What is the extreme f(x,y)=(2y+3)/(xy+1) ?

The extreme f(x,y)=(2y+3)/(xy+1) is Saddle(2/3 ,0)

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extreme f(x,y)=(2y+3)/(xy+1)

extreme points

f (x, y) = \frac{2 y + 3}{x y + 1}

Saddle (\frac{2}{3}, 0)

Solution steps

Find the critical points:

(\frac{2}{3}, 0)

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

D (x, y) = \frac{24 x ^{2} y ^{3} - 16 x y ^{3} + 27 x ^{2} y ^{2} - 16 y ^{2} + 18 x y - 24 y - 9}{( 1 + x y ) ^{6}}

Check the sign of at each critical point

Check critical point

(\frac{2}{3}, 0) :

Saddle

Saddle (\frac{2}{3}, 0)

f (x) = 4 x^{3} - 5 x^{2} + 3

P (x, y) = x^{2} + 6 x + 9 - 4 y^{2}

f (x) = x^{3} + 3 x^{2} - 9 x + 7

f (x) = 2 x^{3} + 9 x^{2} - 24 x + 5

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

What is the extreme f(x,y)=(2y+3)/(xy+1) ?
The extreme f(x,y)=(2y+3)/(xy+1) is Saddle(2/3 ,0)