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

The extreme f(x,y)=x^3-y^2-12x+4y+2 is Maximum(-2,2),Saddle(2,2)

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

extreme

Maximum (- 2, 2), Saddle (2, 2)

Solution steps

Find the critical points:

(- 2, 2), (2, 2)

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

D (x, y) = - 12 x

Check the sign of at each critical point

Check critical point

(- 2, 2) :

Maximum

Check critical point

(2, 2) :

Saddle

Maximum (- 2, 2), Saddle (2, 2)

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