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

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

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

extreme points

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

Minimum (- 1, 0)

Solution steps

Find the critical points:

(- 1, 0)

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

D (x, y) = 4 e^{x^{2} + y^{2} + 2 x} (2 x^{2} + 4 x + 3) (2 e^{x^{2} + y^{2} + 2 x} y^{2} + e^{x^{2} + y^{2} + 2 x}) - 4 e^{2 (x^{2} + y^{2} + 2 x)} y^{2} (2 x + 2)^{2}

Check the sign of at each critical point

Check critical point

(- 1, 0) :

Minimum

Minimum (- 1, 0)

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

4 x^{2} - 2 x + 11

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

f (x, y) = 2 x^{2} + 3 y^{2} - 4 x y + 4 x - 2 y + 3

f (x, y) = 400 - 4 x^{2} - 36 y^{2}

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