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

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

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

extreme points

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

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) = 12 x + ln (x)

ln (27 + 8 x^{3})

x^{3} - 6 x^{2} - 15 x + 7

f (x, y) = x^{3} + y^{2} - 3 x + 1

f (x, y) = x y + \frac{1}{x} + \frac{8}{y}

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