Question

can someone tell me how to find domain and range ??

Answer 1

Any ideas?

Answer 2

You need ideas to get answers @cutestpearl

Answer 3

oh commmon leave me alone .... if u dont know anything and u can't say anything right then just shush up.....

Answer 4

First you must ask yourself "Do you want to learn and understand?"

Answer 5

i do................... duh!! thats why i asked this question

Answer 6

@cutestpearl Well to find domain and range there are limits to them.

Answer 7

what kind of limits ??? can u tell me in detail

Answer 8

Say x = time. And you know domain = x. X cannot be a negative number because time cannot be negative.

Answer 9

So domain is limited to positive numbers only.

Answer 10

Then re-ask your question more thoughtfully @cutestpearl

Answer 11

oh ok

Answer 12

@cutestpearl Try to come up with YOUR OWN example

Answer 13

And lets also say that Bob is tracking how many snacks he eats in a week. x = time and y = snacks. X is also limited to 168 hours since there are 168 hours in a week. y as you know is equal to range. y = snacks. y is also limited to positive numbers because you can't eat negative snacks.

Answer 14

\[g(x)=3\div(x-2)(x+3) whats the domain here ?\]

Answer 15

To understand domain and range, you must understand the basics first.
The domain of a function is the set all its x-values, and the range of a function the set is all the y-values. For example:

State the domain and range of the following relation.
{(2, –3), (4, 6), (3, –1), (6, 6), (2, 3)}

To give the domain and the range, I just list the values without duplication:

domain: {2, 3, 4, 6}

range: {–3, –1, 3, 6}

Answer 16

Exactly ^^

Answer 17

oh ok getting it now

Answer 18

But, in addition, you also have to know whether or not the sets you are dealing with is a function. So you have to know the definition of a function as well.

Answer 19

3÷(x−2)(x+3) so in this question domain is 2 and -3 ???

Answer 20

No

Answer 21

I wasn't done explaining

Answer 22

too much distractions first the @skullpatrol and now @yellowlegoguy99

Answer 23

Lol

Answer 24

What the skull?

Answer 25

Anyway, I think the domain for g(x)=3÷(x−2)(x+3) is All Real Numbers.

Answer 26

how and why ?

Answer 27

@hero sorry for interrupting explain me again :/

Answer 28

x - 2 and x + 3 don't really explain much since you don't know what x is.

Answer 29

You should post your questions separately. The understanding must come in steps. First you need to understand what domain and range is. Then afterwards, you will need to understand how to find domain and range of coordinate pairs. From there, you can then understand how to find the domain and range of algebraic functions.

Answer 30

domain is all the x-values and range is all the y-values

Answer 31

plus domain is always +ve

Answer 32

If you try actually reading some of the resources posted on the internet, you will be able to understand for yourself

http://www.purplemath.com/modules/fcns2.htm

Answer 33

green said: "Anyway, I think the domain for g(x)=3÷(x−2)(x+3) is All Real Numbers."

notice that division sign in there .... when we are dividing by a variable we have to exclude certain real numbers to avoid a divide by 0

Answer 34

@Hero is its answer x neq 2 and x neq -3 ???

Answer 35

in a relation x values represent domain and y represent range.

Answer 36

You're certainly onto something. Allow to me explain.
The domain is the set of numbers that the function can take in without returning an indeterminate form. So, what is an indeterminate form? It is something that cannot be evaluated, like \(\frac {0}{0}, \frac{\infty}{0}, \infty\), so on; we cannot input anything that puts the function in this form. The domain is not always easy to find.
The range, on the other hand, is the set of numbers the function will give us after we input the numbers in the domain. For the function \(f(x) = x + 3\), the range is all real numbers, as is the domain. In order to make any number \(y\) from that function, we just have to input \(f(y-3) = y -3 + 3 = y\). I hope that's not too confusing.
If so, we'll continue on with your problem: