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Math Help - mean value theory infinity proble

  1. #1
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    mean value theory infinity proble

    pprove that for alll 0<c<\frac{{1}}{2} there is x>0 so \frac{1}{x}-\frac{1}{e^{x}-1}=c

    i have tried f(x)=\frac{1}{x}-\frac{1}{e^{x}-1}
    limit of f(x) x->infinity = zero
    limit of f(x) x->0 = pi/2

    so in similar problems i have defined

    g(x)= 0.5 when x=0
    g(x)= f(x) when 0<x<infinity
    g(x)=0 when x=infinity


    and then i have to point one is 0 the other is 1/2
    and i do mean value theorem

    but its not legat to definne such function
    infinity is not a point,what to do?
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  2. #2
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    Re: mean value theory infinity proble

    Quote Originally Posted by transgalactic View Post
    pprove that for alll 0<c<\frac{{1}}{2} there is x>0 so \frac{1}{x}-\frac{1}{e^{x}-1}=c
    f(x)=\frac{1}{x}-\frac{1}{e^{x}-1}
    Please review the statement of this question.
    As written it is not true.
    f is decreasing and on (0,0.5] the minimum is c.0.4585
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  3. #3
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    Re: mean value theory infinity proble

    Quote Originally Posted by transgalactic View Post
    limit of f(x) x->infinity = zero
    limit of f(x) x->0 = pi/2
    The first limit is correct; the second one should be 1/2. Thus, for each 0 < c < 1/2, you can find x1 > 0 such that c <= f(x1) < 1/2 and x2 > 0 such that f(x2) <= c. The function is continuous for x > 0, so by the intermediate value theorem, there exists an x1 <= x <= x2 such that f(x) = c.
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  4. #4
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    Re: mean value theory infinity proble

    yes i had a typo.
    limit of f(x) x->infinity = zero
    limit of f(x) x->0 = 1/2


    my problem is how to explain that there is such x1 c <= f(x1) < 1/2
    from the limit?

    same thing goes for
    f(x2) <= c

    i dont know how to get to these to innequalitties from limit definition

    from the first i got
    for every x>M -epsilon<f(x)<epsilon

    from the second limit definition i get
    for every |x-0|<delta 1/2-epsilon<f(x)<epsilon+1/2


    i dont know how from these limit definition i get these innequlities
    ?
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  5. #5
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    Re: mean value theory infinity proble

    For the limit x -> infinity, take epsilon = c. For the limit x -> 0, take epsilon = 1/2 - c.
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  6. #6
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    Re: mean value theory infinity proble

    1/2-epsilon<f(x)<epsilon+1/2
    take epsilon = 1/2 - c
    so i get

    1+c<f(x)<1-c

    -epsilon<f(x)<epsilon
    epsilon=c
    so i get

    -c<f(x)<c
    so from the second i get f(x2)<c


    how i get c <= f(x1) < 1/2
    from
    1+c<f(x)<1-c
    why its smaller then 1/2
    ?
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  7. #7
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    Re: mean value theory infinity proble

    Quote Originally Posted by transgalactic View Post
    1/2-epsilon<f(x)<epsilon+1/2
    take epsilon = 1/2 - c
    so i get

    1+c<f(x)<1-c
    Check the last line again. Also, it is helpful to think not about algebraic expressions but about distances on the numerical line. Imagine 1/2, then c located below it and some point x whose distance to 1/2 is less than the distance between c and 1/2.
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  8. #8
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    Re: mean value theory infinity proble

    sorry made a mistake
    c<f(x)<1-c

    and it is given that 0<c<1/2

    but stil it doesnt prove f(x)<1/2

    f(x)<t

    t is allways bigger then 1/2

    if c was 1/2 we would get f(x)<1/2

    but its not
    c is bellow 1/2

    ?
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  9. #9
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    Re: mean value theory infinity proble

    Quote Originally Posted by transgalactic View Post
    sorry made a mistake
    c<f(x)<1-c

    and it is given that 0<c<1/2

    but stil it doesnt prove f(x)<1/2
    I guess you are concerned about f(x) < 1/2 because I said:
    Quote Originally Posted by emakarov
    Thus, for each 0 < c < 1/2, you can find x1 > 0 such that c <= f(x1) < 1/2
    According to the graph, f(x) is monotonically decreasing, so f(x) < 1/2. However, it is not important for the current problem. The important thing is that one can find x1 such that f(x1) > c and x2 such that f(x2) < c.
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  10. #10
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    Re: mean value theory infinity proble

    oohh i get it thanks
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  11. #11
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    Re: mean value theory infinity proble

    never mind the question
    i had a question and i solved it
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