Finding max mins point of inflection.

$\displaystyle f(x)= 2x^2 + \frac {8}{x^2} $

$\displaystyle f'(x)= 4x - \frac{16}{x^3} $

$\displaystyle f''(x) = 4+\frac{48}{x^4} $

So taking f''(x)

$\displaystyle \frac{4}{1} + \frac{48}{x^4}=\frac{4x^4}{x^4}+\frac{48}{x^4}$

$\displaystyle \frac{4x^4+48}{x^4} \cdot x^4 = 0 \cdot x^4$

$\displaystyle 4x^4+48 = 0$

$\displaystyle 48=-4x^4 => -12 = x^4 => -1.86 = x$

so i have a point of inflection at -1.86?

then evaluate f(-1.86).

$\displaystyle 2\cdot -1.86^2 + \frac{8}{-1.86^2} = -9.23$

So this is a local maximum?

somthing seems wrong , can anyone point out what it is?