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

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

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

extreme points

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

Minimum (- 2, - 1)

Solution steps

Find the critical points:

(- 2, - 1)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(- 2, - 1) :

Minimum

Minimum (- 2, - 1)

f (x) = 416 - x x

f (x) = x^{\frac{4}{5}} - 4

f (x) = (\frac{5 x ^{4}}{ln ( x )})

f (x) = - (\frac{4}{x ^{2}}), (- 3, - 1)

f (x) = 4 - x^{3} + 5 x

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