What is the extreme f(x,y)=2x^2-8y^2-13 ?

The extreme f(x,y)=2x^2-8y^2-13 is Saddle(0,0)

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extreme f(x,y)=2x^2-8y^2-13

extreme points

f (x, y) = 2 x^{2} - 8 y^{2} - 13

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 16

D (x, y) = - 64

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

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

y = - x^{2} + 5, (- 2, 4)

f (x) = 4 sin (x) + 3 cos (x)

f (x y) = x^{2} + y^{2} + x + y + x y

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

What is the extreme f(x,y)=2x^2-8y^2-13 ?
The extreme f(x,y)=2x^2-8y^2-13 is Saddle(0,0)