What is the extreme f(x)=x^2-4xy+y^2-4 ?

The extreme f(x)=x^2-4xy+y^2-4 is Saddle(0,0)

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extreme f(x)=x^2-4xy+y^2-4

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - 12

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = 15 x^{2} - 60 x + 25

f (x) = 10 sin (2 x), (- 2 π, 2 π)

- \frac{1}{2} x^{4} + \frac{4}{3} x^{3} - 14

f (x, y) = 2 x y - 32 x + 64 y

f (x) = 6 x y - \frac{1}{2} (x^{4} + y^{4}) - 2

What is the extreme f(x)=x^2-4xy+y^2-4 ?
The extreme f(x)=x^2-4xy+y^2-4 is Saddle(0,0)