What is the extreme x^4+2x^2+y^4-2y^2+3 ?

The extreme x^4+2x^2+y^4-2y^2+3 is Saddle(0,0),Minimum(0,-1),Minimum(0,1)

extreme x^4+2x^2+y^4-2y^2+3

extreme points

x^{4} + 2 x^{2} + y^{4} - 2 y^{2} + 3

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = 12 y^{2} - 4

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(0, - 1) :

Minimum

Check critical point

(0, 1) :

Minimum

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

f (x) = 9 \cdot x^{3} - 7 \cdot x^{2} + 3 \cdot x + 10

f (x) = x^{2} - 6, - 2 \leq x \leq 4

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

6 x^{2} - 48 x - 190

f (x) = \frac{e ^{x}}{( 4 + e ^{x} )}

What is the extreme x^4+2x^2+y^4-2y^2+3 ?
The extreme x^4+2x^2+y^4-2y^2+3 is Saddle(0,0),Minimum(0,-1),Minimum(0,1)