What is the extreme f(x,y)=x^3+3x^2+6xy+y^2 ?

The extreme f(x,y)=x^3+3x^2+6xy+y^2 is Saddle(0,0),Minimum(4,-12)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x,y)=x^3+3x^2+6xy+y^2

extreme points

f (x, y) = x^{3} + 3 x^{2} + 6 x y + y^{2}

Saddle (0, 0), Minimum (4, - 12)

Solution steps

Find the critical points:

(0, 0), (4, - 12)

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

D (x, y) = 12 x - 24

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(4, - 12) :

Minimum

Saddle (0, 0), Minimum (4, - 12)

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

f (x) = \frac{x ^{3} + 9}{x ^{2} + 4}

f (x, y) = 2 x^{3} - 9 x^{2} + 12 x + 2 y^{3} + 3 y^{2}

f (x) = x 6 - x^{2}

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

What is the extreme f(x,y)=x^3+3x^2+6xy+y^2 ?
The extreme f(x,y)=x^3+3x^2+6xy+y^2 is Saddle(0,0),Minimum(4,-12)