What is the extreme f(x)=x^3+y^2-3x^2-3y^2-9x ?

The extreme f(x)=x^3+y^2-3x^2-3y^2-9x is Maximum(-1,0),Saddle(3,0)

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extreme f(x)=x^3+y^2-3x^2-3y^2-9x

extreme

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = - 4

D (x, y) = - 4 (6 x - 6)

Check the sign of at each critical point

Check critical point

(- 1, 0) :

Maximum

Check critical point

(3, 0) :

Saddle

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

What is the extreme f(x)=x^3+y^2-3x^2-3y^2-9x ?
The extreme f(x)=x^3+y^2-3x^2-3y^2-9x is Maximum(-1,0),Saddle(3,0)