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

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

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

extreme points

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

Saddle (\frac{1}{2}, \frac{1}{2})

Solution steps

Find the critical points:

(\frac{1}{2}, \frac{1}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = 0

D (x, y) = - 4

Check the sign of at each critical point

Check critical point

(\frac{1}{2}, \frac{1}{2}) :

Saddle

Saddle (\frac{1}{2}, \frac{1}{2})

f (x) = \frac{x}{2} - sin (x), - π \leq x \leq π

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

f (x) = x (20 - 2 x) (12 - 2 x)

f (x) = 5 x^{2} - 40 ln (x)

y = x^{3} - 6 x^{2} + 2

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