What is the extreme P(x,y)=x+y^2-10x^2y^2+9y^6 ?

The extreme P(x,y)=x+y^2-10x^2y^2+9y^6 is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme P(x,y)=x+y^2-10x^2y^2+9y^6

extreme points

P (x, y) = x + y^{2} - 10 x^{2} y^{2} + 9 y^{6}

Solution steps

Find the critical points:

\frac{\partial ^{2} f}{\partial x ^{2}} = 270 y^{4} + 2 - 20 x^{2}

D (x, y) = - 5400 y^{6} - 40 y^{2} - 1200 x^{2} y^{2}

Check the sign of at each critical point

Check critical point

Saddle

Check critical point

Saddle

f (x) = - x^{\frac{2}{3}} (x - 1), - 1 \leq x \leq 1

f (x) = t 25 - t^{2}

f (x) = \frac{x + 6}{x ^{2} + 6}

ln (x^{2} + 49)

f (x, y) = 9 - 2 x^{2} - y^{2}

What is the extreme P(x,y)=x+y^2-10x^2y^2+9y^6 ?
The extreme P(x,y)=x+y^2-10x^2y^2+9y^6 is