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

The extreme f(x,y)=ln(x-y)+x^2+y is Saddle(-1/2 ,-3/2)

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

extreme points

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

Saddle (- \frac{1}{2}, - \frac{3}{2})

Solution steps

Find the critical points:

(- \frac{1}{2}, - \frac{3}{2})

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

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

Check the sign of at each critical point

Check critical point

(- \frac{1}{2}, - \frac{3}{2}) :

Saddle

Saddle (- \frac{1}{2}, - \frac{3}{2})

What is the extreme f(x,y)=ln(x-y)+x^2+y ?
The extreme f(x,y)=ln(x-y)+x^2+y is Saddle(-1/2 ,-3/2)