What is the extreme f(x,y)=x^4-2x^2+y^2-4y ?

The extreme f(x,y)=x^4-2x^2+y^2-4y is Saddle(0,2),Minimum(-1,2),Minimum(1,2)

extreme f(x,y)=x^4-2x^2+y^2-4y

extreme points

f (x, y) = x^{4} - 2 x^{2} + y^{2} - 4 y

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

Solution steps

Find the critical points:

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

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

D (x, y) = 2 (12 x^{2} - 4)

Check the sign of at each critical point

Check critical point

(0, 2) :

Saddle

Check critical point

(- 1, 2) :

Minimum

Check critical point

(1, 2) :

Minimum

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

f (x) = 2 cos (x) + sin^{2} (x)

f (x, y) = x^{4} + y^{4} - 4 x y + 2

f (x) = P (x) = x^{3} + x^{2} - 5 x + 3

2 x^{3} - 3 x^{2} - 12 x + 12

f (x) = x^{3} - 3 x + 1, 0 \leq x \leq 5

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