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

1. 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.

2. Originally Posted by lllll
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 $(x,1), x>0$.

3. 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 ?
$\lim_{x \to 1} f(x)=1> f(\alpha)$
$f(1/2)=\sqrt{2}-\ln(2) \approx 0.721 > f(\alpha)$