sep 57:00 amstuttgarthsmath.weebly.com/uploads/1/3/4/6/...math analysis ch 1 review solutions 2...

Math Analysis Ch 1 Review Solutions

1

September 12, 2012

Sep 57:00 AM

#1. 8x + 10 = 4x 304x 4x4x + 10 = 30

10 104x = 404 4

x = 10


2

September 12, 2012

Sep 57:00 AM

# 2. 4 3(x + 2) = 5x 7(4 x)4 3x 6 = 5x 28 + 7x CLT

2 3x = 12x 28+3x +3x2 = 15x 28

+28+2826 = 15x15 15

x = 2615


3

September 12, 2012

Aug 308:23 AM

Find LCDfor ALL fractions.

LCD = 30

2y + 20 6 = 5y + 5 3 CLT

2y +14 = 5y + 2

14 = 3y + 2 2 2 12 = 3y

2y 2y

3 3 y = 4

y + 10 1 y + 1 1#3. 15 5 6 10 =

2y + 20 6 5y + 5 3 30 30 30 30 =


4

September 12, 2012

Aug 281:34 PM

# 4. 3(2 x) 2 ≤ 2x 1

x ≥ 1

To graph inequalities: TWO WAYS(1) Coordinate System

For x ≥ 1First think ofx ≥ 1 as x = 1Second, what does ≥tell us, solid or dottedline?Third, shade in the solution. In this case all values of x ≥ I

(2) Number line

Unfilled > , <

Filled ≥ , ≤

1 02345 1 2 3 4 5

Remember when graphing the cases: y = #

x = #

6 3x 2 ≤ 2x 14 3x ≤ 2x 1+3x +3x 4 ≤ 5x 1+1 +15 ≤ 5x5 5


5

September 12, 2012

Aug 308:33 AM

The familiar method of breaking the absolute value inequality into TWO inequalities.

TWO WAYS OF SOLVING ABSOLUTE VALUE INEQUALITIES:

#5. |y+9| < 5

(1) |y + 9| ≤ 5

a. y + 9 < 5 b. y + 9 > 5y < 4

(2) |y + 9| < 5

5 < y + 9 < 5 Solve for y.

14 < y < 4

The first inequality stays the same but drops the absolute value bars

The second inequality has the inequality symbol flipped and the number becomes its opposite.

y > 14

In this second form, what I call the compounded form, the inequality sign moves to the front BUT DOES NOT FLIP. The number becomes its opposite again and is placed at the very front.

Whatever you do to the middle part, you also have to do to ALL OTHER PARTS!

9 9 9

To Graph on a coordinate system, remember the three steps from before! Don't remember? Look back!

y < 4y > 14


6

September 12, 2012

Aug 308:39 AM

# 6. |3 2x| ≤ 5

5 ≤ 3 2x ≤ 53

8 ≤ 2x ≤ 22 2 24 ≥ x ≥ 1

(2) Number line

(1) Coordinate System

10 2 3 4 5 6 7 8 9 1012345678910

1 ≤ x ≤ 4Proper compound Inequalities are written with the inequality pointing to the left.

33


7

September 12, 2012

Aug 308:42 AM

# 7. Standard form of a complex number

a + bi

9 4i

Real term: 9 Imaginary: 4iconjugate: opposite of imaginary term

5i Looking at standard forma + bi

a.

9 + 4i

b.

Real term: 0 Imaginary: 5i

conjugate: opposite of imaginary term5i

c. 10Real term: 10 Imaginary: 0

conjugate: 10


8

September 12, 2012

Aug 308:46 AM

#8a. 2+4i

CLT

#8b. (3 + 5i) (4 8i)3 + 5i 4 + 8i

7 + 13i

#8c. (1 2i)(3 + 4i)FOIL

3 + 4i 6i 8i2 Note : i2 = -1

3 2i 8(1)3 2i + 8

11 2i


9

September 12, 2012

Aug 308:51 AM

# 9. 2x2 7 = 0

x all by itself

+7 +72x2 = 72 2

x2 = 72

x = ± 72√

ax2 = ±c

When we have a quadratic equation with bx term missing, then we should convert the remaining terms into


10

September 12, 2012

Aug 308:54 AM

ax2 = ±c# 10. 5x2 + 20 = 020 205x2 = 205 5x2 = 4

x = i√4x = ±2i

x = √-4


11

September 12, 2012

Aug 308:56 AM

# 11. 2x2 = 4x

ax2 + bx + c = 02x2 4x = 0

I can factor, but what?What do they have in common?

