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) = 3 x + y

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

f (x, y) = 2 x^{3} - 3 x^{2} y - 12 x^{2} - 3 y^{2}

f (x) = tan (x)

f (x, y) = x^{2} - y^{2} - 2 x + 4 y + 6

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)