chemmaj88
At what POINTS is this function continuous..y=cosx/x...someone please explain to me how to solve these types of problems? is it whatever numbers make this an undefined problem?
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ChrisV
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well x=/=0
ChrisV
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i would say limit x>0 \[(-\infty,0)\]\[(0,\infty)\]
ChrisV
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but im a nub so dont trust me much
chemmaj88
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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?
ChrisV
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well the denominator can never be 0
ChrisV
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so x cannot be 0
ChrisV
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not much to guess at really
ChrisV
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just means if x were zero
ChrisV
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the value would be undefined
ChrisV
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hence a hole in your graph
chemmaj88
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got it! ok i got one more...
chemmaj88
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y=sqrt(2x+3)
ChrisV
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well since it is a square root is can never be negative
ChrisV
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so it would be \[(0,\infty)\]
ChrisV
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the graph of a square root function looks like
ChrisV
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|dw:1328895361125:dw|
chemmaj88
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the books answer is, all X greater than or equal to -3/2
ChrisV
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yea thats right
ChrisV
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because
ChrisV
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y=sqrt(2x+3)
ChrisV
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to make the square root 0 x would have to be -3/2
ChrisV
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hence its still not a negative
ChrisV
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been up all night sorry
ChrisV
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getting tired
ChrisV
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but since -3/2 make it the sqrt of 0 thats the limit
ChrisV
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all other numbers have toi be greater then -3/2
chemmaj88
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so if u plug in -3/2 "sqrt(2(-3/2)+3" this will equal 0?
ChrisV
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yes because the 2's cancel
ChrisV
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2x-3/2 = -3
ChrisV
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-3+3 = 0
chemmaj88
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got it! so on these types of problems are u really just looking for w/e numbers will make it undefined?
ChrisV
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basically
ChrisV
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and if you cannot make it undefined
ChrisV
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its continuous from -inf to +inf
ChrisV
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ok well you have a good one, gotta get some rest now
chemmaj88
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thanks!