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

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

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{3 ( ( 1 - 2 x ^{2} - 3 y ^{2} ) ^{\frac{3}{2}} + 9 y ^{2} - 3 y ^{2} + 1 - 2 x ^{2} )}{( 1 - 2 x ^{2} - 3 y ^{2} ) ^{3}}

D (x, y) = \frac{- 48 x ^{6} - 216 x ^{4} y ^{2} + 72 x ^{4} - 324 x ^{2} y ^{4} + 216 x ^{2} y ^{2} - 36 x ^{2} + 36 x ^{2} ( - 2 x ^{2} - 3 y ^{2} + 1 ) ^{\frac{3}{2}} - 2 x ^{2} - 3 y ^{2} + 1 + 54 y ^{2} ( - 2 x ^{2} - 3 y ^{2} + 1 ) ^{\frac{3}{2}} - 2 x ^{2} - 3 y ^{2} + 1 - 162 y ^{6} + 162 y ^{4} - 54 y ^{2} + 6}{( 1 - 2 x ^{2} - 3 y ^{2} ) ^{6}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = 6 x^{\frac{4}{3}} - 3 x^{\frac{1}{3}}

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

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

f (x, y) = 4 - 2 x - 3 y

f (x) = 7 - x

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