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

The extreme f(x)=8y^2+x^2-x^2y is Minimum(0,0),Saddle(4,1),Saddle(-4,1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=8y^2+x^2-x^2y

extreme

Minimum (0, 0), Saddle (4, 1), Saddle (- 4, 1)

Solution steps

Find the critical points:

(0, 0), (4, 1), (- 4, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 16

D (x, y) = 16 (2 - 2 y) - 4 x^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(4, 1) :

Saddle

Check critical point

(- 4, 1) :

Saddle

Minimum (0, 0), Saddle (4, 1), Saddle (- 4, 1)

What is the extreme f(x)=8y^2+x^2-x^2y ?
The extreme f(x)=8y^2+x^2-x^2y is Minimum(0,0),Saddle(4,1),Saddle(-4,1)