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

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

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

extreme points

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

Minimum (1, - 1)

Solution steps

Find the critical points:

(1, - 1)

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(1, - 1) :

Minimum

Minimum (1, - 1)

f (x) = 5 - 3 x^{2}, - 5 \leq x \leq 1

f (x) = 16 x + 21 x^{\frac{16}{21}}

e^{- y} (x^{2} + y^{2})

f (x) = x^{\frac{2}{3}} (x - 3), - 3 \leq x \leq 3

f (x) = 2 x^{3} - 27 x^{2} + 84 x + 11

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