What is the extreme f(x,y)=2x^2+4xy^3-2y^2+15 ?

The extreme f(x,y)=2x^2+4xy^3-2y^2+15 is Saddle(0,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x,y)=2x^2+4xy^3-2y^2+15

extreme points

f (x, y) = 2 x^{2} + 4 x y^{3} - 2 y^{2} + 15

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 24 x y - 4

D (x, y) = 4 (24 x y - 4) - 144 y^{4}

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = 2 x^{3} - 3 x^{2} - 12 x - 7

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

f (x) = x x^{2} + 16

f (x) = \frac{x}{x ^{2} + 14 x + 48}

f (x) = 3 x^{4} - 4 x^{3} - 12 x^{2} - 1

What is the extreme f(x,y)=2x^2+4xy^3-2y^2+15 ?
The extreme f(x,y)=2x^2+4xy^3-2y^2+15 is Saddle(0,0)