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

The extreme f(x,y)=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(x,y)=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} + 3 x^{2} + 45 x - 7

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

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

f (x) = 2 y - 2 - x

f (x) = 9 - 8 x - x^{3}

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