What is the extreme f(x,y)=x^4-(18x)/(12y^2)-xy^3-x^2 ?

The extreme f(x,y)=x^4-(18x)/(12y^2)-xy^3-x^2 is Saddle(0,-\sqrt[5]{3/2})

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^4-(18x)/(12y^2)-xy^3-x^2

extreme points

f (x, y) = x^{4} - \frac{18 x}{12 y ^{2}} - x y^{3} - x^{2}

Solution steps

Find the critical points:

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{9 x}{y ^{4}} - 6 x y

D (x, y) = \frac{- 108 x ^{3} + 18 x}{y ^{4}} - 72 x^{3} y + 12 x y - 9 y^{4} + \frac{18}{y} - \frac{9}{y ^{6}}

Check the sign of at each critical point

Check critical point

Saddle

y = x^{2} - z^{2}

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

ln (x) + \frac{1}{x}

f (x) = \frac{x ^{2}}{( x - 4 ) ^{2}}

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

What is the extreme f(x,y)=x^4-(18x)/(12y^2)-xy^3-x^2 ?
The extreme f(x,y)=x^4-(18x)/(12y^2)-xy^3-x^2 is Saddle(0,-\sqrt[5]{3/2})