What is the extreme f(x)=3x^2+12xy+9x^2+y^3 ?

The extreme f(x)=3x^2+12xy+9x^2+y^3 is Minimum(-1,2),Saddle(0,0)

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extreme f(x)=3x^2+12xy+9x^2+y^3

extreme points

f (x) = 3 x^{2} + 12 x y + 9 x^{2} + y^{3}

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

Solution steps

Find the critical points:

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

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

D (x, y) = 144 y - 144

Check the sign of at each critical point

Check critical point

(- 1, 2) :

Minimum

Check critical point

(0, 0) :

Saddle

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

V (u, w) = \frac{1}{2} (3 u + w)

f (x) = (4 - x^{2})^{2}

3 x + 2 y

7 t e^{- \frac{t}{15}}

f (x) = (2 x - 2 y - 10)

What is the extreme f(x)=3x^2+12xy+9x^2+y^3 ?
The extreme f(x)=3x^2+12xy+9x^2+y^3 is Minimum(-1,2),Saddle(0,0)