What is the extreme f(x,y)=2y^2+x^2-x^2y ?

The extreme f(x,y)=2y^2+x^2-x^2y is Minimum(0,0),Saddle(2,1),Saddle(-2,1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x,y)=2y^2+x^2-x^2y

extreme points

f (x, y) = 2 y^{2} + x^{2} - x^{2} y

Minimum (0, 0), Saddle (2, 1), Saddle (- 2, 1)

Solution steps

Find the critical points:

(0, 0), (2, 1), (- 2, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

D (x, y) = 4 (2 - 2 y) - 4 x^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(2, 1) :

Saddle

Check critical point

(- 2, 1) :

Saddle

Minimum (0, 0), Saddle (2, 1), Saddle (- 2, 1)

\frac{2 x ^{2}}{x ^{2} - 25}

f (x) = \frac{x}{x ^{2} + 81}

F (I, J) = (25 I + 12 J) N

f (x, y) = 3 x y - x^{2} y - x y^{2}

y = x^{3} - 3 x^{2} + 3

What is the extreme f(x,y)=2y^2+x^2-x^2y ?
The extreme f(x,y)=2y^2+x^2-x^2y is Minimum(0,0),Saddle(2,1),Saddle(-2,1)