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

The extreme f(x,y)=6x^2-2x^3+3y^2+6xy is Minimum(0,0),Saddle(1,-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)=6x^2-2x^3+3y^2+6xy

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = - 72 x + 36

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(1, - 1) :

Saddle

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

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

f (x) = x^{2} + x

f (x) = \frac{x}{x ^{2} - x + 4}, 0 \leq x \leq 6

f (x) = x^{3} - 6 x^{2} + 11

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

What is the extreme f(x,y)=6x^2-2x^3+3y^2+6xy ?
The extreme f(x,y)=6x^2-2x^3+3y^2+6xy is Minimum(0,0),Saddle(1,-1)