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You're looking at the restrictions, so where the denominator is = 0, so solve for \[x^2 - 1 = 0\]
i got that part. i don't exactly know how to write it in interval notation tho
i don't think i did it right
You could write it as such \[( - \infty, - 1) \cup (1, \infty)\] because open bracket implies it's not included, a square bracket means, it would be included in the domain.
of course I forgot 0 xD
But you get the point right?
I used to hate those write in set notation or interval notation parts to the question because I got confused with ( and ]. This was before OpenStudy popped up.
Yeah I can't remember last time I even used interval notation to be honest
College Algebra had truckloads... but the thing is it doesn't appear again afterwards.
Yea, I used it in calculus 1, then never saw it again haha...
|dw:1434430709134:dw| so this is how it looks like in a number line (random example) \[(- \infty, -1/2] \cup [3, \infty)\] I think this will make it clear on what exactly is going on.
thanks this is going to help a lot for the rest of my hw
No problem :)