What is the extreme f(x,y)=4x^2-8xy+2y^2 ?

The extreme f(x,y)=4x^2-8xy+2y^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 f(x,y)=4x^2-8xy+2y^2

extreme points

f (x, y) = 4 x^{2} - 8 x y + 2 y^{2}

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - 32

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = - 3 x^{2} + 24 x - 200

2 - x^{4} + 2 x^{2} - y^{2}

- x^{2} + 162 ln (x)

f (x) = x^{4} - 8 x^{3} + 18 x^{2} - 16 x + 40

3 x^{2} - x y + 5 y^{2}

What is the extreme f(x,y)=4x^2-8xy+2y^2 ?
The extreme f(x,y)=4x^2-8xy+2y^2 is Saddle(0,0)