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) = (x - 2) (x - 3)^{2} (x - 4)^{3}

x^{3} - 6 x^{2} + 10

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

x^{3} - 3 x + 5

x^{3} - 3 x + 9

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)