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bailey07
04-28-2008, 10:34 PM
f(x) = x/x-3;g(x)=-1/x

i got {xIxcannot=3,xcannot=0}

3/3-3=0
-1/0=0

am i right?

tkhunny
04-28-2008, 10:36 PM
Okay, but work on your notation.

x/x-3 = (x/x)-3 = 1-3 = -2

x/(x-3) is what you want.

bailey07
04-28-2008, 10:54 PM
wait how are u getting that..its suppose to be in set builder notation.

tkhunny
04-28-2008, 10:59 PM
It's called "Order of Operations". There are rules for this sort of thing.

bailey07
04-28-2008, 11:04 PM
alrite so the answer is {xIx-3} ?

tkhunny
04-29-2008, 08:56 AM
3/3-3 = (3/3)-3 = 1-3 = -2 -- This is not particularly meaningful for a Domain Question.

3/(3-3) = 3/0 -- Whoops. I guess x = 3 is NOT in the Domain. So, the Domain is x\neq3, or Real Numbers, excepting x = 3.