What is the extreme f(x)=4x^3-6x^2y+12x^2-12y^2 ?

The extreme f(x)=4x^3-6x^2y+12x^2-12y^2 is Saddle(0,0)

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extreme f(x)=4x^3-6x^2y+12x^2-12y^2

extreme points

f (x) = 4 x^{3} - 6 x^{2} y + 12 x^{2} - 12 y^{2}

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - 24 (24 x + 24 - 12 y) - 144 x^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = 7 + 54 x - 2 x^{3}

f (x) = (2 + x)^{2} (1 - x)^{2}

y = 6 x - ln (6 x)

f (x) = 7 + 81 x - 3 x^{3}

f (x, y) = 7 x - 8 y + 2 x y - x^{2} + y^{1}

What is the extreme f(x)=4x^3-6x^2y+12x^2-12y^2 ?
The extreme f(x)=4x^3-6x^2y+12x^2-12y^2 is Saddle(0,0)