What is the extreme f(x)=3+8x^3-y^3-12xy ?

The extreme f(x)=3+8x^3-y^3-12xy is Saddle(0,0),Maximum(-1,2)

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extreme f(x)=3+8x^3-y^3-12xy

extreme points

f (x) = 3 + 8 x^{3} - y^{3} - 12 x y

Saddle (0, 0), Maximum (- 1, 2)

Solution steps

Find the critical points:

(0, 0), (- 1, 2)

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

D (x, y) = - 288 x y - 144

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 1, 2) :

Maximum

Saddle (0, 0), Maximum (- 1, 2)

y (x, t) = x (3 t - 9)

f (x) = x^{4} + 8 x^{3} - 2

f (x) = 2 x^{2} + 3 x y + 4 y^{2} + 5 x - 2 y

- \frac{1}{x}

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

What is the extreme f(x)=3+8x^3-y^3-12xy ?
The extreme f(x)=3+8x^3-y^3-12xy is Saddle(0,0),Maximum(-1,2)