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

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

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

extreme

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 4 (2 e^{x^{2}} x^{2} + e^{x^{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)=e^{x^2}+y^2+1 ?
The extreme f(x,y)=e^{x^2}+y^2+1 is Minimum(0,0)