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

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

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

extreme points

f (x, y) = \frac{1}{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 y ^{2} - x ^{2} + 1}{( 1 - x ^{2} - y ^{2} ) ^{\frac{5}{2}}}

D (x, y) = \frac{- 2 x ^{4} - 4 x ^{2} y ^{2} + x ^{2} - 2 y ^{4} + y ^{2} + 1}{( - x ^{2} - y ^{2} + 1 ) ^{5}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

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

a^{3} y = x^{3} (4 a - 3 x)

f (x) = x y

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

f (x) = e^{x^{2} - 8 x - 1}

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