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

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

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

extreme points

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

Minimum (0, 0), Saddle (0, 2), Saddle (\frac{2}{3}, 0), Maximum (\frac{2}{3}, 2)

Solution steps

Find the critical points:

(0, 0), (0, 2), (\frac{2}{3}, 0), (\frac{2}{3}, 2)

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

D (x, y) = (- 6 x + 2) (- 6 y + 6)

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(0, 2) :

Saddle

Check critical point

(\frac{2}{3}, 0) :

Saddle

Check critical point

(\frac{2}{3}, 2) :

Maximum

Minimum (0, 0), Saddle (0, 2), Saddle (\frac{2}{3}, 0), Maximum (\frac{2}{3}, 2)

3 x^{5} - 10 x^{3}

f (x) = 8 x^{\frac{1}{3}} - x^{\frac{4}{3}}

f (x, y) = x^{2} + 3 x y + 2 y^{4}

f (x) = x^{2} + 6 x + 15

f (x) = ln (x)

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