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?

i got {xIxcannot=3,xcannot=0}

3/3-3=0

-1/0=0

am i right?

View Full Version : finding the domain: f(x) = x/x-3;g(x)=-1/x

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?

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.

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.

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.

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