Posted by **Billy Baston** on Monday, March 25, 2013 at 7:49pm.

x^2/x-3 on the interval from 3 until infinity. How would you do absolute minimum on this?

- absolute minimum math -
**Reiny**, Monday, March 25, 2013 at 8:26pm
y = x^2/(x-3)

dy/dx = ( (x-3)(2x) - x^2) /(x-3)^2

= (2x^2 - 6x - x^2)/(x+3)^2

= (x^2 - 6x)/(x-3)^2

for a max/min, dy/dx = 0

x^2 - 6x = 0

x(x-6) = 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

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