What is the x coordinate of the absolute minimum of 6x + [12/x] when x > 0?

Answer 1

let y = 6x + 12/x
dy/dx = 6 - 12/x^2
= 0 for a max/min
6 = 12/x^2
x^2 = 2
x = √2, since x>0

when x = 2, y = 12 - 12/2 = 6

so the min is 6 when x = 2

Answer 2

I tried that answer but it wasn't right. I tried sqrt 2 and it was right.