What is the extreme f(x,y)=x^3-2xy+y^2+4 ?

The extreme f(x,y)=x^3-2xy+y^2+4 is Minimum(2/3 , 2/3),Saddle(0,0)

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extreme f(x,y)=x^3-2xy+y^2+4

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = 12 x - 4

Check the sign of at each critical point

Check critical point

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

Minimum

Check critical point

(0, 0) :

Saddle

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

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

f (x, y) = (x - y) (25 - x y)

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

f (x) = x y + \frac{3}{x} + \frac{9}{y}

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

What is the extreme f(x,y)=x^3-2xy+y^2+4 ?
The extreme f(x,y)=x^3-2xy+y^2+4 is Minimum(2/3 , 2/3),Saddle(0,0)