What is the extreme f(x)=e^{x^2+3y^2-2x} ?

The extreme f(x)=e^{x^2+3y^2-2x} is Minimum(1,0)

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

extreme points

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

Minimum (1, 0)

Solution steps

Find the critical points:

(1, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 (6 e^{x^{2} + 3 y^{2} - 2 x} y^{2} + e^{x^{2} + 3 y^{2} - 2 x})

D (x, y) = 12 e^{x^{2} + 3 y^{2} - 2 x} (2 x^{2} - 4 x + 3) (6 e^{x^{2} + 3 y^{2} - 2 x} y^{2} + e^{x^{2} + 3 y^{2} - 2 x}) - 36 e^{2 (x^{2} + 3 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) = 4 - (8 + 7 x)^{\frac{2}{7}}

\frac{3}{( x - 1 ) ( x ^{2} - 4 )}

f (x, y) = 2 x^{2} - 8 y^{2} - 4 x + 9 y + 2

f (x) = - 6 x^{2} - 12 x + 18

\frac{2 x + 2}{x - 2}

What is the extreme f(x)=e^{x^2+3y^2-2x} ?
The extreme f(x)=e^{x^2+3y^2-2x} is Minimum(1,0)