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

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - 20

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

g (x, y) = \frac{x ^{2} + y}{x ^{2} + y ^{2} + 1}

f (x) = - 2 x^{3} + 24 x^{2} - 42 x + 9

y = 9 x^{3} - 7 x^{2} + 3 x + 10

x^{2} - 1 + 2

f (x) = - \frac{7}{3} x^{3} + 14 x^{2} + 147 x + 2

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