What is the extreme f(x,y)=x^3+y^2-2xy-x+3 ?

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

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extreme f(x,y)=x^3+y^2-2xy-x+3

extreme points

f (x, y) = x^{3} + y^{2} - 2 x y - x + 3

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = 2

D (x, y) = 12 x - 4

Check the sign of at each critical point

Check critical point

(1, 1) :

Minimum

Check critical point

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

Saddle

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

f (x, y) = 2 x y - 3 y - 5

f (x) = (x - y) (36 - x y)

f (x) = 5 x - 60 x^{\frac{1}{3}}

f (x) = \frac{1}{6} (x^{3} - 6 x^{2} + 9 x + 6)

f (x) = x^{3} + 3 x^{2} - 24 x + 12

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