Question

Domain and Range

Answer 1

what is the function

Answer 2

?

Answer 3

Thats what the question is :)

Answer 4

one or more of the four mentioned..

Answer 5

Alright so the domain is all the x values the function can have and the range is all the y values.

Answer 6

Right

Answer 7

alright so now you have to figure out what makes a function undefined

Answer 8

so, you'd set them to 0 and solve for it, right?

Answer 9

It's the 1st and the 3rd option.
For the 2nd and 4th choice the value of y < - 2 which is beyond the range.

Answer 10

1st and 3rd

Answer 11

The first equation and the third equation on a graph are typically reflected.. if I graphed each of these are right.. Im just not sure how to find domain and range in general.

Answer 12

each of these right*

Answer 13

Well it depends on the function.
If there is no denominator, the domain is most probably all real numbers.
If there is a denominator, the domain would be all except that number which when subtracted from the number of the denominator would give you a 0.

Answer 14

Because it is too much to be spoken here, http://www.purplemath.com/modules/fcns2.htm

Answer 15

I had all the four functions graphed, reading " y < - 2 which is beyond the range." made perfect sense ._." lol

Answer 16

Why the .__.?

Answer 17

ln x

Answer 18

I didn't expect anything I was going to read to make sense with what I graphed

Answer 19

domain usually depends on ln(x) log(x) or fraction functions

Answer 20

usually all real unless you have one of those

Answer 21

also square roots factor into domain

Answer 22

Finding domains:

1) look for what will make the function undefined
2) find the value that will make the function undefined (c/0, ln(-c),negative square root,etc)

For example say you have

\[y=\sqrt{x+1}\]
if x+1 is less than 0, then the value is undefined

\[x+1\geq0\]
\[x\geq -1\]

Answer 23

so your domain would be \[(-1,\infty]\]

Answer 24

Right....

Answer 25

and you're havinga problem finding the range?

Answer 26

Mhm, finding both in general.

Answer 27

do you know how to find the inverse of afunction

Answer 28

Yeah (:

Answer 29

well think about what you're doing when you take the inverse of the function, you're interchanging values .. also think of what inverse functions look like

Answer 30

so if you were to find an inverse function , the domain of that inverse will be your range

Answer 31

Okay, that does make sense :3

Answer 32

however we know that the domain cuts it off at -1 so your lowest value will be at x=-1

Answer 33

so you have

\[\sqrt{x+1}\]

\[D:[-1,\infty]\]
\[R:[f(-1),\infty]=[0,\infty]\]

Answer 34

so what is the function you're using and show me what you'd do first

Answer 35

There was a mistakee in my work... when you deal with square roots don't square, it messes up the equality leave it as such, and see that xcan only be >= 0

\[y=\sqrt{x+1}\]


\[x=\sqrt{y+1}\]

Answer 36

make sense?

Answer 37

a bit, not fully.. and I had to go earlier for a few minuets - I apologize.

Answer 38

it's ok so can you type your equation you are trying to find the domain and range for please =]

Answer 39

From earlier?

Answer 40

one that was done by someone on here or one not done yet =]

Answer 41

well there was the four up there, two of them met the question's answer. What about this one?
y=-(3)^x-2

Answer 42

alright so you have
\[y=-(3)^x-2\]

first the domain is there any x value that will make y be undefined

Answer 43

think about what happens with \[3^x\] as x is negative and positive

Answer 44

Hint:\[x^{-1}=\frac{1}{x}\]

Answer 45

so \[3^{-1}=\frac{1}{3}\]

Answer 46

correct?.... meaning that x can be ______