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

The extreme f(x,y)=(sqrt(x-x^2))/(1-y^2) is Saddle(1/2 ,0)

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

extreme points

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

Saddle (\frac{1}{2}, 0)

Solution steps

Find the critical points:

(\frac{1}{2}, 0)

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

D (x, y) = - \frac{- 8 x ^{2} y ^{2} + 8 x y ^{2} - 5 y ^{2} - 1}{2 x ( y + 1 ) ^{4} ( y - 1 ) ^{4} ( x - 1 )}

Check the sign of at each critical point

Check critical point

(\frac{1}{2}, 0) :

Saddle

Saddle (\frac{1}{2}, 0)

f (x) = x^{2} + x y + \frac{1}{2} y^{2} - 4 x + y

f (x) = \frac{1}{3} x^{3} - 2 x^{2} - 5 x + 2

x y e^{x + 2 y}

1 - 2 x^{2}

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

What is the extreme f(x,y)=(sqrt(x-x^2))/(1-y^2) ?
The extreme f(x,y)=(sqrt(x-x^2))/(1-y^2) is Saddle(1/2 ,0)