What is the extreme f(x,y)= 7/(-91x^2-3y^2+9) ?

The extreme f(x,y)= 7/(-91x^2-3y^2+9) is Minimum(0,0)

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extreme f(x,y)= 7/(-91x^2-3y^2+9)

extreme points

f (x, y) = \frac{7}{- 91 x ^{2} - 3 y ^{2} + 9}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{42 ( 9 - 91 x ^{2} + 9 y ^{2} )}{( - 91 x ^{2} - 3 y ^{2} + 9 ) ^{3}}

D (x, y) = \frac{- 1329299244 x ^{4} - 87646104 x ^{2} y ^{2} + 87646104 x ^{2} - 1444716 y ^{4} + 2889432 y ^{2} + 4334148}{( 9 - 91 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, y) = (2 x + 3) (y^{- 1} + 2)

f (x, y) = 11 x y + 5 x^{2}

ln (x^{2} - 36)

\frac{x + 2}{x - 2}

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

What is the extreme f(x,y)= 7/(-91x^2-3y^2+9) ?
The extreme f(x,y)= 7/(-91x^2-3y^2+9) is Minimum(0,0)