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

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

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extreme f(x,y)=(2xy)/((x-2))

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 0

D (x, y) = - \frac{16}{( x - 2 ) ^{4}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = - 7 (x + 1)^{2} + 3 x + 7

y = 2 x^{3} - 3 x + 11

y = - x^{2} + 1, - 2 \leq x \leq 1

f (x) = x^{3} - 6 x^{2} - 15 x

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

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