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

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

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

extreme points

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

Saddle (0, 1), Maximum (0, - 2), Minimum (3, 1), Saddle (3, - 2)

Solution steps

Find the critical points:

(0, 1), (0, - 2), (3, 1), (3, - 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 12 y + 6

D (x, y) = (12 y + 6) (12 x - 18)

Check the sign of at each critical point

Check critical point

(0, 1) :

Saddle

Check critical point

(0, - 2) :

Maximum

Check critical point

(3, 1) :

Minimum

Check critical point

(3, - 2) :

Saddle

Saddle (0, 1), Maximum (0, - 2), Minimum (3, 1), Saddle (3, - 2)

f (x) = x^{5} - 15 x^{3}

f (x, y) = x y - \frac{1}{x} - \frac{1}{y}

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

f (x) = e^{x} (15 - x^{2})

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

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