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})

f (x) = e^{x^{2} - 4 x - 1}

y = e^{x} + I n x - x^{2} + \frac{3}{x ^{5}} + 3

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

- 3 x^{5} + 5 x^{3}

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

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)