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