2x+1)(x−1)

So the other roots are −12

and 1.

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2.2 −3 −11 6

−2 −4 14 −6

2 −7 3 0

This means:2x3−3x2−11x+6 = (x+2)(2x2−7x+3)

Now:

2x2−7x+3 = 2x2−6x− x+3

= 2x(x−3)−1(x−3)

= (2x−1)(x−3)

So the other roots are12

and 3.

3.12 16 −5 −3

− 32 −18 3 3

12 −2 −2 0

This means:

12x3 +16x2−5x−3 = (x+32)(12x2−2x−2)

Now:

12x2−2x−2 = 12x2−6x+4x−2

= 6x(2x−1)+2(2x−1)

= (6x+2)(2x−1)

So the other roots are −13

and13

.

4.3 7 −22 −8

− 13 −1 −2 8

3 6 −24 0

This means:

3x3 +7x2−22x−8 = (x+13)(3x2 +6x−24)

Now:

3x2 +6x−24 = 3(x2 +2x−8)

= 3(x+4)(x−2)

So the other roots are −4 and 2.

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4.5.5 Long division

Tips: for polynomial long division, we write the polynomial being divided under the bar, and the divisoroutside. At every stage, we look only at what needs to be multiplied by the leading coefficient of thedivisor to make the leading coefficient of the dividend. Pitfalls: remember to put blanks where there arepowers of x missing (see number 3). Also be careful when subtracting in intermediate steps – you’resubtracting all the terms.

1.2x2 +3x+5

3x−1)

6x3 +7x2 +12x−5−6x3 +2x2

9x2 +12x−9x2 +3x

15x−5−15x+5

0

This one came out even, so the answer is 2x2 +3x+5.

2.3x +7

x−3)

3x2−2x +5−3x2 +9x

7x +5−7x+21

26

So3x2−2x+5

x−3= 3x+7+

26x−3

3.4x3 +16x2 +60x+246

x−4)

4x4 −4x2 +6x−4x4 +16x3

16x3 −4x2

−16x3 +64x2

60x2 +6x−60x2 +240x

246x−246x+984

984

That was a long one! Remember the blanks!4x4−4x2 +6x

x−4= 4x3 +16x2 +60x+246+

984x−4


1. This is just “multiplying out”:

y = 2(x−1)2 +4 = 2(x2−2x+1)+4 = 2x2−4x+2+4 = 2x2−4x+6

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2. We can ignore the last term as a start. The first two terms are 2x2 +4x = 2(x2 +2x). The perfectsquare closest to the expression in parentheses is x2 +2x+1 = (x−1)2. If we were to replace that,we would have: 2(x−1)2 = 2x2 +4x+2. But what we have is 2x2 +4x+5, which is three more. So:

y = 2x2 +4x+5 = 2(x−1)2 +3

3. The vertex of the standard parabola y = x2 is (0,0). This parabola is shifted right by 1 and up by 4.So its vertex is at (1,4).

4. We wrote this parabola in vertex form above. So the vertex is (−1,3).

5. The vertex form will look like y = a(x+1)2 +3 if the vertex is at (−1,3). If the y-intercept is(0,−2), then the equation at that point reads:

−2 = a(0+1)2 +3⇒−2 = a+3

So a =−5, and the equation is y =−5(x+1)2 +3.

6. If a polynomial has a root of x = a, it has a factor of (x−a). So a parabola with these roots couldhave equation y = (x+2)(x−3). Multiplying this by a constant is not going to change where itequals zero, so any multiple of this would be fine as well.

7. See the answers for the graph. General strategy for graphing in vertex form is to plot the vertex.Then plug in x = 0 to get the y-intercept and plot that. Reflect that point over the axis of symmetry(the vertical line through the vertex) to get a third point on the graph. Then draw the best parabolayou can through those three points.

8. We have already done the work to put this in vertex form, so using that as in the previous example isthe best bet.


Graphing higher-degree polynomial functions is something of an art. But it’s not too bad to get a grasp onsome basics.

1. Polynomial functions of even degree have their ends pointing the same direction (both up or bothdown); polynomial functions of odd degree have their ends pointing in opposite directions (one up,one down).

2. The up and down is decided by the leading coefficient. Positive leading coefficients mean the graphgoes up to the right (and on both sides for a polynomial function of even degree). Negativecoefficients in polynomials of odd degree mean the left end goes up and the right end goes down.

Those are the absolute basics. What else do you have to go on?

