Point on curve thats closest to the origin

**nubshat** · November 6th 2012, 02:02 PM

Find the point on the curve y=sqrt(1/2ln(1/x)) that is closest to the origin

I have attempted to do this question a couple of times, but I can't get the right answer. Please help.
Thanks in advance.

**skeeter** · November 6th 2012, 03:09 PM

$D^2 = (x - 0)^2 + (y - 0)^2$

let $D^2 = u$ ... minimizing $u$ will minimize $D^2$

$u = x^2 + \frac{1}{2} \ln\left(\frac{1}{x}\right)$

$u = x^2 - \frac{1}{2} \ln{x}$

$\frac{du}{dx} = 2x - \frac{1}{2x} = 0$

$4x^2 - 1 = 0$

$x = \frac{1}{2}$

$\frac{d^2u}{dx^2} = 2 + \frac{1}{2x^2} > 0 \implies$ critical value is a minimum

I'll leave you to find the y-coordinate of the point.

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