What is the extreme P(x,y)=2x^2-xy-3y^2 ?

The extreme P(x,y)=2x^2-xy-3y^2 is Saddle(0,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme P(x,y)=2x^2-xy-3y^2

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 6

D (x, y) = - 25

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

x^{4} + (1 - x)^{4} + 10 x^{2} (1 - x)^{2}

- 7 x + 6

f (x, y) = 2 x^{3} - 3 y^{4} + 2 x y

f (x) = cos (x) - 5 x

y = \frac{x ^{2}}{x - 1}

What is the extreme P(x,y)=2x^2-xy-3y^2 ?
The extreme P(x,y)=2x^2-xy-3y^2 is Saddle(0,0)