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

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

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

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

f (x) = t 4 - t

x^{2} + \frac{1}{x}

f (x) = x^{3} - x^{2} - x + 8, - 1 \leq x \leq 0

f (x) = \frac{12 x}{9 + x ^{2}}

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