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

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

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

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

f (x) = \frac{x}{x ^{2} - 9}

f (x) = x ln (x)

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

f (t) = e^{a t}

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