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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = 4 x - 4

Check the sign of at each critical point

Check critical point

(2, 1) :

Minimum

Check critical point

(0, - 1) :

Saddle

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

f (x) = 1 + \frac{7}{x} - \frac{9}{x ^{2}}

2 x^{2} - x^{4}

f (x, y) = - 3 x^{2} + 6 x + 12 x y - 3 y^{2} + 2 y^{3} + 6

f (x) = Y (x) = x^{3} - 3 x^{2} - 9 x + 20

f (x) = - 3 x^{\frac{2}{3}}, - 1 \leq x \leq 1

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