What is the extreme f(x)=3x^2-4y^3-6x+48y-1 ?

The extreme f(x)=3x^2-4y^3-6x+48y-1 is Saddle(1,2),Minimum(1,-2)

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extreme f(x)=3x^2-4y^3-6x+48y-1

extreme

Saddle (1, 2), Minimum (1, - 2)

Solution steps

Find the critical points:

(1, 2), (1, - 2)

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

D (x, y) = - 144 y

Check the sign of at each critical point

Check critical point

(1, 2) :

Saddle

Check critical point

(1, - 2) :

Minimum

Saddle (1, 2), Minimum (1, - 2)

What is the extreme f(x)=3x^2-4y^3-6x+48y-1 ?
The extreme f(x)=3x^2-4y^3-6x+48y-1 is Saddle(1,2),Minimum(1,-2)