Answer
$x=\frac{1}{\sqrt2}$
Work Step by Step
We use the equation for the period to find:
$T = \sqrt{\frac{I}{mgx}}$
$T = \sqrt{\frac{m(x^2+\frac{1}{2}R^2)}{mgx}}$
$T = \sqrt{\frac{(x^2+\frac{1}{2}R^2)}{gx}}$
We take the derivative of the inside of the radical and set it equal to zero to find:
$T = \sqrt{\frac{(x+\frac{R^2}{2x})}{g}}$
$0=\frac{1}{g}(1+\frac{-2R^2}{4x^2})$
$0=(1+\frac{-2R^2}{4x^2})$
$-1=(\frac{-2R^2}{4x^2})$
$x =\sqrt{\frac{R^2}{2}}$
Using the first derivative test, we obtain that the minimum will occur at:
$x=\frac{1}{\sqrt2}$
