absolute minimum math
posted by Billy Baston .
x^2/x3 on the interval from 3 until infinity. How would you do absolute minimum on this?

y = x^2/(x3)
dy/dx = ( (x3)(2x)  x^2) /(x3)^2
= (2x^2  6x  x^2)/(x+3)^2
= (x^2  6x)/(x3)^2
for a max/min, dy/dx = 0
x^2  6x = 0
x(x6) = 0
x = 0 or x = 6
so in the interval given, we need x = 6
if x = 6
y = 36/3 = 12
The minimum is 12