I need to find the absolute minimum value of $\displaystyle f(x)= 8x+ (1/16x^2)$ on the interval (0, infinity).

I took the first derivative, $\displaystyle f'(x)= 8-(1/8x^3)$ and set it equal to zero.

I found that x= 4, but I don't think that's the absolute minimum. Please help!!!