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i would say limit x>0 \[(-\infty,0)\]\[(0,\infty)\]
but im a nub so dont trust me much
the book says all x except x=0 so yes ur on it, but like, do u just see this problem n knw, or do u go thru a guess n check process, what process do u go thru to get an answer for these types of problems?
well the denominator can never be 0
so x cannot be 0
not much to guess at really
just means if x were zero
the value would be undefined
hence a hole in your graph
got it! ok i got one more...
well since it is a square root is can never be negative
so it would be \[(0,\infty)\]
the graph of a square root function looks like
the books answer is, all X greater than or equal to -3/2
yea thats right
to make the square root 0 x would have to be -3/2
hence its still not a negative
been up all night sorry
but since -3/2 make it the sqrt of 0 thats the limit
all other numbers have toi be greater then -3/2
so if u plug in -3/2 "sqrt(2(-3/2)+3" this will equal 0?
yes because the 2's cancel
2x-3/2 = -3
-3+3 = 0
got it! so on these types of problems are u really just looking for w/e numbers will make it undefined?
and if you cannot make it undefined
its continuous from -inf to +inf
ok well you have a good one, gotta get some rest now