What is the extreme f(x,y)=x^3-5x^2+3y^2+3x+1 ?

The extreme f(x,y)=x^3-5x^2+3y^2+3x+1 is Saddle(1/3 ,0),Minimum(3,0)

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extreme f(x,y)=x^3-5x^2+3y^2+3x+1

extreme points

f (x, y) = x^{3} - 5 x^{2} + 3 y^{2} + 3 x + 1

Saddle (\frac{1}{3}, 0), Minimum (3, 0)

Solution steps

Find the critical points:

(\frac{1}{3}, 0), (3, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6

D (x, y) = 6 (6 x - 10)

Check the sign of at each critical point

Check critical point

(\frac{1}{3}, 0) :

Saddle

Check critical point

(3, 0) :

Minimum

Saddle (\frac{1}{3}, 0), Minimum (3, 0)

What is the extreme f(x,y)=x^3-5x^2+3y^2+3x+1 ?
The extreme f(x,y)=x^3-5x^2+3y^2+3x+1 is Saddle(1/3 ,0),Minimum(3,0)