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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

extreme

Saddle (0, - \frac{4}{3})

Solution steps

Find the critical points:

(0, - \frac{4}{3})

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

D (x, y) = - 12 x^{2} y^{2} - 24 x y - 9

Check the sign of at each critical point

Check critical point

(0, - \frac{4}{3}) :

Saddle

Saddle (0, - \frac{4}{3})

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