What is the extreme f(x)=4y^3+x^2-12y-36y+2 ?

The extreme f(x)=4y^3+x^2-12y-36y+2 is Minimum(0,2),Saddle(0,-2)

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extreme f(x)=4y^3+x^2-12y-36y+2

extreme

Minimum (0, 2), Saddle (0, - 2)

Solution steps

Find the critical points:

(0, 2), (0, - 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 24 y

D (x, y) = 48 y

Check the sign of at each critical point

Check critical point

(0, 2) :

Minimum

Check critical point

(0, - 2) :

Saddle

Minimum (0, 2), Saddle (0, - 2)

What is the extreme f(x)=4y^3+x^2-12y-36y+2 ?
The extreme f(x)=4y^3+x^2-12y-36y+2 is Minimum(0,2),Saddle(0,-2)