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

The extreme f(x,y)=2x^2-2x^2y+y^2 is Minimum(0,0),Saddle(1,1),Saddle(-1,1)

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

extreme points

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

Minimum (0, 0), Saddle (1, 1), Saddle (- 1, 1)

Solution steps

Find the critical points:

(0, 0), (1, 1), (- 1, 1)

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

D (x, y) = 2 (4 - 4 y) - 16 x^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(1, 1) :

Saddle

Check critical point

(- 1, 1) :

Saddle

Minimum (0, 0), Saddle (1, 1), Saddle (- 1, 1)

f (x, y) = x^{2} + y^{2} + x^{2} y + 5

f (x) = 3 x^{2} + 5 x - 1

f (x) = x^{3} + 12 x^{2} - 27 x + 8, - 10 \leq x \leq 0

x^{2} e^{- 3 x}

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

What is the extreme f(x,y)=2x^2-2x^2y+y^2 ?
The extreme f(x,y)=2x^2-2x^2y+y^2 is Minimum(0,0),Saddle(1,1),Saddle(-1,1)