What is the extreme f(x,y)=(x^2+y)/(xy-1) ?

The extreme f(x,y)=(x^2+y)/(xy-1) is Saddle(-1,2)

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extreme f(x,y)=(x^2+y)/(xy-1)

extreme points

f (x, y) = \frac{x ^{2} + y}{x y - 1}

Saddle (- 1, 2)

Solution steps

Find the critical points:

(- 1, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{2 x ( - x ^{3} - 1 )}{( x y - 1 ) ^{3}}

D (x, y) = \frac{- x ^{6} y ^{2} + 6 x ^{5} y + 4 x ^{4} y ^{3} - 5 x ^{4} + 4 x ^{3} y ^{2} - 12 x ^{2} y + 4 x y ^{3} + 4 x - 4 y ^{2}}{( - 1 + x y ) ^{6}}

Check the sign of at each critical point

Check critical point

(- 1, 2) :

Saddle

Saddle (- 1, 2)

f (x, y) = x^{2} + x y + y^{2} - 3 x - 6 y + 1

\frac{x - 5}{x ^{2}}

x^{4} - 18 x^{2}

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

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

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