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

The extreme f(x)=3x^2+12xy+9y^2+y^3 is Minimum(-4,2),Saddle(0,0)

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

extreme points

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

Minimum (- 4, 2), Saddle (0, 0)

Solution steps

Find the critical points:

(- 4, 2), (0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y + 18

D (x, y) = 36 y - 36

Check the sign of at each critical point

Check critical point

(- 4, 2) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (- 4, 2), Saddle (0, 0)

f (x) = 6 x^{2} + 2, 1 \leq x \leq 2

f (x) = x^{4} + 2 x^{2} + 1

f (y) = 20.85 x y + 10.7 x^{2} + 51.75 y^{2}

3 x^{2} + 6 x - 6

x^{3} - 12 + 3

What is the extreme f(x)=3x^2+12xy+9y^2+y^3 ?
The extreme f(x)=3x^2+12xy+9y^2+y^3 is Minimum(-4,2),Saddle(0,0)