What is the extreme f(x)=x^2+xy+y^2-2x+2y-8 ?

The extreme f(x)=x^2+xy+y^2-2x+2y-8 is Minimum(2,-2)

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extreme f(x)=x^2+xy+y^2-2x+2y-8

extreme points

f (x) = x^{2} + x y + y^{2} - 2 x + 2 y - 8

Minimum (2, - 2)

Solution steps

Find the critical points:

(2, - 2)

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(2, - 2) :

Minimum

Minimum (2, - 2)

f (x) = x^{2} - y^{2} - 2 x + 4 y + 6

f (x) = 2 x^{3} - 9 x^{2} + 12 x + 6, 0 \leq x \leq 3

y = \frac{x}{x ^{2} - x + 4}

\frac{4 x}{1 + 9 x ^{2}}

f (x, y) = y x - y^{2} - x + 6 y

What is the extreme f(x)=x^2+xy+y^2-2x+2y-8 ?
The extreme f(x)=x^2+xy+y^2-2x+2y-8 is Minimum(2,-2)