What is the extreme f(x,y)=x^2y+4xy-y^2+3 ?

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

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^2y+4xy-y^2+3

extreme points

f (x, y) = x^{2} y + 4 x y - y^{2} + 3

Saddle (0, 0), Saddle (- 4, 0), Maximum (- 2, - 2)

Solution steps

Find the critical points:

(0, 0), (- 4, 0), (- 2, - 2)

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

D (x, y) = - 4 y - 4 x^{2} - 16 x - 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 4, 0) :

Saddle

Check critical point

(- 2, - 2) :

Maximum

Saddle (0, 0), Saddle (- 4, 0), Maximum (- 2, - 2)

f (x) = x^{4} - 2 x^{3} + 1

\frac{2 x}{x ^{2} - 9}

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

f (x, y) = x^{2} - 2 x + y^{2} + 2 y + 1

f (x) = 2 x + \frac{1}{x}

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