|February 27th, 2015, 03:54 PM||#1|
Joined: Feb 2015
From: United States
Local max, local min, or saddle point
I have a function g(x,y) = x^2*y - x - 1 for which my partial derivatives set equal to 0 are 2xy - 1 = 0 and x^2 = 0. Solving the 2nd equation, I get x = 0. If I plug in 0 into the first equation, both x and y goes away because they are being multiplied by 0. What to do in this case ? I have to find critical points, and also local max, local min, or saddle point.
|February 27th, 2015, 04:14 PM||#2|
Joined: Dec 2013
Math Focus: Mainly analysis and algebra
So the second equation has no solutions when $x=0$, but the first tells you that any critical points must be on $x=0$. What is the logical conclusion?
|local, max, min, point, saddle|
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