Question

Find the Domain and Range of the following sets of ordered pair:
{(x,y) | y-2x=4}
{(x,y) | y=1-x/x-2}
{(x,y) | y=x^2-2}
{(x,y) | y=√ 4-x}

Any Ideas guys?

Answer 1

those are separate questions right?

Answer 2

Nope. They are on the same number.

Answer 3

Uhhh wait. Yeah they are. Just on the same number.

Answer 4

It's like 1. a) b) etc.

Answer 5

ohhh

Answer 6

are you familiar with domain restrictions?

Answer 7

Uhmm nope. I think we did study that before. But it' been a year since then. But I do know that the Domain is the X and the Range is Y. It's just that the set are quite complicated with all the roots and division going on.

Answer 8

Some of them are even Quadratic Functions right?

Answer 9

Let me just search for that first. So what does that got to do with the Sets?

Answer 10

domain restrictions happen ONLY when:

1) \(\frac{1}{f(x)}\) f(x) means an expression with x variables

2) \(\sqrt{f(x)}\) same meaning of f(x)

now look at a) is it in the form of either one of these?

Answer 11

Yes? So I have to rewrite it into slope form with is y=mx+b?

Answer 12

look at a...is there any fraction?

Answer 13

NOpe

Answer 14

is there any square root symbol?

Answer 15

Nope. So does that mean. That it doesn't have Domain Restriction?

Answer 16

yes..that means the domain for a is all real numbers

Answer 17

now...to find range you have to solve for x...what do you think you should do to isolate x?

Answer 18

we're still talking about a) btw

Answer 19

Uhhhm. Move Y to the right side and then Divide both side with 2?

Answer 20

how will it look like? can you type it?

Answer 21

x=4+y/2

Answer 22

now..is there a fraction?

Answer 23

Yes

Answer 24

is there a y in the denominator?

Answer 25

No

Answer 26

There's 2

Answer 27

therefore first restriction is crossed out...because there has to be a variable in the denominator

is there a square root symbol?

Answer 28

Nope

Answer 29

therefore the range for a is all real numbers

Answer 30

now...you may ask what if there is a restriction...

Answer 31

Yeah

Answer 32

let's take b for example

Answer 33

is there a fraction?

Answer 34

Yes there is, and there's also a variable

Answer 35

in the denominator

Answer 36

therefore there is arestriction...

Answer 37

Yes

Answer 38

to find this restriction equate the denominator to 0 (note this is only done for the first restriction)

what do you get when you equate the denominator to 0?

Answer 39

Uhmmm ..wait

Answer 40

sure

Answer 41

Does Y become 0?

Answer 42

LIke 0 = 1-x/x-2?

Answer 43

dont mind anything else...just the denomiantor

Answer 44

Ohhh!

Answer 45

It become the opposite

Answer 46

x=2, x=-1

Answer 47

just the \(\mathbf{denominator}\)

x - 2 = 0

there fore x = 2

that is the restriction

the domain is all real numbers except 2

Answer 48

got it?

Answer 49

Yep same goes for the Range right?

Answer 50

for range...you have to solve for x first before attempting to find restircitions

Answer 51

*domain sorry

Answer 52

What happens with the Numerator? I don't touch it?

Answer 53

but that is the domain

Answer 54

when finding for restrictions in restriction #1 you just look at the denominator

Answer 55

So every time I have a equation such as that. I equate the denominator only to get the Domain?

Answer 56

yep

Answer 57

Ohhh . Okay now I understand. So how do I solve for the Range?

Answer 58

Since I already have X do I just substitute?

Answer 59

that's usually the tricky part..start by cross multiplying

Answer 60

no you dont sub...2 is not a value of x...it's the opposite actually...it's the not value of x...x is true for all numberS EXCEPT 2

Answer 61

Okay. So what Do I cross multiply?

Answer 62

y(x-2) = x - 1

got that?

Answer 63

Ohhhhhhh. Okay . So I equate them?

Answer 64

nope...distribute y

what do you get??

Answer 65

Ohh wait. so Multiply both side with the denominator to transfer it to the other side of the equation?

Answer 66

note this kind of manipulation..it is very common ;)

Answer 67

yep

Answer 68

that's what we call cross multiply

Answer 69

Okay. So after cross multiplying>

Answer 70

distribute y into x - 2

Answer 71

xy-2y=1-x

Answer 72

Then?

Answer 73

Solve for Y?

Answer 74

1-x(-xy)/2 ?

Answer 75

range means solving for x remembeR?

Answer 76

Wait, Isn't Range Y?

Answer 77

so xy - 2y = 1 - x

put all x on one side..all non x to the other

Answer 78

range is the set of y yes...look at when we were solving for domain..it was y = f(x) <---F(x) is an expression of x

Answer 79

OHh okay.

Answer 80

so put all x on one side?

Answer 81

Yep it would look like. -1+y-2y= x-x ?

Answer 82

or is it 2x = 1-y+2y?

Answer 83

so Divide by 2 ir would be x= 1+y/2?

Answer 84

it's actually neither

Answer 85

what? Then what is it?

Answer 86

im gonna save you the solving

Answer 87

That would really help.

Answer 88

put all the x on one side so it would just be

xy +x = 1 - 2y

Answer 89

got that?

Answer 90

OHhhh. What I did i, that I removed X from XY! Silly me! So you can't do that right?

Answer 91

Yep I got it

Answer 92

But isn't 2y negative? So when you move it , it becomes positive right?

Answer 93

now...you "remove" x by factoring it

x(y +1) = 1 - 2y

Answer 94

yeah my mistake

Answer 95

x(y+1) = 1 + 2y

Answer 96

Where did you get 1? x=1?

Answer 97

Ohh now I get it

Answer 98

you good?