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Math Help - need to show abs(ln(x)) <= 1/sqrt(x) on (0,1]

  1. #1
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    need to show abs(ln(x)) <= 1/sqrt(x) on (0,1]

    I'm having a great of difficulty showing |\ln(x)| \leq \frac{1}{\sqrt{x}} on the interval (0,1]

    I was thinking that |\ln(x)| can be written as -\ln(x) since the given values will simply be reflected, which would give me

    0 \leq \underbrace{\frac{1}{\sqrt{x}}}_{\mbox{always}\ >0}+\ln(x)

    but I'm stuck with the natural logarithm.


    another think that I was thinking about was to show that the function never cross on the given interval, but I don't know how
    that would be so since the both tend to infinity as they approach 0.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by lllll View Post
    I'm having a great of difficulty showing |\ln(x)| \leq \frac{1}{\sqrt{x}} on the interval (0,1]

    I was thinking that |\ln(x)| can be written as -\ln(x) since the given values will simply be reflected, which would give me

    0 \leq \underbrace{\frac{1}{\sqrt{x}}}_{\mbox{always}\ >0}+\ln(x)

    but I'm stuck with the natural logarithm.


    another think that I was thinking about was to show that the function never cross on the given interval, but I don't know how
    that would be so since the both tend to infinity as they approach 0.
    hmm, you can use MVT on f(x)=\frac{1}{\sqrt{x}}+\ln(x) on the interval (x,1), x>0.
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  3. #3
    Moo
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    Hello,

    Consider f(x)=\frac{1}{\sqrt{x}}+\ln(x)

    Compute its derivative, and set it =0.
    You'll get the equation x-\frac{\sqrt{x}}{2}=0, after multiplying the equation by x^2 (which is possible since x \neq 0)
    and this will give only one solution, \alpha

    note that f(\alpha)>0
    now is it a minimum ?
    \lim_{x \to 1} f(x)=1> f(\alpha)
    f(1/2)=\sqrt{2}-\ln(2) \approx 0.721 > f(\alpha)
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