What is the extreme f(x)=x^3+y^3+3x^2-3y^2-8 ?

The extreme f(x)=x^3+y^3+3x^2-3y^2-8 is Minimum(0,2),Saddle(0,0),Saddle(-2,2),Maximum(-2,0)

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extreme f(x)=x^3+y^3+3x^2-3y^2-8

extreme points

f (x) = x^{3} + y^{3} + 3 x^{2} - 3 y^{2} - 8

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

Solution steps

Find the critical points:

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

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

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

Check the sign of at each critical point

Check critical point

(0, 2) :

Minimum

Check critical point

(0, 0) :

Saddle

Check critical point

(- 2, 2) :

Saddle

Check critical point

(- 2, 0) :

Maximum

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

f (x) = (x - 4)^{2} + 1 + 3

x^{2} y + y^{3} - 75 y

y = \frac{x ^{2}}{x ^{2} - 1}

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

f (x) = x^{4} - 6 x^{2} + 9

What is the extreme f(x)=x^3+y^3+3x^2-3y^2-8 ?
The extreme f(x)=x^3+y^3+3x^2-3y^2-8 is Minimum(0,2),Saddle(0,0),Saddle(-2,2),Maximum(-2,0)