What is the extreme f(x,y)=x^3+y^3+3x^2-6y^2 ?

The extreme f(x,y)=x^3+y^3+3x^2-6y^2 is Minimum(0,4),Saddle(0,0),Saddle(-2,4),Maximum(-2,0)

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extreme f(x,y)=x^3+y^3+3x^2-6y^2

extreme points

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

Minimum (0, 4), Saddle (0, 0), Saddle (- 2, 4), Maximum (- 2, 0)

Solution steps

Find the critical points:

(0, 4), (0, 0), (- 2, 4), (- 2, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y - 12

D (x, y) = (6 x + 6) (6 y - 12)

Check the sign of at each critical point

Check critical point

(0, 4) :

Minimum

Check critical point

(0, 0) :

Saddle

Check critical point

(- 2, 4) :

Saddle

Check critical point

(- 2, 0) :

Maximum

Minimum (0, 4), Saddle (0, 0), Saddle (- 2, 4), Maximum (- 2, 0)

- 2 + 2 x - x^{2}

f (x) = \frac{x ^{3}}{3} + 8 x^{2} - 18 x + 7

- \frac{x + 2}{( x - 1 ) ( x - 3 ) ^{2}}

f (x) = 1 - 2 x + 4 y - x^{2} - 4 y^{2}

x^{2} = - y + 4

What is the extreme f(x,y)=x^3+y^3+3x^2-6y^2 ?
The extreme f(x,y)=x^3+y^3+3x^2-6y^2 is Minimum(0,4),Saddle(0,0),Saddle(-2,4),Maximum(-2,0)