Question

how to find the range and domain of any function.

Answer 1

Do you have any example ??

Answer 2

See for example:

\[f(x) = ax + b\]

Here x in input and f(x) or y is output..


Domain is to find the input or you can say values of x..

And Range is to find the values of f(x) or y or you can output of the function..

Answer 3

@waterineyes , I just want to know how to work the range and domain of any function, I know that they are the y and x-values respectively.
i can find the domain and range of a linear function but the others like quadratic,rational,logarithmic etc are hard.

Answer 4

@bronzegoddess Do u know what is a function ?

Answer 5

For quadratic we are having @experimentX here with us..
So don't worry..

Answer 6

Ha ha ha..

Answer 7

oh ... not really

Answer 8

He he he..

Answer 9

@bronzegoddess can you give one example here so that we can work on it..??

Answer 10

sorry, i was doing my other assignments, @hba yes, i know what a function is.
@waterineyes for example y=2-x-x^2

Answer 11

no are no restriction on that example so it would be all real numbers or (-infinity, infinity), it just depends on the function

Answer 12

I think you can find the domain easily there @bronzegoddess yes or no ??

Answer 13

@nickhouraney that's what i want to know when is it not all real numbers, how can you tell?

Answer 14

do you know number inside the square root brackets must be greater than equal to 0 ??

Answer 15

@waterineyes is the domain all real numbers?

Answer 16

Yep..

Answer 17

sorry didnt see your response waterineyes

Answer 18

There are certain conditions:

1. In case of square root : The number inside the brackets must be \(\ge 0\)

2. In case of Fraction: Denominator must not be 0..

3. In case of Logs: The number must be positive..

Answer 19

ah because a square root can't be equal to a negative..

Answer 20

@bronzegoddess are you seeing any of these conditions here in that example??

Answer 21

Yep..

Answer 22

1, so the range is ]infinity,0]

Answer 23

How did you find the range ??

Answer 24

well if y is the output, and the function is a square and squares result only in + integers then the range is from 0 to infinity....

Answer 25

Really ??

Answer 26

no

Answer 27

if you graph it youll see

Answer 28

i don't know.. i was just guessing from your 3 cases..

Answer 29

Simply put x = 2 there..

Answer 30

That are the cases I have told you for your question that what will the domain be not all real numbers..

They are the three cases of Domain..

Not range..

Answer 31

as far as i know graphing is the only way to find the range any other ways water?

Answer 32

I mean to say that they also apply to Find the range also..

Answer 33

Yes I have one other way beside graphing..

Answer 34

ah, sorry i mistoke them for range.

Answer 35

graphing calculator? lol

Answer 36

we aren't allowed graphing calcs in the exam :(

Answer 37

Ha ha ha..

No @nickhouraney

No graphs..

Answer 38

@waterineyes your case number 1 can also have negative integers right? why is it only positives?

Answer 39

@bronzegoddess this much you got that Domain is values of x and here Domain is All Reals ???

Answer 40

In square root brackets ???

Answer 41

i don't understand your questions.. yes the domain is all real number but you said that the number in the bracket has to be \[\ge 0\] which means 0 and positive intergers but negative's can also apply..

Answer 42

More clearly:

\[\sqrt{-2}\]
Can you find its value ??

Answer 43

the greater than or equal to 0 is only applied to his first case

Answer 44

\[i \sqrt{2}\]

Answer 45

I mean to say that whatever be the number inside the brackets must be greater than or equal to zero...


No in case of Relations and Function we do't include Complex Numbers..

Answer 46

@waterineyes what do you mean by square root brackets?

Answer 47

he just meant inside square root

Answer 48

inside the square root, you can only have positives and zero but in brakcets you can have both negative, zero and positve

Answer 49

@waterineyes you are a guy?

Answer 50

See if you question is like this:

Find the domain of : \(y = \sqrt{x + 2}\)

Here to find domain you know \(x+ 2\) must be greater than or equal to 0:

So:

\[x + 2 \ge 0 \implies x \ge -2\]

Answer 51

okay.. then?

Answer 52

This is domain..

Answer 53

okay and the range?

Answer 54

If you find its range then:

you have to find y in term of x:

Square both sides;

\[y^2 = x+ 2\]

\[x = y^2 - 2\]

As there are no three cases so y is all reals..

Answer 55

ahh i see

Answer 56

*Sorry find x in terms of y..

Answer 57

as there are no three cases?

Answer 58

That I have mentioned earlier square root , logs fractions ..

Answer 59

i know what you are saying though, to find the domain, solve for y and to find the range solve for x..right?

Answer 60

but those are for domains..

Answer 61

No no they equally apply to range also..

Answer 62

oh okay :) thank you!
so i need to remember the three cases; and when i am solving for y then i am find the domain and x is range

Answer 63

Remember:

For domain you function should be like this : y = something

For range you function should be : x = something..

Answer 64

x is not range..



x is domain..

y is range..

Answer 65

\[y = \sqrt{x + 2}\]
Here you are finding values of x and this is domain..

Answer 66

yes, i know i mean solve for x to find the range and solve for y to find the domain.. x=something and y=something

Answer 67

yes, thats what i meant :)

Answer 68

\[x = y^2 - 2\]

Here you are finding y that is called Range..

Answer 69

But we have not found range for this function : \(y = 2-x-x^2\)

Answer 70

Do you know about completing the square method ??

Answer 71

yes

Answer 72

For simplicity multiply -1 both the sides first..

Answer 73

-y=-2+x+x^2

Answer 74

\[x^2 +x -2 = -y\]

\[(x)^2 + x + (\frac{1}{2})^2 - (\frac{1}{2})^2 + 2= -y\]

Answer 75

x^2 +x+2=-y

Answer 76

@waterineyes we didn't finish completing the square..

Answer 77

First, Check page 39: Chapter 1.4 Functions
http://finedrafts.com/files/Larson%20PreCal%208th/Larson%20Precal%20CH1.pdf

Then check out a great Domain and Range topic initiated by UnkleRhaukus a couple of weeks ago.

http://openstudy.com/users/panlac01#/updates/501644f3e4b04dfc808a90e7

Additionally, there's a lot of tutorial videos look just like in the link below. Salman Khan is all over the place :) so I just happen to pick his tutorial.
http://www.youtube.com/watch?v=VhokQhjl5t0

Answer 78

thank you very much, i'll look at the links :)

Answer 79

Sorry there will come -2..

Answer 80

\[(x)^2 + x + (\frac{1}{2})^2 - (\frac{1}{2})^2 - 2= -y\]

Answer 81

how? do you have another case for irrational functions?

Answer 82

\[-y = (x + \frac{1}{2})^2 - \frac{9}{4}\]

Now again multiply -1:
\[y =- (x + \frac{1}{2})^2 + \frac{9}{4}\]

Answer 83

See you have \(y = 2 - x - x^2\)

When you multiply -1, then 2 will become negative..

Answer 84

\[\implies -y = x^2 + x - 2\]

Answer 85

By comparing that will the vertex equation:

\[y = (x - h)^2 + k\]

\[(h,k) = (-\frac{1}{2}, \frac{9}{4})\]

Answer 86

As there is negative x in our equation so range will be:

\[Range : (-\infty, k]\]