What is the extreme f(x,y)=9x^2-2x^3+3y^2+6xy ?

The extreme f(x,y)=9x^2-2x^3+3y^2+6xy is Minimum(0,0),Saddle(2,-2)

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extreme f(x,y)=9x^2-2x^3+3y^2+6xy

extreme points

f (x, y) = 9 x^{2} - 2 x^{3} + 3 y^{2} + 6 x y

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

Solution steps

Find the critical points:

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

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

D (x, y) = - 72 x + 72

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(2, - 2) :

Saddle

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

f (x) = 395 x^{2} - 2370 x^{3}, 0 \leq x \leq 0.16

C (t) = \frac{t}{4 t ^{2} + 11}

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

f (x) = \frac{2 x}{x ^{2} + 4}, - 4 \leq x \leq 4

\frac{5 x}{x ^{2} - 9}

What is the extreme f(x,y)=9x^2-2x^3+3y^2+6xy ?
The extreme f(x,y)=9x^2-2x^3+3y^2+6xy is Minimum(0,0),Saddle(2,-2)