What is the extreme f(x,y)=ln(xy-1) ?

The extreme f(x,y)=ln(xy-1) is Saddle(0,0)

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extreme f(x,y)=ln(xy-1)

extreme points

f (x, y) = ln (x y - 1)

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = \frac{x y + 1}{( - 1 + x y ) ^{3}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = 3 x^{3} - 6 x^{2} + 1

f (x) = x^{4} - x^{3} - 6 x^{2}

f (x) = \frac{1}{x ^{2} + 2 x + 9}

f (x) = (x - 3)^{4}

f (x, y) = 1 - x - y

What is the extreme f(x,y)=ln(xy-1) ?
The extreme f(x,y)=ln(xy-1) is Saddle(0,0)