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 points

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

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} + 1} y^{2} + e^{x^{2} + y^{2} + 1})

D (x, y) = 4 (2 e^{x^{2} + y^{2} + 1} x^{2} + e^{x^{2} + y^{2} + 1}) (2 e^{x^{2} + y^{2} + 1} y^{2} + e^{x^{2} + y^{2} + 1}) - 16 e^{2 (x^{2} + y^{2} + 1)} x^{2} y^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

F (x) = x^{3} - 6 x^{2} + 9 x + 15

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

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

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

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

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)