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Epo9
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 Posted: Sun, 16 Oct 2005 02:45:11 UTC Post subject: Continuous Function |
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How can f(x) = 1/sqrt(x-2) be continuous on the interval (-oo, +oo)?? Isn't the function continuous only when x>2?
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stapel
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| Posted: Sun, 16 Oct 2005 02:51:06 UTC Post subject: |
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Last edited by stapel on Tue, 18 Oct 2005 15:14:47 UTC; edited 2 times in total |
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Matt
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| Posted: Sun, 16 Oct 2005 03:16:54 UTC Post subject: Re: Continuous Function |
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| Epo9 wrote: | | How can f(x) = 1/sqrt(x-2) be continuous on the interval (-oo, +oo)?? |
It isn't.
| Epo9 wrote: | | Isn't the function continuous only when x>2? |
Absolutely correct. |
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prowl92
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| Posted: Sun, 16 Oct 2005 04:14:38 UTC Post subject: |
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Well, the range can be continuous if you take the square root to be plus or minus.
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Matt
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| Posted: Sun, 16 Oct 2005 04:15:36 UTC Post subject: |
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| prowl92 wrote: | | Well, the [function] can be continuous if you take the square root to be plus or minus. |
What about x=2? |
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prowl92
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| Posted: Sun, 16 Oct 2005 04:33:44 UTC Post subject: |
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Well x=2 is going to be the asymptote, but that has little to do with the range. The function can never equal 0 though, so I was wrong anyway.
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stapel
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| Posted: Sun, 16 Oct 2005 04:53:54 UTC Post subject: |
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Last edited by stapel on Tue, 18 Oct 2005 15:14:53 UTC; edited 2 times in total |
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prowl92
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| Posted: Sun, 16 Oct 2005 05:06:09 UTC Post subject: |
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Sometimes we assume it's plus or minus, especially with function inverses. I've gotten so many different responses from many different teachers, I don't know what to believe. I've heard that in engineering and many practical applictions they just take the positive root. Either way, my response was pointless.
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math4ever
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| Posted: Sun, 16 Oct 2005 05:33:25 UTC Post subject: |
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[quote="prowl92"]Sometimes we assume it's plus or minus, [quote]
There is no minus sign, the answer is the postive root only ... i.e.  (even though when you square -3 you get 9) ... that is how the square root is defined.
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prowl92
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| Posted: Sun, 16 Oct 2005 06:10:23 UTC Post subject: |
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I am thinking along these lines.
The range is  and the domain is all real number. By definition, the inverse should have the range of all real numbers and domain should be , which only happens when the inverse is  .
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math4ever
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| Posted: Sun, 16 Oct 2005 06:18:38 UTC Post subject: |
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| prowl92 wrote: | I am thinking along these lines.
The range is and the domain is all real number. By definition, the inverse should have the range of all real numbers and domain should be , which only happens when the inverse is . |
I see what you are saying but how is that relevant to the poster's problem ? ;)
PS: Are you sure that  ?
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prowl92
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| Posted: Sun, 16 Oct 2005 06:29:31 UTC Post subject: |
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It isn't directly, but it's related to the positive and negative square root debate we were having.
As for your P.S., as far as I know-- it's true. It's clear you have counterexamples though 
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math4ever
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| Posted: Sun, 16 Oct 2005 06:48:41 UTC Post subject: |
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| prowl92 wrote: |
As for your P.S., as far as I know-- it's true. It's clear you have counterexamples though |
well, the way you defined your function f (the domain being R) then f does not have an inverse because it is not one to one.... example f(-1)=f(1)=1 so if you apply f inverse to that you get and  which shows that  is not a function...
However, if you restrict f to  then f has an inverse...
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