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

The extreme f(x)=x^2-y^2sqrt(1-x^2-y^2) is Saddle(0,0),Minimum(0,-sqrt(2/3)),Minimum(0,sqrt(2/3))

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

extreme points

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

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = - \frac{6 y ^{4} + 9 x ^{2} y ^{2} - 9 y ^{2} + 2 x ^{4} - 4 x ^{2} + 2}{( 1 - x ^{2} - y ^{2} ) 1 - x ^{2} - y ^{2}}

D (x, y) = \frac{6 y ^{8} + 8 x ^{2} y ^{6} + 12 y ^{6} - x ^{2} - y ^{2} + 1 - 15 y ^{6} - 2 x ^{4} y ^{4} + 30 x ^{2} y ^{4} - x ^{2} - y ^{2} + 1 - 9 x ^{2} y ^{4} - 30 y ^{4} - x ^{2} - y ^{2} + 1 + 11 y ^{4} - 4 x ^{6} y ^{2} + 22 x ^{4} y ^{2} - x ^{2} - y ^{2} + 1 + 6 x ^{4} y ^{2} - 44 x ^{2} y ^{2} - x ^{2} - y ^{2} + 1 + 22 y ^{2} - x ^{2} - y ^{2} + 1 - 2 y ^{2} + 4 x ^{6} - x ^{2} - y ^{2} + 1 - 12 x ^{4} - x ^{2} - y ^{2} + 1 + 12 x ^{2} - x ^{2} - y ^{2} + 1 - 4 - x ^{2} - y ^{2} + 1}{( - x ^{2} - y ^{2} + 1 ) ^{3}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

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

Minimum

Check critical point

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

Minimum

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

What is the extreme f(x)=x^2-y^2sqrt(1-x^2-y^2) ?
The extreme f(x)=x^2-y^2sqrt(1-x^2-y^2) is Saddle(0,0),Minimum(0,-sqrt(2/3)),Minimum(0,sqrt(2/3))