What is the extreme f(x)=6y^2+4x^2-12xy-4y ?

The extreme f(x)=6y^2+4x^2-12xy-4y is Saddle(-1,-2/3)

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extreme f(x)=6y^2+4x^2-12xy-4y

extreme points

f (x) = 6 y^{2} + 4 x^{2} - 12 x y - 4 y

Saddle (- 1, - \frac{2}{3})

Solution steps

Find the critical points:

(- 1, - \frac{2}{3})

\frac{\partial ^{2} f}{\partial x ^{2}} = 12

D (x, y) = - 48

Check the sign of at each critical point

Check critical point

(- 1, - \frac{2}{3}) :

Saddle

Saddle (- 1, - \frac{2}{3})

f (x) = x^{3} + 4 x^{2} - 3 x + 6

f (x, y) = x^{3} + y^{3} + 2 x^{2} + 4 y^{2} + 6

f (x) = 3 x^{2} - 18 x + 23

f (x, y) = 5 + 3 x^{2} + 3 y^{2} + 2 y^{3} + x^{3}

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

What is the extreme f(x)=6y^2+4x^2-12xy-4y ?
The extreme f(x)=6y^2+4x^2-12xy-4y is Saddle(-1,-2/3)