What is the extreme 2x^2-2xy+y^2-2x ?

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

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extreme 2x^2-2xy+y^2-2x

extreme points

2 x^{2} - 2 x y + y^{2} - 2 x

Minimum (1, 1)

Solution steps

Find the critical points:

(1, 1)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(1, 1) :

Minimum

Minimum (1, 1)

f (x) = \frac{y - 1}{y ^{3} - y + 1}

f (x, y) = 2 x^{2} + 2 y^{2} - 16 x + 12 y

f (x) = x^{3} - 15 x^{2} + 63 x - 15

f (x, y) = \frac{8460}{x} + \frac{8460}{y} + 5 x y

f (x) = (85 - 0.05 x) - (600 + 35 x)

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