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

The extreme f(x,y)=(x^2-y^2)/(sqrt(4x^2+y^2-4)) is

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extreme f(x,y)=(x^2-y^2)/(sqrt(4x^2+y^2-4))

extreme points

f (x, y) = \frac{x ^{2} - y ^{2}}{4 x ^{2} + y ^{2} - 4}

Saddle (0, - 22), Saddle (0, 22), Saddle (- 2, 0), Saddle (2, 0), Saddle (- 2, 0), Saddle (2, 0)

Solution steps

Find the critical points:

(0, - 22), (0, 22), (- 2, 0), (2, 0), (- 2, 0), (2, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{2 ( - 18 x ^{4} + 3 x ^{2} y ^{2} + 34 x ^{2} - 2 y ^{2} - 16 )}{( 4 x ^{2} + y ^{2} - 4 ) ^{\frac{5}{2}}}

D (x, y) = \frac{- 576 x ^{6} + 528 x ^{4} y ^{2} - 64 x ^{4} + 72 x ^{2} y ^{4} - 1472 x ^{2} y ^{2} + 1664 x ^{2} - 24 y ^{6} - 64 y ^{4} + 896 y ^{2} - 1024}{( y ^{2} - 4 + 4 x ^{2} ) ^{5}}

Check the sign of at each critical point

Check critical point

(0, - 22) :

Saddle

Check critical point

(0, 22) :

Saddle

Check critical point

(- 2, 0) :

Saddle

Check critical point

(2, 0) :

Saddle

Check critical point

(- 2, 0) :

Saddle

Check critical point

(2, 0) :

Saddle

Saddle (0, - 22), Saddle (0, 22), Saddle (- 2, 0), Saddle (2, 0), Saddle (- 2, 0), Saddle (2, 0)

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

P (a, c) = a + 7 + c

f (x) = \frac{3}{x ^{2} - 2 x}

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

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

What is the extreme f(x,y)=(x^2-y^2)/(sqrt(4x^2+y^2-4)) ?
The extreme f(x,y)=(x^2-y^2)/(sqrt(4x^2+y^2-4)) is