Check. 2x2 4x

When you are missing "c" from ax2 + bx + c = 0Just keep it in this form→ ax2 + bx = 0

2x(x 2) = 0

2x(x 2) = 0

If you aren't sure about your factoring, CHECK IT by distributing.

2x(x 2) = 0So what is our solution?

Remember this?Ex. x2 5x + 6 = 0 (x 2)(x 3) = 0

x = 2; x = 3We could easily factor that and just say the following, right?

But why can we do that?

(x 2)(x 3) = 0 (x 2) = 0 => x = 2 (x 3) = 0 => x = 3

2x(x 2) = 0Returning to our problem, what is our solution?

2x(x 2) = 0

(x 2) = 0 x = 2

2x = 0 x = 0


12

September 12, 2012

Aug 309:01 AM

# 12. 2x2 = 7x - 32x2 - 7x + 3 = 0

x = 1/2, x = 3(2x - 1)(x - 3) = 0

Solution if you are good at factoring

If you are not good at factoring, you will have to use the quadratic formula!

x = b ± √b2 4ac2a

x =(7) ± √(7)2 4(2)(3)

2(2)

ALWAYS PUT YOUR a, b, c VALUES IN PARENTHESES!It helps you avoid making silly errors!

x = 7 ± √49 244

x = 7 ± √254

x = 7 ± 54

x = 7 + 54 = 12

4 = 3 x = 7 - 54 = 2

4 = 12


13

September 12, 2012

Aug 309:05 AM

# 13. m2 + m + 1 = 0

x = b ± √b2 4ac2a

x =(1) ± √(1)2 4(1)(1)

2(1)

x = 1 ± √1 42

x = 1 ± √32

x = 1 ± i√32

If the m's are bothering you, change the variables to something you are more used to, like x's

x2 + x + 1 = 0


14

September 12, 2012

Aug 281:44 PM

# 14. y2 = (y+1)32

If you have a x2

you want to thinkquadratic function

ax2 + bx + c = o

Why work with fractions though? There are TWO EASIER WAYS OF SOLVING THIS:

(1) Find a common denominator

(2) Finding LCD and multiply(You have to remember this trick!)

LCD = 2

32

y2 = y +32

32

y2 = y +32

22

32

y2 = y +32

32

y2 = y +32

2( )2y2 = 3y + 3

2y2 - 3y - 3 = 0

2y2 = 3y + 3 => 2y2 - 3y - 3 = 0

x = b ± √b2 4ac2a

Now let's find the solution:

x =(3) ± √(3)2 4(2)(3)

2(2)

x =3 ± √9 + 24

4

x = 3 ± √334

√33 4

34 ±or

32

y2 - y - = 032


15

September 12, 2012

Aug 309:11 AM

#15. √5x 6 x = 0√5x 6 = x (√5x 6)2 = (x)2

5x 6 = x25x + 6 5x + 6

0 = x2 5x + 6 0 = (x 2)(x 3)x = 2; x = 3

Note:ax2 + bx + c = 0try to keep ax2

positive


16

September 12, 2012

Aug 309:13 AM

#16. √15 + 6xThe question is when is the expression a real number? Conversely, when do we have imaginary numbers?

