What is the extreme f(x,y)=x^3+y^3-3x-12y+13e ?

The extreme f(x,y)=x^3+y^3-3x-12y+13e is Saddle(-1,2),Maximum(-1,-2),Minimum(1,2),Saddle(1,-2)

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extreme f(x,y)=x^3+y^3-3x-12y+13e

extreme points

f (x, y) = x^{3} + y^{3} - 3 x - 12 y + 13 e

Saddle (- 1, 2), Maximum (- 1, - 2), Minimum (1, 2), Saddle (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) = 36 x y

Check the sign of at each critical point

Check critical point

(- 1, 2) :

Saddle

Check critical point

(- 1, - 2) :

Maximum

Check critical point

(1, 2) :

Minimum

Check critical point

(1, - 2) :

Saddle

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

f (x) = - \frac{2}{3} x^{3} - \frac{11}{2} x^{2} - 5 x + 9

f (x) = x^{3} - 9 x^{2} + 27 x - 26

f (x) = \frac{x ^{2} - 10 x + 25}{x - 8}

f (x) = \frac{- 2}{3} x^{3} + 10 x^{2} - 48 x - 1

x^{\frac{2}{3}} (x^{3} - 44)

What is the extreme f(x,y)=x^3+y^3-3x-12y+13e ?
The extreme f(x,y)=x^3+y^3-3x-12y+13e is Saddle(-1,2),Maximum(-1,-2),Minimum(1,2),Saddle(1,-2)