1. The constant term gives the y-intercept.

2. If you can factor a little, the roots give you some x-intercepts.

Go on to calculus – you’ll learn even more techniques. For now, if you need more, check the onlinelectures, section 2.2.

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1.

2x2 +5x−3x2 +2x−3

=2x2 +6x− x−3(x+3)(x−1)

=(2x−1)(x+3)(x+3)(x−1)

=2x−1x−1

,x 6=−3

2.

−4x3y2

2x2y3 ·−4xy3

10xy2 =−2x

y· −2y

5

=4xy5y

=4x5,x,y 6= 0

3.

3x−2

+3x

2x+3=

3(2x+3)(x−2)(2x+3)

+3x(x−2)

(2x+3)(x−2)

=3(2x+3)+3x(x−2)

(2x+3)(x−2)

=6x+9+3x2−6x(2x+3)(x−2)

=3(x2 +3)

(2x+3)(x−2)

4.

4x−48x2−144

=4(x−12)

(x+12)(x−12)

=4

x+12,x 6= 12

5.

x2−92x+2

· x2 +2x+1(x−3)(x+1)

=(x+3)(x−3)

2(x+1)· (x+1)(x+1)(x−3)(x+1)

=x+3

2,x 6=−1,1,3

6.

x2 + x−23x2−9x

÷ x2 +4x−56x−18

· x2 +2x−15x2− x−6

=(x+2)(x−1)

3x(x−3)· 6(x−3)(x+5)(x−1)

· (x−3)(x+5)(x−3)(x+2)

=2x,x 6=−5,−2,1,3

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1.

x+53x2

x2−256x3

=x+53x2 ·

6x3

x2−25

=(x+5) ·6x3

3x2(x+5)(x−5)

=2x

x−5,x 6= 0,−5

2.4+

12

13+

16

=

9212

= 9

3.

5x−5 +

3x+3

5x+3 +

3x−5

=

5(x+3)(x−5)(x+3) +

3(x−5)(x+3)(x−5)

5(x−5)(x+3)(x−5) +

3(x+3)(x−5)(x+3)

=

5x+15+3x−15(x+3)(x−5)

5x−25+3x+9(x+3)(x−5)

=8x

8x−16=

8x8(x−2)

=x

x−2,x 6=−3,5


General principles: to graph a rational function:

1. Factor top and bottom completely.

2. If possible, cancel any common factors. If any factors cancel, take note.

3. If, after the previous step, the degree of the numerator is one more than the degree of thedenominator, there is a slant asymptote. To find the slant asymptote, perform long division of thepolynomials, and ignore the remainder. The result will be linear, the equation of your slantasymptote. Plot it as a dotted line.

4. If the degree of the denominator is greater than the degree of the numerator, the denominator is“more powerful”, so the graph has a horizontal asymptote of y = 0. This means that the left and rightends of the graph will head towards zero.

5. If the degree of the numerator and denominator are the same, the graph will approach (horizontally)the quotient of their leading coefficients. Plot this as a dotted (horizontal) line.

6. Any roots of the numerator (after cancelling) are zeros, or x-intercepts, of the function. Plot these onthe graph.

7. Any roots of the denominator (after cancelling) are vertical asymptotes. Plot vertical dashed lines forthese. Your final graph will not cross these lines.

8. Plug in x = 0 to find the y-intercept, and plot it.

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9. At this point, dotted lines will divide your graph into segments. You now need to sketch in the graph,not crossing the vertical asymptotes, and at the end approaching the horizontal (or slant) asymptote.You need a point in every vertical region to decide whether the graph is above or below thehorizontal (or slant) asymptote. Your x-intercepts and y-intercept will be a good starting point. Plot afew more points, if necessary, and make sure that your graph does not have new zeros that you didn’tfind above.

10. If a factor disappeared when simplifying, you need to plot an empty circle in the graph over thatx-value, whose y-value corresponds to the value that x produces when plugged in to the simplifiedfunction.

Tips for the specific problems:

1. This function has a zero at x =−2, a vertical asymptote at x = 3, a horizontal asymptote of y = 1. Atx = 4, y = 6. These points allow you to tell where the graph lies. Then just follow the asymptotes.

2.x2−4x+3

xx−2=

(x−3)(x−1)(x+2)(x−1)

=x−3x+2

,x 6= 1

The simplified function has x-intercept (3,0), vertical asymptote x =−2, horizontal asymptote y = 1,and a “hole” at (1,−2

3).