√15 + 6x ≥ 0(√15 + 6x)2 ≥ (0)215 + 6x ≥ 015 15

6x ≥ 156 6x ≥ 5

2


17

September 12, 2012

Sep 127:31 AM


18

September 12, 2012

Sep 127:35 AM


19

September 12, 2012

Sep 48:16 AM

3u 9 = u 1 2 + 2u

No Solution

3u 9 = 3u 33u 3u

9 = 3

#18. u 3 1 1 u 2u 2 6 3u 3 =

What can we do? Find a common denominator, but what is it?If we had the fractions

1 1 1 2 6 3 = => LCD = 6 3 1 2

6 6 = 6Then it would be easy.So 6 is part of the common denominator, but what about the u's?

u 3 1 1 u 2u 2 6 3u 3 = Factor

u 3 1 1 u 2(u 1) 6 3(u 1) =

We see two of the three denominator also have (u 1) as part of the denominator. Therefore our LCD should be 6(u 1). Note 6u 6 could also work just as well.

3(u 3) 1 2(1 u) 3*2(u 1) 6 2*3(u 1) = (u 1)

(u 1)

Remember, whatever you do to the bottom (denominator) you also have to do to the top (numerator).

3(u 3) 1 2(1 u) 3*2(u 1) 6 2*3(u 1) = (u 1)

(u 1)

3u 9 (2 2u) 6(u 1) 6(u 1) = (u 1)

6(u 1)

Once you have a common denominator, you can cancel/cross it out and just work with the numerator.


20

September 12, 2012

Sep 47:11 AM

LCD = 24#19. x + 3 2 x 5 8 ≤ 3


21

September 12, 2012

Sep 48:29 AM


22

September 12, 2012

Sep 48:37 AM


23

September 12, 2012

Sep 48:41 AM

#22a. d(A, B) = 15b. d(A, C) = 13c. d(B, C) = 28

First, "d" means distanceSecond, distance is always positive

There are at least two ways of solving this problem. Either are fine. Our goal should be familiarizing ourselves with the notation d( , ).

d = |b a| = |a b|

1. Since distance is always positive a formula is given written as an absolute value. Either are fine as both will produce the same solution

2. The other, simpler "oldschool", method involves just drawing a number line, plotting the locations of the points, and counting the tick marks.

20181614121086420


24

September 12, 2012

Sep 48:46 AM

#23a. (3 + i)2 2(3 + i) + 39 + 3i + 3i + i2 6 2i + 3

9 + 6i + i2 3 2i9 + 6i + i2 3 2i

6 + 4i + i2 Note : i2 = -1

6 + 4i + (1)5 + 4i

Be very careful when foiling.Common mistake:

(3 + x)2 => when foiled ≠ 9 + i2 (3 + x)2 => (3 + x)(3 + x) = 9 + 3x + 3x + x2


25

September 12, 2012

Sep 48:49 AM


26

September 12, 2012

Sep 48:55 AM

#24a. (2 √4) (3 √9)

2 i√13 + i√4

(2 i√4) (3 i√9)(2 2i) (3 3i)2 2i 3 + 3i

1 + i

2 √13 + √4

2 i3 + 2i

Again, when dividing complex numbers, we have to multiply the denominator by the conjugate of the denominator!

What is the conjugate?Real part stays the sameImaginary becomes its opposite

(2 i)(3 + 2i) (3 2i)

(3 2i)

(2 i)(3 + 2i) (3 2i)

(3 2i)

6 4i 3i + 2i29 + 6i 6i 4i2

6 7i + 2i29 4i2

Note : i2 = -1

6 7i + 2(1)9 4(1)

6 7i 29 + 44 7i13

#24b.


27

September 12, 2012

Sep 49:05 AM

One more note: Our answers should always be in form of

#24c. 4 + √-25√-4

4 + i√25i√44 + 5i2i

Until now, we have used the conjugate of the complex number denominator to help us divide complex numbers into each other.

There is an exception to the rule:When there is no REAL term of the complex number, just the IMAGINARY term, we multiply by

STOP! Are we done yet?

NO! We shouldn't have i's in the denominator.

