What is the extreme f(x)=x^3+y^2+2xy-4x-3y+5 ?

The extreme f(x)=x^3+y^2+2xy-4x-3y+5 is Minimum(1, 1/2),Saddle(-1/3 , 11/6)

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extreme f(x)=x^3+y^2+2xy-4x-3y+5

extreme points

f (x) = x^{3} + y^{2} + 2 x y - 4 x - 3 y + 5

Minimum (1, \frac{1}{2}), Saddle (- \frac{1}{3}, \frac{11}{6})

Solution steps

Find the critical points:

(1, \frac{1}{2}), (- \frac{1}{3}, \frac{11}{6})

\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, \frac{1}{2}) :

Minimum

Check critical point

(- \frac{1}{3}, \frac{11}{6}) :

Saddle

Minimum (1, \frac{1}{2}), Saddle (- \frac{1}{3}, \frac{11}{6})

What is the extreme f(x)=x^3+y^2+2xy-4x-3y+5 ?
The extreme f(x)=x^3+y^2+2xy-4x-3y+5 is Minimum(1, 1/2),Saddle(-1/3 , 11/6)