3.x−1

x2− x−2=

x−1(x+1)(x−2)

This function has an x-intercept of (1,0) and two vertical asymptotes: x =−1 and x = 2. Itshorizontal asymptote is y = 0, since the degree of the denominator is larger. A little more pointplotting fills out the graph.

4.x2 +3x+2

x−1=

(x+1)(x+2)x−1

The degree of the numerator is one greater than the degree of the denominator. The result of longdivision is x+4 plus a remainder. So y = x+4 is a slant asymptote. The zeros are x =−1 andx =−2, and there is a vertical asmptote of x+1. Plugging in x = 2 gives enough information tofinish the plot.


To start: the rules of exponents with the corresponding rules of logarithms:

1. am ·an = am+n

2. am/an = am−n

3. (am)n = amn

4. a0 = 1, if a 6= 0.

1. loga mn = loga m+ loga n

2. loga(m/n) = loga m− loga n

3. loga xm = m loga x

4. loga 1 = 0

1.82/3 = (81/3)2 = (

3√

8)2 = 22 = 4

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2. Negative exponents indicate taking a reciprocal. They do not make the answer negative!

8−2/3 =1

82/3 =14

3.(−8)2/3 = [(−8)1/3]2 = (−2)2 = 4

4.

(−8)−2/3 =1

(−8)2/3 =14

5. Order of operations here: exponents take precedence over the negative sign.

−82/3 =−4

6. Powers are applied to numerator and denominator:(49

)3/2

=43/2

93/2 =

√4

3

√9

3 =23

33 =8

27

7. This logarithm asks the question, “what power of 2 is 12 ?” Since 1

2 is the reciprocal of 2, the answeris -1.

8. This one asks, “what power of 3 is 27?” The answer is 3.

9. What power of 9 is 3? Since 3 is the square root of 9, the answer is 12 .

1. Rewrite this one in exponential form. When log has no shown base, the base is ten:

x = 10−3 = 0.001

2.

x = 2−1/2 =1√2=

√2

2

3.

e2x−2 = 0 ⇒ e2x = 2

⇒ 2x = ln2

⇒ x =12

ln2

4.

8−3 ·20.5x =−40 ⇒ −3 ·20.5x =−48

⇒ 20.5x = 16

⇒ 0.5x = 4⇒ x = 8

5.log3 9x = 3⇒ 9x = 33⇒ 9x = 27⇒ x = 3

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6. x occurs twice in this problem in places that cannot be combined. So we must find a way to factor:

e2x−3ex = 28⇒ (ex)2−3(ex)−28 = 0

The left side looks like y2−3y−28, which factors as (y−7)(y+4), but with y replaced by ex. So:

(ex−7)(ex +4) = 0,

which means that ex = 7 or ex =−4. Since ex is never negative, the only possiblity is ex = 7, whichmeans x = ln7.

7. We want to use rules of logarithms to combine the terms on the left:

log(x+1)− log(x−1) = log3⇒ logx+1x−1

= log3

Now we have an equality of two logarithms. Since the logarithm function is one-to-one, thearguments of the logarithms must be equal:

x+1x−1

= 3⇒ x+1 = 3(x−1)⇒ x+1 = 3x−3⇒ 4 = 2x⇒ x = 2

Note that any time we solve a logarithm equation, we must check that the solution can be plugged into the original equation. In this case, x = 2 is fine. If plugging it in made the argument of a logarithmnegative, we have a false solution, which must be discarded.

8.

ln(x2−4)− ln(x+2) = ln(3− x) ⇒ lnx2−4x+2

= ln(3− x)

⇒ x2−4x+2

= 3− x

⇒ (x+2)(x−2)x+2

= 3− x

⇒ x−2 = 3− x⇒ x =52

And x = 52 does not make any logarithm arguments in the original equation negative (nor any

denominators zero!). So it is a valid solution.

Question: why combine logs? It makes equation solving easier.

1. Note here that the log terms cannot be combined until the coefficient 12 has been brought inside the

first as a power. In a sense, the rules of logarithms should be applied in the same order as order ofoperations (the 1

2 becomes an exponent, so it’s first).

12

log2 x+ log2 y = log2 x1/2 + log2 y = log2√

x+ log2 y = log2 y√

x

2. Same thing here:

3 logx−2logy− logz = logx3− logy2− logz

= logx3− (logy2 + logz)

= logx3− logy2z

= logx3

y2z

Why did I factor out the negative sign in the second line? Remember that order of operations foraddition and subtraction goes right to left. So I would have to do two divisions if I didn’t factor. Thismakes more room for error.

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Question: why do we even learn about separating out logs? Answer: You need to do it in calculus.

