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Math Help - functions

  1. #1
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    functions

    let me ask you this one..

    we got f(x), g(x), in both there is differentiation.

    for all x, we know:  f'(x)*g(x)  g'(x)*f(x)

    we also know there are: f(a)=f(b)=0

    we need to prove that there is a point 'c', in zone [a,b], which gives us g(c)=0.

    please help !
    Last edited by dannee; June 30th 2010 at 02:14 PM.
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  2. #2
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    Hi

    Suppose there exists no c in zone [a,b], which gives us g(c)=0

    Consider h(x) = \frac{f(x)}{g(x)}

    h is therefore defined and differentiable for all x in [a,b] ...
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  3. #3
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    firstly thanks for your response.

    let's see if i got it right.

    based on Rolle, we say there is f'(d)=0 on [a,b]

    then we say, h(x)=f(x)/g(x)  ,  h(x)  f'(x)/g'(x)

    but this could not be, because we have f(a)=0 , f'(d)=0 both in zone [a,b]

    am i right ??
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  4. #4
    Senior Member jakncoke's Avatar
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    Quote Originally Posted by dannee View Post
    firstly thanks for your response.

    let's see if i got it right.

    based on Rolle, we say there is f'(d)=0 on [a,b]

    then we say, h(x)=f(x)/g(x)  ,  h(x)  f'(x)/g'(x)

    but this could not be, because we have f(a)=0 , f'(d)=0 both in zone [a,b]

    am i right ??
    sorry i don't follow ur argument. But here is another hint adding to what running-gag said, if  h(x) = f(x)/g(x) then  h'(x) = \frac{g(x)f'(x) - g'(x)f(x)}{g(x)^2} what does the numerator remind you of?
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  5. #5
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    it means h'(x) is a monotonic function, but i don't know how to go from there
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  6. #6
    Senior Member jakncoke's Avatar
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    Quote Originally Posted by dannee View Post
    it means h'(x) is a monotonic function, but i don't know how to go from there
    h(a)=0=h(b) use a special theorem to derive a contradiction.
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  7. #7
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    got it, thank you very much
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