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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = - 36 y

Check the sign of at each critical point

Check critical point

(0, 1) :

Saddle

Check critical point

(0, - 1) :

Minimum

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

f (x) = 12 - 3 x^{2}, (- 2, 5)

f (x) = x^{3} - x^{2} - x - 1

f (x) = 4 x y + 120 y - 20 y^{2} - \frac{1}{10} x^{2} y - 80

\frac{d}{d x} (x^{3} y^{2})

y = \frac{x ^{3}}{3} - x^{2} - 3 x

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