Question

find domain of

Answer 1

\[y=\sqrt{\frac{3x-2}{1-2x}}\]

Answer 2

is it \[(x:3x-2\ge0 \cup 1-2x>0) intersection (x:3x-2\le0 \cup 1-2x<0 ) \]

Answer 3

oh i got it i think its \[(x:3x-2\ge0 \cap1-2x>0) \cup (x:3x-2\le0 \cap1-2x<0)\]

Answer 4

@hfw ??

Answer 5

The domain must have x-values that makes the equation defined. The denominator cannot be zero as some constant / 0 is undefined ( 1/ 0 = undefined). Also, this is a sqrt equation so the numerator cannot be 0 as the sqrt of 0 is undefined as well

The domain in interval notation is:

(-infinity, 1/2) u (1/2, 2/3) u (2/3, infinity)

Sorry for delaying. Computer problems

Answer 6

yeah i understand can you check my working right?

Answer 7

Your work seems correct but some would say it is incomplete since you did not completely define what x is greater than or less than

Answer 8

yeah im just showing how to start .. will evaluate later

Answer 9

Check what x is by isolating it

Answer 10

wait but is it the intersection or union between the 2 restrictions?

Answer 11

For example,
(-2,4) U ( 5, 6) means: \[-2 < x <4\]
\[5<x <6\]

Answer 12

() means the values show the variable cannot be those values

Answer 13

so its the union of the 2 cases: case1 top, bottom are both positive so top>0 intersection bottom>0 and case2: top<0 intersection bottom<0

Answer 14

yes i understand the intervals can you check my intersectin and union signs are correct

Answer 15

by the way when you wanna find intersection between 2 intervals can you find it by drawing number line and labelling ?

Answer 16

Okay, it is right although the (x:3x−2≥0∩1−2x>0) should be (x:3x−2 >0∩1−2x>0) as 3x -2 cannot be 0. Your original expression makes it able to be 0. And there is a easier and clean answer.

Answer 17

why cant top be 0?

Answer 18

sqrt(0) is defined isnt it

Answer 19

Oh wait, you are right, sqrt of zero is valid, so your original equation is right. Sorry, I am a bit tired

Answer 20

thanks its okay

Answer 21

and another quesotn

Answer 22

if its f(x)=sqrt(a)+sqrt(b) where a and b are functions, is domain of f
intersection of a>=0 b>=0 or is it union a>=0 b>=0

Answer 23

intersection right

Answer 24

I believe it is union. a and b can be valid for your function except for negative numbers. I am not entirely sure. Here is a link that may give a better explaination.

http://answers.yahoo.com/question/index?qid=20081102143652AAZkOc2

Answer 25

are you sure its union

Answer 26

cause if you use a x that only fulfils a>=0, b might be negative still?

Answer 27

I think so now you mention it. Intersection states that both functions have some term in common. In this case, a and b can be any number that is the same and greater or equal to zero.