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f(x,y)=x^2+y^2-3xy for (x,y) in R2.
Explain why the function f:R2→R has no local extreme points.
不知道該如何解題才好 所以希望各位大大能幫個忙 非常的急
- 教書的Lv 61 decade agoFavorite Answer
This f function is differentiable everywhere on R2 with
fx=2x-3y, fy=2y-3x. If this f has a local extrema at a point (a,b) then fx(a,b)=0 and fy(a,b)=0. In fact the system of equations fx=0 and fy=0 has only one solution (a,b)=(0,0).
To classify this only critical point of f we appy the so-called second derivative test. We calculateD=fxx*fyy-fxy^2=(2)(-3)-0=-6 for every (x,y) hence for (0,0). By the second derivative test, D negative at (a.b) means that (a,b) is just a saddle point, not extremal point of f.