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

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

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

extreme points

P (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) = x \cdot ln^{2} (x)

f (x) = f (x, y) = x^{3} + y^{3} - 3 x^{2} - 9 y^{2} - 9 x

\frac{1}{x ^{3}}

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