What is the extreme f(x)=ln(x^2+y^2-4) ?

The extreme f(x)=ln(x^2+y^2-4) is Maximum(0,0)

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extreme f(x)=ln(x^2+y^2-4)

extreme points

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

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - \frac{4 ( x ^{2} + y ^{2} + 4 )}{( x ^{2} + y ^{2} - 4 ) ^{3}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x, y) = x^{3} - y^{3} + 43 x y - 7

f (x, y) = 2^{2} \cdot x^{4} + 2^{2} \cdot y^{4} - 4 \cdot x \cdot y + 13

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

f (x) = \frac{( ln ( x ) ) ^{2}}{x}

f (x) = \frac{x ^{4}}{4} - x^{3}

What is the extreme f(x)=ln(x^2+y^2-4) ?
The extreme f(x)=ln(x^2+y^2-4) is Maximum(0,0)