What is the extreme f(x)=x^2(2-y)-y^3+3y^2+9y ?

The extreme f(x)=x^2(2-y)-y^3+3y^2+9y is Minimum(0,-1),Maximum(0,3),Saddle(3,2),Saddle(-3,2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=x^2(2-y)-y^3+3y^2+9y

extreme

Minimum (0, - 1), Maximum (0, 3), Saddle (3, 2), Saddle (- 3, 2)

Solution steps

Find the critical points:

(0, - 1), (0, 3), (3, 2), (- 3, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 6 y + 6

D (x, y) = 2 (- y + 2) (- 6 y + 6) - 4 x^{2}

Check the sign of at each critical point

Check critical point

(0, - 1) :

Minimum

Check critical point

(0, 3) :

Maximum

Check critical point

(3, 2) :

Saddle

Check critical point

(- 3, 2) :

Saddle

Minimum (0, - 1), Maximum (0, 3), Saddle (3, 2), Saddle (- 3, 2)

What is the extreme f(x)=x^2(2-y)-y^3+3y^2+9y ?
The extreme f(x)=x^2(2-y)-y^3+3y^2+9y is Minimum(0,-1),Maximum(0,3),Saddle(3,2),Saddle(-3,2)