What is the extreme f(x)=x^3+y^3+3x^2-9y^2-1 ?

The extreme f(x)=x^3+y^3+3x^2-9y^2-1 is Minimum(0,6),Saddle(0,0),Saddle(-2,6),Maximum(-2,0)

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extreme f(x)=x^3+y^3+3x^2-9y^2-1

extreme

Minimum (0, 6), Saddle (0, 0), Saddle (- 2, 6), Maximum (- 2, 0)

Solution steps

Find the critical points:

(0, 6), (0, 0), (- 2, 6), (- 2, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y - 18

D (x, y) = (6 x + 6) (6 y - 18)

Check the sign of at each critical point

Check critical point

(0, 6) :

Minimum

Check critical point

(0, 0) :

Saddle

Check critical point

(- 2, 6) :

Saddle

Check critical point

(- 2, 0) :

Maximum

Minimum (0, 6), Saddle (0, 0), Saddle (- 2, 6), Maximum (- 2, 0)

What is the extreme f(x)=x^3+y^3+3x^2-9y^2-1 ?
The extreme f(x)=x^3+y^3+3x^2-9y^2-1 is Minimum(0,6),Saddle(0,0),Saddle(-2,6),Maximum(-2,0)