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

The extreme f(x,y)=-3+x^2+y^2 is Minimum(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)=-3+x^2+y^2

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

y = 4 x^{2} - 2 x^{4}

f (x) = (x^{2} - 1) e^{- a x}

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

\frac{x ^{2} - 24}{x - 5}

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

What is the extreme f(x,y)=-3+x^2+y^2 ?
The extreme f(x,y)=-3+x^2+y^2 is Minimum(0,0)