What is the extreme f(x)=2x^2+3y^2-4x-12y+13 ?

The extreme f(x)=2x^2+3y^2-4x-12y+13 is Minimum(1,2)

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extreme f(x)=2x^2+3y^2-4x-12y+13

extreme points

f (x) = 2 x^{2} + 3 y^{2} - 4 x - 12 y + 13

Minimum (1, 2)

Solution steps

Find the critical points:

(1, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6

D (x, y) = 24

Check the sign of at each critical point

Check critical point

(1, 2) :

Minimum

Minimum (1, 2)

f (x) = 5 - 6 x^{2} - 2 x^{3}

\frac{3}{2} x^{2} + x^{3} + 3 y^{2}

f (x) = \frac{3}{x}

f (x, y) = (9 + x y) (x + y)

y = u (t - 1) - u (t - 4)

What is the extreme f(x)=2x^2+3y^2-4x-12y+13 ?
The extreme f(x)=2x^2+3y^2-4x-12y+13 is Minimum(1,2)