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

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

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

extreme points

f (x, y) = 2 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) = - 8

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = x (85 - 0.05 x) - (600 + 35 x)

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

P (x, y) = x^{2} - y^{2} + 6 y - 9

f (x, y) = \frac{x ^{3}}{3} - y^{2} - 5 x + x y - 3

h (x) = - 2 x^{2} - 1, - 4 \leq x \leq 3

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