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

The extreme f(x,y)=ln(1+x^2+y^2) is Minimum(0,0)

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extreme f(x,y)=ln(1+x^2+y^2)

extreme points

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

Minimum (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) :

Minimum

Minimum (0, 0)

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

f (x) = \frac{x ^{5}}{5} - \frac{8}{3} x^{3}

f (x) = 4 x^{2} - 5 x^{3}

x^{3} - 6 x^{2} + 15

\frac{4 x}{25 - x ^{2}}

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