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

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

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

extreme points

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

Saddle (- 1, 1), Minimum (1, 1)

Solution steps

Find the critical points:

(- 1, 1), (1, 1)

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

D (x, y) = 12 x

Check the sign of at each critical point

Check critical point

(- 1, 1) :

Saddle

Check critical point

(1, 1) :

Minimum

Saddle (- 1, 1), Minimum (1, 1)

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

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

f (x) = x^{\frac{2}{3}} (x + 2)

y

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

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