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

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

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

extreme points

f (x, y) = y^{2} - x^{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^{3} - 12 x + 4

y = e^{r x}

\frac{- 2 x ^{2} + 5 x - 1}{2 x - 1}

f (x) = x^{3} - 6 x^{2} + 9 x + 2

f (x) = - x^{4} + 6 x^{2} - 4