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

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

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

extreme points

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

D (x, y) = \frac{16 ln ^{4} ( 3 ) \cdot 3 ^{- 4 x^{2} - 4 y^{2}} x ^{2} y ^{2} + 16 ln ^{4} ( 3 ) \cdot 3 ^{- 2 x^{2} - 2 y^{2}} x ^{2} y ^{2} - 16 ln ^{4} ( 3 ) \cdot 9 ^{- 2 x^{2} - 2 y^{2}} x ^{2} y ^{2} - 16 ln ^{4} ( 3 ) \cdot 9 ^{- x^{2} - y^{2}} x ^{2} y ^{2} + 8 ln ^{3} ( 3 ) \cdot 3 ^{- 4 x^{2} - 4 y^{2}} x ^{2} - 8 ln ^{3} ( 3 ) \cdot 3 ^{- 2 x^{2} - 2 y^{2}} x ^{2} - 8 ln ^{3} ( 3 ) \cdot 3 ^{- 2 x^{2} - 2 y^{2}} y ^{2} + 8 ln ^{3} ( 3 ) \cdot 3 ^{- 4 x^{2} - 4 y^{2}} y ^{2} + 4 ln ^{2} ( 3 ) \cdot 3 ^{- 2 x^{2} - 2 y^{2}} + 8 ln ^{2} ( 3 ) \cdot 3 ^{- 3 x^{2} - 3 y^{2}} + 4 ln ^{2} ( 3 ) \cdot 3 ^{- 4 x^{2} - 4 y^{2}}}{( 3 ^{- 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) = - sin^{2} (x), - π \leq x \leq π

\frac{- 6}{x - 7}

f (x) = x (22 - 40 + 2 x) (\frac{40}{2} - x)

f (x, y) = (y - 8) \cdot x^{2} - y^{2}

y = 3 x^{3} - 36 x^{2} + 7

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