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

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

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

extreme points

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

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = sin (x) - cos (x)

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

f (x) = x^{2} + y^{2} + 5

x^{3} - 3 x^{2} + 1

f (x) = \frac{8}{135} x^{3} - \frac{1}{20} x^{2} - \frac{7}{16} x

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