What is the extreme f(x,y)=(3x^2-1)/(x^2+y^2+1) ?

The extreme f(x,y)=(3x^2-1)/(x^2+y^2+1) is Minimum(0,0)

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extreme f(x,y)=(3x^2-1)/(x^2+y^2+1)

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = \frac{- 36 x ^{6} y ^{2} + 144 x ^{6} - 72 x ^{4} y ^{4} + 36 x ^{4} y ^{2} + 48 x ^{4} - 36 x ^{2} y ^{6} - 144 x ^{2} y ^{4} - 204 x ^{2} y ^{2} - 80 x ^{2} - 36 y ^{6} - 72 y ^{4} - 20 y ^{2} + 16}{( x ^{2} + y ^{2} + 1 ) ^{6}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = 3 x + \frac{27}{x}

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

f (x) = \frac{2}{3} x^{3} - 4 x^{2} - 45 x

f (x, y) = ln (x + y + 4)

2 x^{3} - 3 x^{2} - 12 x + 5

What is the extreme f(x,y)=(3x^2-1)/(x^2+y^2+1) ?
The extreme f(x,y)=(3x^2-1)/(x^2+y^2+1) is Minimum(0,0)