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

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

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

extreme points

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

Minimum (- 3, 2)

Solution steps

Find the critical points:

(- 3, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(- 3, 2) :

Minimum

Minimum (- 3, 2)

f (x) = x^{3} - \frac{3}{2} x^{2}, - 3 \leq x \leq 6

f (x, y) = e^{2 x} - x^{3} y^{5}

f (x) = 3 x^{4} + x - 1

x + \frac{108}{x ^{2}}

f (x) = \frac{x ^{4}}{4} - 3 x^{3} + 4 x^{2}

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