What is the extreme f(x,y)=x^2+y^2+ln(xy-2) ?

The extreme f(x,y)=x^2+y^2+ln(xy-2) is Saddle(sqrt(3/2),sqrt(3/2)),Saddle(-sqrt(3/2),-sqrt(3/2))

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extreme f(x,y)=x^2+y^2+ln(xy-2)

extreme points

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

Saddle (\frac{3}{2}, \frac{3}{2}), Saddle (- \frac{3}{2}, - \frac{3}{2})

Solution steps

Find the critical points:

(\frac{3}{2}, \frac{3}{2}), (- \frac{3}{2}, - \frac{3}{2})

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

D (x, y) = \frac{- 2 x ^{2} - 2 y ^{2}}{( - 2 + x y ) ^{2}} + 4 + \frac{x y + 2}{( - 2 + x y ) ^{3}}

Check the sign of at each critical point

Check critical point

(\frac{3}{2}, \frac{3}{2}) :

Saddle

Check critical point

(- \frac{3}{2}, - \frac{3}{2}) :

Saddle

Saddle (\frac{3}{2}, \frac{3}{2}), Saddle (- \frac{3}{2}, - \frac{3}{2})

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

f (x) = - x^{4} + 50 x^{2} - 625, - 6 \leq x \leq 6

f (x) = \frac{x ^{2}}{x + 2}

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

f (x) = x^{2} - 12 x - 13

What is the extreme f(x,y)=x^2+y^2+ln(xy-2) ?
The extreme f(x,y)=x^2+y^2+ln(xy-2) is Saddle(sqrt(3/2),sqrt(3/2)),Saddle(-sqrt(3/2),-sqrt(3/2))