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

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

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extreme f(x)=x^2-xy+y^2-1x+1y

extreme points

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

Minimum (\frac{1}{3}, - \frac{1}{3})

Solution steps

Find the critical points:

(\frac{1}{3}, - \frac{1}{3})

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(\frac{1}{3}, - \frac{1}{3}) :

Minimum

Minimum (\frac{1}{3}, - \frac{1}{3})

x + f (x) = x^{2} + 13

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

f (x) = 3 x - 1

f (x) = (60 - 2 x) (30 - x) x

f (x) = f (x) = 3 x - x^{3}

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