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

The extreme f(x,y)=x^2-y^2-2x-4y-4 is Saddle(1,-2)

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extreme f(x,y)=x^2-y^2-2x-4y-4

extreme points

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

Saddle (1, - 2)

Solution steps

Find the critical points:

(1, - 2)

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

D (x, y) = - 4

Check the sign of at each critical point

Check critical point

(1, - 2) :

Saddle

Saddle (1, - 2)

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

f (x) = x 7 - x

y = 5 t - 2 u (t - 1) + 3 u (t - 5)

T (x, y) = 8 - x^{2} - 4 y^{2}

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

What is the extreme f(x,y)=x^2-y^2-2x-4y-4 ?
The extreme f(x,y)=x^2-y^2-2x-4y-4 is Saddle(1,-2)