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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = - 108 x y

Check the sign of at each critical point

Check critical point

(- 1, 2) :

Maximum

Check critical point

(- 1, - 2) :

Saddle

Check critical point

(1, 2) :

Saddle

Check critical point

(1, - 2) :

Minimum

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

f (x) = \frac{x}{5} + \frac{4 + ( 6 - x ) ^{2}}{3}

f (x) = - x^{3} + 3 x^{2} + 9 x - 1

f (t) = 120 - 120 h (t - 2)

f (x) = x^{6} (1 - x)^{18}

f (x) = x^{3} - x^{2} - x + 4, - 1 \leq x \leq 2

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