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Math Help - Surjuctivity of a function

  1. #1
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    Surjuctivity of a function

    Why is the following function surjective(onto) ?

    f: (1,\infty)\longrightarrow (\ln 2,\infty)
    f(x)=\int_{x^2}^{x^4}\! \frac{1}{\ln t}\, dt
    Last edited by bigli; June 2nd 2010 at 02:18 AM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by bigli View Post
    Why is the following function surjective(onto) ?

    f: (1,\infty)\longrightarrow (\ln 2,\infty)
    f(x)=\int_{x^2}^{x^4}\! \frac{1}{\ln t}\, dt
    Try using the IVT and creating some nice inequalities
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  3. #3
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    is this IVT? : intermediate value theorem. anyway, How do I use it?
    Last edited by bigli; June 1st 2010 at 07:39 PM.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by bigli View Post
    What is IVT?
    Intermediate value theorem.
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  5. #5
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    How? please hint me.
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  6. #6
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    The interemdiate value theorem says that if f(x) is continous on the interval [a, b] then it takes on all values between f(a) and f(b) on that interval.

    Here, f(x) is \int_{x^2}^{x^4}\frac{1}{ln t} dt which is continous on [tex][1, \infty) because ln t is continuous for all positive t and the integral of a continuous function is continuous.

    You will still need to show that f(x) is unbounded. That is, that if y is any positive number there exist x such that f(x)> y. That, I think, is why Drexel28 mentions "some nice inequalities".
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  7. #7
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    In fact, how do I show \lim_{x\rightarrow 1^+ }\int_{x^2}^{x^4}\!\frac{1}{\ln t}\, dt=\ln 2 ?
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