What is the extreme f(x,y)=1+e^{(x^2+y^2)} ?

The extreme f(x,y)=1+e^{(x^2+y^2)} is Minimum(0,0)

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extreme f(x,y)=1+e^{(x^2+y^2)}

extreme

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 4 (2 e^{x^{2} + y^{2}} x^{2} + e^{x^{2} + y^{2}}) (2 e^{x^{2} + y^{2}} y^{2} + e^{x^{2} + y^{2}}) - 16 e^{2 (x^{2} + y^{2})} x^{2} y^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

What is the extreme f(x,y)=1+e^{(x^2+y^2)} ?
The extreme f(x,y)=1+e^{(x^2+y^2)} is Minimum(0,0)