What is the extreme f(x,y)=x^3+y^3-3x^2-3y+10 ?

The extreme f(x,y)=x^3+y^3-3x^2-3y+10 is Saddle(0,1),Maximum(0,-1),Minimum(2,1),Saddle(2,-1)

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extreme f(x,y)=x^3+y^3-3x^2-3y+10

extreme points

f (x, y) = x^{3} + y^{3} - 3 x^{2} - 3 y + 10

Saddle (0, 1), Maximum (0, - 1), Minimum (2, 1), Saddle (2, - 1)

Solution steps

Find the critical points:

(0, 1), (0, - 1), (2, 1), (2, - 1)

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

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

Check the sign of at each critical point

Check critical point

(0, 1) :

Saddle

Check critical point

(0, - 1) :

Maximum

Check critical point

(2, 1) :

Minimum

Check critical point

(2, - 1) :

Saddle

Saddle (0, 1), Maximum (0, - 1), Minimum (2, 1), Saddle (2, - 1)

2 t^{4} + 6 t^{3} - 5

f (x, y) = arcsec (x^{2} - ∣ y ∣)

f (x) = x^{\frac{2}{5}}

f (x) = 2 x^{2} - 4 x y + 3 y^{2} - 8 x + 8 y - 1 = 0

f (x) = x^{3} + 3 y^{3} + 3 x^{2} + 3 y^{2} + 24

What is the extreme f(x,y)=x^3+y^3-3x^2-3y+10 ?
The extreme f(x,y)=x^3+y^3-3x^2-3y+10 is Saddle(0,1),Maximum(0,-1),Minimum(2,1),Saddle(2,-1)