How do we get rid of the i?

ii

(4 + 5i)(2i)

ii

*

Let's try it out

4i + 5i22i2

Note : i2 = -1

4i + 5(1)2(1)

4i 52

or 2i + 52

a + bi

5 4i2

2i 52

or

with the real part first and the bi, or imaginary part, secondTrue solution can either be


28

September 12, 2012

Sep 49:11 AM

#25. (y + 113 )2 = 20(y + 113 )2 = √20√y + = √2011

3 √20 = √4 * 5= √4 * √5= 2√5

113

y + 113 = 2√5113 113y = 2√5

or

y = 6√5 113


29

September 12, 2012

Sep 49:16 AM

LCD = u2

Perform the Trick:Find LCD and multiply

#26. 1 + 3u2 = 2

u

u2 1 + 3u2 = 2

u )(u2 + 3 = 2uu2 2u + 3 = 0

u = b ± √b2 4ac2a

u =(2) ± √(2)2 4(1)(3)

2(1)

u = 2 ± √4 122

u = 2 ± √82

u = 2 ± i√82

√8 = √4 * 2= √4 * √2= 2√2u = 2 ± 2i√2

2

u = 1 ± i√2

1 + 342

= 24

1 + 316 = 2

4

Just in case you don't see why u2 is the common denominator and not u

With numbers its easy. You would say 16 is the LCD, but isn't 16 really just 42?


30

September 12, 2012

Sep 412:51 PM

LCD = (x 3)(x + 2)

Factoring we have (x 3)(x + 2)

or

#27. xx2 x 6

2x 3 = 3

When looking at this problem, you might want to think finding a common denominator is a good way to start

But... before we multiply the two denominators together, we also might be interested in factoring

x2 x 6

x(x 3)(x 2)

2x 3 = 3

Notice, we do actually have a factor in common.

Using our trick of finding the LCD and then multiplying, we have

(x 3)(x + 2) x(x 3)(x + 2)

2x 3 = 3( )

x 2(x + 2) = 3(x 3)(x + 2)x 2x 4 = 3(x2 x 6)x 4 = 3x2 3x 18+x +4 +x +4

0 = 3x2 2x 14We have a Quadratic Function, Looks Like we will need the Quadratic Formula

x = b ± √b2 4ac2a

x =(2) ± √(2)2 4(3)(14)

2(3)

x = 2 ± √4 + 1686

x = 2 ± √1726

x =2 ± 2√43

6

√172 = √4 * 43= √4 * √43= 2√43

x =1 ± 1√43

3

x = 1 1√43± 3 3


31

September 12, 2012

Sep 41:01 PM

= x(x3)13

13x = ∛x


32

September 12, 2012

Sep 68:36 AM

(1) We "pretend" for a split second that we really do havea quadratic function and either factor or use the quadratic formula to find our solutions. THERE'S ONLY ONE CATCH!!!! Our solutions are not x = ...., but rather x2 = ....

#29. m4 + 5m2 36 = 0If the m's bother you, we can also change this to

x4 + 5x2 36 = 0Before we solve this, let's take a quick look at a standard quadratic function

ax2 + bx1 + c = 0Notice in a quadratic function the powers share a 2:1 ratio. The same as in the original problem. When this happens, we have a special quadratic function case. This means, as long as there is a 2:1 ratio between the highest power and the next highest power in a function, we can think of it as a quadratic function.

2:1

x4 + 5x2 36 = 02:1

Therefore we have one of two options:

Factoring x4 + 5x2 36 = 0(x2 + 9)(x2 4) = 0

Our solutions are:

x2 + 9 = 0x2 = 9x = ±√9x = ±i√9x = ±3i

x2 4 = 0x2 = 4x = ±√4x = ±2

(2) We could use our newly learned usubstitution.Let u = x2

Our problem can then be rewritten as u2 + 5u 36 = 0

Either factoring or using quadratic formula we would have (u + 9)(u 4) = 0

Our solutions according to u are:

u + 9 = 0 u 4 = 0u = 9 u = 4

If we choose to use the usubstitution method, we just have to remember to sub back in our original parts

u = 9 u = 4x2 = 9 x2 = 4x = ±√9x = ±i√9x = ±3i

x = ±√4x = ±2


33

September 12, 2012

Sep 68:43 AM

when put together, the other needed factor

y = 9 4

y = 3 FINAL SOLUTION:


