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

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

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

extreme points

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

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 2

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x) = \frac{1}{3 x ^{3} - 9 x + 2}

y = \frac{( \frac{1}{x - 1} - \frac{2}{x} + z )}{2 + \frac{1}{x}}

f (x) = x^{4} - 8 x^{2} + 7

f (x) = - 2 x^{3} + 45 x^{2} - 300 x

f (x) = - x^{2} - 4 x + 5

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