What is the extreme f(x)=x^3+3xy^2-15x-12y ?

The extreme f(x)=x^3+3xy^2-15x-12y is Saddle(1,2),Saddle(-1,-2),Minimum(2,1),Maximum(-2,-1)

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extreme f(x)=x^3+3xy^2-15x-12y

extreme points

f (x) = x^{3} + 3 x y^{2} - 15 x - 12 y

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

Solution steps

Find the critical points:

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

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

D (x, y) = 36 x^{2} - 36 y^{2}

Check the sign of at each critical point

Check critical point

(1, 2) :

Saddle

Check critical point

(- 1, - 2) :

Saddle

Check critical point

(2, 1) :

Minimum

Check critical point

(- 2, - 1) :

Maximum

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

x^{3} - 6 x^{2} + 9 x + 1

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

x^{4} + y^{4} - 4 x y + 1

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

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

What is the extreme f(x)=x^3+3xy^2-15x-12y ?
The extreme f(x)=x^3+3xy^2-15x-12y is Saddle(1,2),Saddle(-1,-2),Minimum(2,1),Maximum(-2,-1)