I think for the first one there are no restrictions, because so the implied domain should be .
Incorrect. If then you have just committed the cardinal sin of dividing by zero when you do this. The domain is all real numbers EXCEPT . The graph of has a 'hole' at .
Originally Posted by Raoh
hi
for the first one :
[snip]
Incorrect. See above.
Originally Posted by Raoh
[snip]
for the second :
Your bracket notation needs some minor corrections:
Last edited by mr fantastic; Jun 14th 2009 at 05:33 AM.
Hmm. I'm definitely doing something wrong here. I'm getting Mr F says: Correct. Now cancel the common factor of (x + 1) to get x - 1. You can do this provided (otherwise you're dividing by zero).
Mr F says: Wrong.
Which must be wrong. But the answer is definitely R/{-1}
Originally Posted by Stroodle
Originally Posted by Raoh
hi
for the first one :
for the second :
Thanks Raoh.
How'd you get the second answer?
Originally Posted by mr fantastic
[snip]
For the second one, you require all values of such that .
Incorrect. If then you have just committed the cardinal sin of dividing by zero when you do this. The domain is all real numbers EXCEPT . The graph of has a 'hole' at .
My judgement was that we don't actually divide by since the final expression is , but I understand what you mean.
So, for example, will not be defined in even though it's equal to ??
Oh, nevermind, I think I got it.
I think I was saying something like "we don't need any restrictions for the functionbecause it's equal to ", which is obviously false
Sorry about that..