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

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = 12 x

Check the sign of at each critical point

Check critical point

(- 1, 1) :

Saddle

Check critical point

(1, 1) :

Minimum

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

f (x) = 7 x - x^{2}

f (x) = ln (4 - ln (x))

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

f (x) = - x^{3} + 6 x^{2} + 63 x

f (x) = - 6 x^{2} + 7 x y - 2 y^{2} + x + y

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