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

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

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

extreme

Saddle (0, 3), Saddle (0, - 3), Minimum (1, 0), Maximum (- 1, 0)

Solution steps

Find the critical points:

(0, 3), (0, - 3), (1, 0), (- 1, 0)

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

D (x, y) = 12 x^{2} - 4 y^{2}

Check the sign of at each critical point

Check critical point

(0, 3) :

Saddle

Check critical point

(0, - 3) :

Saddle

Check critical point

(1, 0) :

Minimum

Check critical point

(- 1, 0) :

Maximum

Saddle (0, 3), Saddle (0, - 3), Minimum (1, 0), Maximum (- 1, 0)

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