1.

log(

x√

yz

)= logx

√y− logz = logx+ log

√y− logz = logx+

12

logy− logz

2.log4 4x2y3 = log4 4+ log4 x2 + log4 y3 = 1+2log4 x+3log4 y


4.8.1 Matrix Multiplication

The size of a matrix is given by its dimensions, in the form (rows) × (columns). For example, the matrix1 7 3−2 0 34 3 1−2 −2 1

is 4×3.Two matrices may be multiplied left-to-right if the number of columns of the left matrix is equal to thenumber of rows of the right matrix.Matrix multiplication is done like this: the result has, in row m and column n, the number gotten bymultiplying each entry in row m of the left hand matrix by the corresponding entry in column n of the righthand matrix and adding the results. The resulting matrix has the same number of rows as the left matrixand the same number of columns as the right matrix. In brief, a m×n matrix may be multiplied by an n× pmatrix and the result is m× p.See the work below to learn how this works in practice. Also note that the order in which matrices aremultiplied is important. Some pairs of matrices can only be multiplied in one direction.

1.


1. A circle is divided into 360 equal pieces of arc called degrees. Another way is to measure the outsidein terms of the length of the radius. One radius measured out on the outside of a circle is called aradian. Since the circumference of a circle is equal to 2πr, there are 2π radians in a circle. Thus theratio of radians to degrees is 2π

360 . So to convert radians to degrees, we multiply by 180π

. To go theother way, multiply by π

180 .

2. See previous item.

3. Coterminal angles look the same when drawn. In other words, they differ by multiples of 2π (inradians) or 360◦. So 390◦ is coterminal to 390◦−360◦ = 30◦. Subtracting another 360 gives anegative angle of −330◦. To do part b) in radians, you should write 2π = 12π

6 , so it can be easilyadded or subtracted with 17π

6 . So a positive coterminal angle is 5π

6 , and a negative would be −7π

6 .

4. The complement of an angle is what needs to be added to make 90◦ or π

2 radians. The supplement iswhat needs to be added to make a straight line (180◦ or π radians).

a) 90◦−72◦ = 18◦;

b) 180◦−72◦ = 108◦;

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c) Write π

2 = 6π

12 . Then 6π

12 −π

12 = 5π

12 .

d) Write π = 6π

6 . Then 6π

6 −5π

5 = π

6 .


Here’s a good way to remember the first quadrant trig function values:

θ(rad) θ(deg) sinθ cosθ

0 0◦ 0 =√

02 1 = sqrt4

2

π/6 30◦ 12 =

√1

2sqrt3

2

π/4 45◦ 0 =√

22

sqrt22

π/3 60◦√

32

12 = sqrt1

2

π/2 90◦ 1 =√

42 0 = sqrt0

2

For the angles in the other quadrants, we need to find a reference angle. For example, 5π

6 is π

6 above thenegative x-axis. So its sine and cosine will be the same as that of π

6 , but with different signs. In the secondquadrant (where 5π

6 lives), sine is positive and cosine is negative.

1. Standard first quadrant.


3. 135◦ is 45◦ above the negative x-axis. secx = 1cosx , and cosx is negative in the second quadrant. So

sec(135◦) = 1cos(135◦) =

1√2/2

=√

2.

4. 300◦ is 60◦ below the positive x-axis (in the fourth quadrant). There, sinx < 0 and cosx > 0, sotanx < 0.

tan(300◦) =− sin60◦

cos60◦=−√

3/21/2

=√

3.

5. cotx = cosxsinx . Since sin0 = 0. cot0 is not defined.


7. Write 2π = 6π

3 . Adding this to −5π

3 to get a coterminal angle gives π

3 . This is a first quadrant angle.

8. 3π

4 is between π

2 and π , so in the second quadrant. It is π

4 above the negative x-axis. In the secondquadrant, cosine is negative and sine is positive, so

tan(3π

4) =− sin(π/4)

cos(π/4)=−1.

9. Since 4 > 3, 4π

3 is a bit more than π , so in the third quadrant. The angle is π

3 below the x-axis. There,sinx < 0, so

csc(4π

3) =− 1

sin(π/3)=

2√3=

2√

33

10. cosπ =−1, so secπ = 1cosπ

=−1.

Once you know the sine of an angle, you know almost all the trig functions. This is becausesin2 x+ cos2 x = 1. This means that

cosx =±√

1− sin2 x

To decide plus or minus, we need geometric information about the angle.

