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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

\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, 0) :

Saddle

Check critical point

(0, 2) :

Maximum

Check critical point

(2, 0) :

Minimum

Check critical point

(2, 2) :

Saddle

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

f (x) = \frac{ln ( x )}{3 x}, 1 \leq x \leq 4

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

3 + 6 x^{2} - 4 x^{3}

f (x, y) = x^{2} + y^{2} + 8 x - 4 y + 7

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

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