Question

f(x) = x/(x-3)
please state how to find the domain of this problem thanks so much

Answer 1

Find the non-permissible values of the denominator, ie the denominator cannot be zero, this can be represented by the following (x-3)=0 *not equal to zero*

Answer 2

thus, the NPV is +3

Answer 3

So the domain is: {X e R | x =/ 0}

Answer 4

so would that be all real numbers greater than 0??

Answer 5

sorry... {X e R | x =/ 3}

Answer 6

what does X and R stand for in the formula

Answer 7

no just above zero, but also below

Answer 8

so the absolute value of x greater and less than 0

Answer 9

"x" reffering to the variable, the 'e' means "Element of" and the 'R' means "Real Numbers"

Answer 10

When dealing with absolute values, all results are positive, could you post your question pls?

Answer 11

so would it be lxl>0

Answer 12

true

Answer 13

greater than or equal to**

Answer 14

okay yeah thanks so lxl≥0

Answer 15

\[lxl \ge0\]

Answer 16

then with it, add that x is not equal to 3 and that's your domain. Now notations vary and I may be formatting differently than what your teacher expects

Answer 17

but that's the gist of it: {|x| >_ 0} U {x =/ 3} That's set builder notation, try interpreting it and convert it into whatever format you use.

Answer 18

so then wouldnt it be lxl>3, lxl<3
D: (-∞, 3) U (3,∞)

Answer 19

\[D: (-\infty,3) \cup(3,\infty)\]

Answer 20

not quite, see, at x=3, there is a vertical seperation between two 'broken' parts of the graph (called a vertical asymptote) this means that x exists at all values (including 2.9999999...) but not at 3

Answer 21

actually yes, that last re is correct

Answer 22

isnt that what ")" and not "]" is for

Answer 23

I was replying to a more recent one, you got it

Answer 24

thanks

Answer 25

glad to help ^_^

Answer 26

have a good one m8

Answer 27

you too :)