What is the extreme 3x^2-12xy+y^3+3y^2 ?

The extreme 3x^2-12xy+y^3+3y^2 is Minimum(12,6),Saddle(0,0)

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extreme 3x^2-12xy+y^3+3y^2

extreme points

3 x^{2} - 12 x y + y^{3} + 3 y^{2}

Minimum (12, 6), Saddle (0, 0)

Solution steps

Find the critical points:

(12, 6), (0, 0)

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

D (x, y) = 36 y - 108

Check the sign of at each critical point

Check critical point

(12, 6) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (12, 6), Saddle (0, 0)

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

1 + \frac{7}{x} - \frac{3}{x ^{2}}

f (x) = \frac{x}{x ^{2} + 1}, - 2 \leq x \leq 2

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

f (x) = 5 + 54 x + 2 x^{3}, 0 \leq x \leq 4

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