What is the extreme f(x,y)=5x^4-x^2+6y^2 ?

The extreme f(x,y)=5x^4-x^2+6y^2 is Saddle(0,0),Minimum(-1/(sqrt(10)),0),Minimum(1/(sqrt(10)),0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x,y)=5x^4-x^2+6y^2

extreme

Saddle (0, 0), Minimum (- \frac{1}{10}, 0), Minimum (\frac{1}{10}, 0)

Solution steps

Find the critical points:

(0, 0), (- \frac{1}{10}, 0), (\frac{1}{10}, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 12

D (x, y) = 12 (60 x^{2} - 2)

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- \frac{1}{10}, 0) :

Minimum

Check critical point

(\frac{1}{10}, 0) :

Minimum

Saddle (0, 0), Minimum (- \frac{1}{10}, 0), Minimum (\frac{1}{10}, 0)

What is the extreme f(x,y)=5x^4-x^2+6y^2 ?
The extreme f(x,y)=5x^4-x^2+6y^2 is Saddle(0,0),Minimum(-1/(sqrt(10)),0),Minimum(1/(sqrt(10)),0)