What is the extreme f(x,y)=2x^2-8y^2-4x+9y+2 ?

The extreme f(x,y)=2x^2-8y^2-4x+9y+2 is Saddle(1, 9/16)

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extreme f(x,y)=2x^2-8y^2-4x+9y+2

extreme points

f (x, y) = 2 x^{2} - 8 y^{2} - 4 x + 9 y + 2

Saddle (1, \frac{9}{16})

Solution steps

Find the critical points:

(1, \frac{9}{16})

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

D (x, y) = - 64

Check the sign of at each critical point

Check critical point

(1, \frac{9}{16}) :

Saddle

Saddle (1, \frac{9}{16})

f (x) = - 6 x^{2} - 12 x + 18

\frac{2 x + 2}{x - 2}

- x^{2} + 2 x

f (x) = x y + \frac{50}{x} + \frac{20}{x}

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

What is the extreme f(x,y)=2x^2-8y^2-4x+9y+2 ?
The extreme f(x,y)=2x^2-8y^2-4x+9y+2 is Saddle(1, 9/16)