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Math Help - indermideate value therem question

  1. #1
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    intermidiate value therem question

    prove that for c<0
    there is only one solution to
    xe^{\frac{1}{x}}=c

    ??
    i tookk the left side as f(x)
    for x=1 we have f(1)>0

    the limit as x->-infinity is -infinity

    what to do?
    ?
    Last edited by transgalactic; September 11th 2011 at 11:32 AM.
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  2. #2
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    Re: intermidiate value therem question

    Quote Originally Posted by transgalactic View Post
    prove that for c<0
    there is only one solution to
    xe^{\frac{1}{x}}=c
    If f(x)=xe^{x^{-1}} can you show that if x<0
    then f '(x)>0.
    If so then f is increasing and continuous on (-\infty,0).
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  3. #3
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    Re: intermidiate value therem question

    ok even if its encreasing
    i need t show function vules bellow and above c
    to use the intermidiete theorem

    could you guide me regardnig how do i get thos values

    because saying that its increasing is not sufficient
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    Re: intermidiate value therem question

    Quote Originally Posted by transgalactic View Post
    ok even if its encreasing
    i need t show function vules bellow and above c
    to use the intermidiete theorem
    could you guide me regardnig how do i get thos values
    \lim _{x \to  - \infty } f(x) =  - \infty
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    Re: intermidiate value therem question

    ok so from the limit definition
    for x<-M f(x)<-N

    i can choose N to be C

    but its not continues in zero
    i cant take x=1 as the other point of intermidiete value theorem
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    Re: intermidiate value therem question

    Quote Originally Posted by transgalactic View Post
    ok so from the limit definition
    for x<-M f(x)<-N
    but its not continues in zero
    Zero has absolutely nothing to do with this question.
    Look at the OP.
    Quote Originally Posted by transgalactic View Post
    prove that for c<0
    there is only one solution to

    xe^{\frac{1}{x}}=c
    ?
    \lim _{x \to 0^ -  } f(x) = 0

    c<0
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    Re: intermidiate value therem question

    yes so by limit definition
    |x|<delta
    -epsilon<f(x)<epsilon

    what epsilon to chose to get a value smaller then c
    ?
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    Re: intermidiate value therem question

    Quote Originally Posted by transgalactic View Post
    yes so by limit definition
    |x|<delta-epsilon<f(x)<epsilon what epsilon to chose to get a value smaller then c
    ?
    At this point, do you even know what you are talking about?
    \delta/\varepsilon has absolutely nothing to do with it, as far as I can tell.

    If c<0 then 2c<c<\frac{c}{2} then because f is increasing f(2c)<f(c)<f\left(\frac{c}{2}\right).
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    Re: intermidiate value therem question

    Quote Originally Posted by Plato View Post
    At this point, do you even know what you are talking about?
    \delta/\varepsilon has absolutely nothing to do with it, as far as I can tell.

    If c<0 then 2c<c<\frac{c}{2} then because f is increasing f(2c)<f(c)<f\left(\frac{c}{2}\right).
    ok i see it now .
    if its decreasing we take smaller values and they get a bigger function value

    so from the decreasing property we can directly write
    f(2c)<f(c)<f\left(\frac{c}{2}\right)

    from here its straight forward indermidiate value theorem

    so why we need the limit
    when x goes to zero the limit is sero

    ?
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    Re: intermidiate value therem question

    Quote Originally Posted by transgalactic View Post
    from here its straight forward indermidiate value theorem
    so why we need the limit
    when x goes to zero the limit is zero?
    We want to know that f to map (-\infty,0) one-to-one and onto (-\infty,0).
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    Re: intermidiate value therem question

    ok so we need to havea value at some point then say that its decreasing asnd we get one to one

    but here we need to prove the solution is c too
    how
    ?
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    Re: intermidiate value therem question

    Quote Originally Posted by transgalactic View Post
    ok so we need to havea value at some point then say that its decreasing asnd we get one to one

    but here we need to prove the solution is c too
    how
    ?
    I simply give up. Good luck.
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  13. #13
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    Re: intermidiate value therem question

    i need to show that there is x1 f(x1)<c
    f(x2)>c

    from the limit when x goes to sero we get zero soby limit definition -e<f(x)<e
    from the limit when x goes minus infinity f(x)<-N
    what e to chhose?
    what N to choose?
    Last edited by transgalactic; September 11th 2011 at 10:42 PM.
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  14. #14
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    Re: intermidiate value therem question

    Quote Originally Posted by transgalactic View Post
    prove that for c<0
    there is only one solution to
    xe^{\frac{1}{x}}=c
    As stated, the problem is to prove that the equation xe^{1/x}=c has 0 or 1 solution for each c < 0. That's what "only one" means: there cannot be more than one. If the problem said "there exists one solution to this equation," that would mean "at least one, but possibly more." To stress that there exists exactly one solution, people sometimes say, "there exists one and only one solution."

    Now, this interpretation is arguable and possibly not every everybody would agree that "only one" means <= 1 and "one" means ">= 1." I would therefore avoid statements containing "only one." However, since it is given, the default interpretation of this problem for me is to prove that there cannot be two solutions.

    So, what exactly do you need to prove here? You may need to contact your instructor for clarification.
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