What is the extreme f(x,y)=e^{((x^2+y^2))/2} ?

The extreme f(x,y)=e^{((x^2+y^2))/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)=e^{((x^2+y^2))/2}

extreme points

f (x, y) = e^{\frac{( x ^{2} + y ^{2} )}{2}}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = e^{\frac{x ^{2} + y ^{2}}{2}} y^{2} + e^{\frac{x ^{2} + y ^{2}}{2}}

D (x, y) = e^{x^{2} + y^{2}} x^{2} + e^{x^{2} + y^{2}} y^{2} + e^{x^{2} + y^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

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

P (x, y) = 3 x + 2 y

3 y^{2} - 2 y^{3} - 3 x^{2} + 6 x y

\frac{5 x}{x ^{2} - 16}

f (x, y) = 3 x^{3} - 9 x y + 3 y^{3}

What is the extreme f(x,y)=e^{((x^2+y^2))/2} ?
The extreme f(x,y)=e^{((x^2+y^2))/2} is Minimum(0,0)