Posted by Billy Baston on Monday, March 25, 2013 at 7:49pm.
x^2/x3 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/(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
Answer This Question
Related Questions
 Calculus (pleas help!!!)  Find the absolute maximum and absolute minimum ...
 Calculus (pleas help!!!)  Find the absolute maximum and absolute minimum values...
 Calculus  Find the absolute maximum and absolute minimum values of the function...
 calculus  2) Let g(s)= t(4−t)^1/2 on the interval [0,2]. Find the ...
 Caluclus  Find the absolute maximum and absolute minimum of f on the interval...
 Calculus  Let f(t)=t\sqrt{4t} on the interval [1,3]. Find the absolute ...
 calculus  Let g(x)=(4x)/(x^2+1) on the interval [4,0]. Find the absolute ...
 calculus  Let g(s)=1/(s2) on the interval [0,1]. Find the absolute maximum and...
 calculus  Let f(x)=x^2+3x on the interval [1,3]. Find the absolute maximum and...
 Calculus  Find the absolute extrema of the function on the interval [2, 9]. (...
More Related Questions