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)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

extreme

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)

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)