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

The extreme f(x,y)=(4-xy)/(2+x^2y^2) is Saddle(0,0)

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{2 x ^{2} ( - x ^{3} y ^{3} + 12 x ^{2} y ^{2} + 6 x y - 8 )}{( 2 + x ^{2} y ^{2} ) ^{3}}

D (x, y) = \frac{3 x ^{8} y ^{8} - 64 x ^{7} y ^{7} + 296 x ^{6} y ^{6} + 192 x ^{5} y ^{5} + 248 x ^{4} y ^{4} + 512 x ^{3} y ^{3} - 672 x ^{2} y ^{2} - 256 x y - 16}{( 2 + x ^{2} y ^{2} ) ^{6}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = 6 + 5 x - 5 x^{2}

f (x, y, z) = ln (4 - x - y)

\frac{0.5 x ^{2} + 4 x - 10}{x - 6}

f (x) = 2 x^{2} - 8 x - 7

f (x) = 3 cos (x) + 3 sin (x)

What is the extreme f(x,y)=(4-xy)/(2+x^2y^2) ?
The extreme f(x,y)=(4-xy)/(2+x^2y^2) is Saddle(0,0)