What is the extreme f(x,y)=sqrt(1-x^2)-sqrt(4-y^2) ?

The extreme f(x,y)=sqrt(1-x^2)-sqrt(4-y^2) is Saddle(0,0)

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extreme f(x,y)=sqrt(1-x^2)-sqrt(4-y^2)

extreme points

f (x, y) = 1 - x^{2} - 4 - y^{2}

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = x - 2 cos (x), 0 \leq x \leq 2 π

f (x) = x^{4} - 20 x^{3} + 28 x^{2} - 4

1448.2 x^{2} - 81667 x + 1000000

f (x, y) = (x - 8)^{2} + (y - 5)^{2} + 16

f (x) = x e^{3 x^{2}}

What is the extreme f(x,y)=sqrt(1-x^2)-sqrt(4-y^2) ?
The extreme f(x,y)=sqrt(1-x^2)-sqrt(4-y^2) is Saddle(0,0)