What is the extreme f(xy)=3x^4+8x^3-18x^2+6y^2+12y-4 ?

The extreme f(xy)=3x^4+8x^3-18x^2+6y^2+12y-4 is Saddle(0,-1),Minimum(1,-1),Minimum(-3,-1)

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extreme f(xy)=3x^4+8x^3-18x^2+6y^2+12y-4

extreme points

f (x y) = 3 x^{4} + 8 x^{3} - 18 x^{2} + 6 y^{2} + 12 y - 4

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = 12

D (x, y) = 12 (36 x^{2} + 48 x - 36)

Check the sign of at each critical point

Check critical point

(0, - 1) :

Saddle

Check critical point

(1, - 1) :

Minimum

Check critical point

(- 3, - 1) :

Minimum

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

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

e^{x - y} (y^{2} - 2 x^{2})

f (x, y) = \frac{y}{x ^{2} + y ^{2}}

\frac{d}{d x} (x y^{2})

f (x) = x^{3} - x^{2} - x - 10

What is the extreme f(xy)=3x^4+8x^3-18x^2+6y^2+12y-4 ?
The extreme f(xy)=3x^4+8x^3-18x^2+6y^2+12y-4 is Saddle(0,-1),Minimum(1,-1),Minimum(-3,-1)