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Math Help - Fonction Help

  1. #1
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    Fonction Help

    Hey friends

    i got a Hard problem about Fonction:

    We consider the function set to]0; + infinity[ g (x) = -lnx + 1 / x

    A. Studying the variations of g
    B. Show that the equation g (x) = 0, admits one and only one solution note (alfa) between 3/2 and 2

    C. Consider the sign of g (x) sure everyone intervals]0;alfa[ and ]alfa; + infinity [

    D. F is the function of R to R defined by f (x) = e ^ -x . lnx
    what the sign of f '

    sorry for my english (i am french educated ) , Just i need your Help to solve this ,

    Thanks in advance
    Last edited by iceman1; January 13th 2008 at 04:25 AM.
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  2. #2
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    Quote Originally Posted by iceman1 View Post
    Hey friends

    i got a Hard problem about Fonction:

    We consider the function set to]0; + infinity[ g (x) = ln-x + 1 / x

    A. Studying the variations of g
    B. Show that the equation g (x) = 0, admits one and only one solution note (alfa) between 3/2 and 2

    C. Consider the sign of g (x) sure everyone intervals]0;alfa[ and ]alfa; + infinity [

    D. F is the function of R to R defined by f (x) = e ^ x-lnx
    what the sign of f '

    sorry for my english (i am french educated ) , Just i need your Help to solve this ,

    Thanks in advance
    g (x) = ln-x + 1 / x can be read in a number of ways - none that make sense within the context of the question. Please give the rule again - carefully formatted.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    g (x) = ln-x + 1 / x can be read in a number of ways - none that make sense within the context of the question. Please give the rule again - carefully formatted.
    g(x) = -Lnx + 1/x Like this==> ( -Lnx) + (1/x) good now?
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  4. #4
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    Quote Originally Posted by iceman1 View Post
    g(x) = -Lnx + 1/x Like this==> ( -Lnx) + (1/x) good now?
    Yes. - ln(x) is quite different to ln(-x).

    Quote Originally Posted by iceman1 View Post
    [snip]B. Show that the equation g (x) = 0, admits one and only one solution note (alfa) between 3/2 and 2
    [snip]
    g(3/2) > 0 and g(2) < 0. Therefore there's at least one solution between 3/2 and 2. Assume there's more than one solution, x = a and x = b, say. Apply Rolle's Theorem. Get a contradiction. Therefore only one solution.
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