What is the extreme f(x,y)=14x^2-2x^3+4xy ?

The extreme f(x,y)=14x^2-2x^3+4xy is Saddle(0,0)

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extreme f(x,y)=14x^2-2x^3+4xy

extreme

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 0

D (x, y) = - 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

What is the extreme f(x,y)=14x^2-2x^3+4xy ?
The extreme f(x,y)=14x^2-2x^3+4xy is Saddle(0,0)