34

September 12, 2012

Sep 69:05 AM

#31. 2.15x 3.73(x 0.930) = 6.11x

2.15x 3.73x + 3.469 = 6.11x1.58x + 3.469 = 6.11x+1.58x +1.58x

3.469 = 7.69x7.697.69

x = 0.451


35

September 12, 2012

Sep 69:10 AM

Note about inequality notation:The inequality signs, WHEN WRITTEN AS A COMPOUND

INEQUALITY LIKE ABOVE, should always point left

#32. 0.770 2.04x ≤ 5.331.52 ≤Solve for x.

0.770 2.04x ≤ 5.331.52 ≤0.770 0.770 0.770

2.04x 2.29 4.56 ≤ ≤ 2.04 2.04 2.04

1.12 ≥ x ≥ 2.24

2.24 ≤ x ≤ 1.12


36

September 12, 2012

Sep 69:10 AM

We said as a compound inequality, the inequality signs should point left.

2.24 ≤ x ≤ 1.12

This is a good time to talk aboutINTERVAL NOTATION

Until now, we have written intervals as inequalities like in the last problem

And we are used to graphing these types of inequalities on a number line using and and as or lines on a coordinate system. As we are getting closer to the college level, we also have to begin using "college" notation. This isn't something new or completely different. We are just going to modify what we already know.

2.24 ≤ x ≤ 1.12Let's take a look at the last solution, we had:

The way I think about the new notation is simple. Look at the symbol ≤, it has a bar underneath the < sign. For me, the bar

≤ | [ Which kind of looks like

Brackets, [ ], tell us the interval we are working with includes the point.

So our solution, 2.24 ≤ x ≤ 1.12, can be written as:

[ 2.24, 1.12]


37

September 12, 2012

Sep 69:16 AM

#33. | 9.71 3.62x | > 5.48

9.71 3.62x > 5.485.48 >

9.719.71 9.7115.19 3.62x 4.233.62 3.62 3.62< <

4.20 < x < 1.17 Notice the "solution" may be pointing to the left as we want it to, but the statement tells us that x is greater than 4.20 but less than 1.17. It would be better if we wrote our solution as:

x > 4.20 x < 1.17orWith the x's first

9.71 3.62x > 5.485.48 >

We can rewrite the absolute value inequality as a compound inequality.

| 9.71 3.62x | > 5.48the number becomes its opposite

the inequality sign stays the same

To solve for x, we can start by subtracting 9.71. Whatever we do to one part, we have to do for all parts.


38

September 12, 2012

Sep 69:20 AM

95 24t 65

LCD = 30

#34. | 8 3 4 5 | t ≤ 1 2

≤ 8 3 4 5 t ≤

1 2

1 2 Now we can perform our trick:Find the LCD and multiply it into ALL PARTS

≤ 8 3 4 5 t ≤

1 2

1 2 30( )

15 ≤ 80 24t ≤ 158080 80

24 24 24≥ ≥

9524 24≥ ≥t 65

9524 24≤ ≤t65 True solution when

written in compound form.


39

September 12, 2012

Sep 129:40 AM

#35. 6.09x2 - 4.57x - 8.86 = 0

x = b ± √b2 4ac2a

x =(4.57) ± √(4.57)2 4(6.09)(8.86)

2(6.09)

x = 4.57 ± √20.885 + 215.8312.18

x = 4.57 ± √236.71512.18

x = 4.57 ± 15.38612.18

x = 1.64 x = 0.888Your solutions may vary based on how you rounded. That's ok for now. As long as your solutions are close to those numbers, its fine.

sep 57:00 amstuttgarthsmath.weebly.com/uploads/1/3/4/6/...math analysis ch 1 review solutions 2...

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