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

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

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

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

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

    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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    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by lllll View Post
    I'm having a great of difficulty showing $\displaystyle |\ln(x)| \leq \frac{1}{\sqrt{x}}$ on the interval $\displaystyle (0,1]$

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

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

    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 $\displaystyle f(x)=\frac{1}{\sqrt{x}}+\ln(x)$ on the interval $\displaystyle (x,1), x>0$.
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  3. #3
    Moo
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    Hello,

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

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

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