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

The extreme f(x,y)=x^3-3x+y^2 is Saddle(-1,0),Minimum(1,0)

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

extreme points

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

Saddle (- 1, 0), Minimum (1, 0)

Solution steps

Find the critical points:

(- 1, 0), (1, 0)

\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, 0) :

Saddle

Check critical point

(1, 0) :

Minimum

Saddle (- 1, 0), Minimum (1, 0)

f (x, y) = 2 x^{2} + y^{2} + 3 x y - 3 y - 5 x + 8

4 x^{3} - 3 x^{2} - 6 x + 3

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

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

f (x) = x^{4} - 6 x^{3} + 12 x^{2} - 8 x + 1

What is the extreme f(x,y)=x^3-3x+y^2 ?
The extreme f(x,y)=x^3-3x+y^2 is Saddle(-1,0),Minimum(1,0)