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

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

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

extreme points

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

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - \frac{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) :

Maximum

Maximum (0, 0)

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

88 - \frac{1}{2} x - \frac{3200}{x}

f (x) = 2 x - 4 sin (x)

f (m, a) = m \cdot a

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