What is the extreme f(x,y)=x^2-2xy+1 ?

The extreme f(x,y)=x^2-2xy+1 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)=x^2-2xy+1

extreme points

f (x, y) = x^{2} - 2 x y + 1

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 0

D (x, y) = - 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

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

f (x) = x^{4} + 4 x

f (x, y) = - 3 x y + x^{3} - y^{3} + \frac{1}{8}

f (x) = x^{2} - 8 x + 1

f (x) = x^{3} - 12 x, 0 \leq x \leq 4

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