Name 45

1. θ is acute, so in first quadrant, where all trig functions are positive. If sinθ = 0.6,cosθ =

√1−0.62 =

√1−0.36 =

√0.64 = 0.8. Then tanθ = 0.6

0.8 = 34 .

2. Here, tangent and sine are both negative. That means cosine is positive. Since tangent is oppositeover adjacent, we can work from a right triangle with legs 8 and 15. 82 +152 = 172, so thehypotenuse is 17. Taking signs into account, sinθ =−15

17 and cosθ = 817 .

3. θ is in the second quadrant, so sinθ is positive and cosθ is negative. Since sine is opposite overhypotenuse, we can work with a right triangle with hypotenuse 5 and one leg 4. Since 32 +42 = 52,the other (adjacent) leg is 3. Thus cosθ =−3

5 and tanθ =−43 .

4.11 Inverse trigonometric functions

Don’t be confused, sin−1 1 doesn’t mean1

sin1. It is asking the question, what angle has a sine of 1? There

are multiple answers, since at least we can add a full rotation to the angle without changing the trigfunctions. So to have a clearly defined inverse function, we choose the angle with the given value that isclosest to 0, and positive when possible. This makes the inverse trig functions work like:

function returns values insin−1 [−π/2,π/2]cos−1 [0,π]tan−1 [−π/2,π/2]

1. Sine gives the y-coordinate of a point on theunit circle. There’s only one point on the unitcircle with y-coordinate 1, π/2.

2. If an angle has tangent 1, it meanssinx = cosx, since tanx = sinx

cosx . This happensat π/4.

3. Negative cosines are in quadrant two or three.Inverse cosine doesn’t take values in quadrantthree. We need an x-coordinate of −

√3

2 . That

is π/6 above the x-axis, so 6π

6 −π

6 = 5π

6 .

4. Sine is negative in the fourth quadrant, for thepurposes of computing sin−1. A sine of 1/2 isan angle of π/6, so the answer here is −π/6.

5. For inverse tangents of negative angles, welook in the fourth quadrant (negative angles).√

3 is steeper than 1, so the answer is −π/3.

6. See number 3. Here the answer will be π/4above the negative x-axis, or 3π

4 .


1. sinx = cosx. This one is probably easiest to see by superimposing the graphs or looking at the unitcircle. Which points on the unit circle have the same x− and y−coordinates? Where the line y = xmeets the circle. So we are solving x2 + x2 = 1, or x =±

√2/2. These happen at the angles π/4 and

5π/4.

2. 3sinx−2 = 5sinx−1. Strategy here is to begin by isolating the trig function:

−1 2sinx⇒ sinx =−1/2

This happens at x = π/6 or 5π/6.

3. 2cos2 x+ cosx−1 = 0. This equation is like a quadratic equation in cosx. Factor it as:

(2cosx−1)(cosx+1) = 0

So we have either 2cosx−1 = 0 or cosx+1 = 0. The first equation means that cosx = 1/2, whichhappens when x = π/3,5π/3. The second equation means cosx =−1, which happens at x = π .

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4. 2cos2 x+3sinx = 0. This one has a mix of trig functions. To make it an equation of a singlefunction, we use an identity: cos2 x+ sin2 x = 1, so cos2x = 1− sin2 x. Plugging this in to the originalequation gives:

2(1− sin2 x)+3sinx = 0⇒ 2−2sin2 x+3sinx = 0⇒ 2sin2 x−3sinx−2 = 0

Factor:(2sinx+1)(sinx−2) = 0

which yields sinx =−1/2 or sinx = 2. The first equation gives x = 7π/6,11π/6. The secondequation cannot be solved, since sinx is never larger than 1.


The rules for modifying the graphs of trig functions are the same as for other functions. However, somespecial terminology is involved.For example, for the function Asin(Bx+C)+D,

1. A is called the amplitude (this term is not used for the tangent function). Since sine and cosine takevalues only between -1 and 1, when multiplied by A, they take values between −A and A.

2. D is a shift up (D positive) or down (D negative). It doesn’t have a special name.

3. Remember that inside of a function, horizontal shifts are performed before stretching. So thefunction shifts left (C positive) or right (C negative). After this, the graph is stretched (orcompressed) by a factor of 1/B. The former value of the function at 0 is now shifted to −C, thenstretched to −C/B. This number is called the phase shift. The sine and cosine functions have aperiod of 2π (meaning they repeat every 2π). The period of the tangent function is π . Because of thestretch factor, for sine and cosine functions, the period is 2π

B , and for tangent, it is π

B .

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