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

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

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extreme f(x,y)=x^2+y^2-2

extreme points

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

f (x, y) = 3 x^{3} - 2 x^{2} y + 5 x y - y^{3} - 1

f (x) = (3 - x) e^{x - 3}

f (x) = 1 + x^{2} e^{5 - 3 x}

y = - 4 e^{- 4 t} + 5 e^{- 5 t}

x^{3} + y^{3} + 9 x^{2} - 3 y^{2} + 15 x - 9 y

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