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

The extreme f(x)=x^3+3x^2+y^2-2y+3 is Minimum(0,1),Saddle(-2,1)

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

extreme points

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

Minimum (0, 1), Saddle (- 2, 1)

Solution steps

Find the critical points:

(0, 1), (- 2, 1)

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

D (x, y) = 2 (6 x + 6)

Check the sign of at each critical point

Check critical point

(0, 1) :

Minimum

Check critical point

(- 2, 1) :

Saddle

Minimum (0, 1), Saddle (- 2, 1)

f (x) = - x^{3} + 8 x^{2} - 15 x

y = 2 x^{3} - 15 x^{2} + 24 x - 5

f (x) = - 7 x^{2} + 42 x + 5

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

f (x, y) = 42 x^{4} + 42 y^{4} - 4 x y + 19

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