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

The extreme 2x^2-2y^2-x^4-2x^2y^2-y^4 is

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extreme 2x^2-2y^2-x^4-2x^2y^2-y^4

extreme points

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

Saddle (0, 0), Maximum (- 1, 0), Maximum (1, 0), Saddle (0, 0), Maximum (- 1, 0), Maximum (1, 0), Saddle (0, 0), Maximum (- 1, 0), Maximum (1, 0)

Solution steps

Find the critical points:

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

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

D (x, y) = 48 x^{4} + 96 x^{2} y^{2} + 32 x^{2} + 48 y^{4} - 32 y^{2} - 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 1, 0) :

Maximum

Check critical point

(1, 0) :

Maximum

Check critical point

(0, 0) :

Saddle

Check critical point

(- 1, 0) :

Maximum

Check critical point

(1, 0) :

Maximum

Check critical point

(0, 0) :

Saddle

Check critical point

(- 1, 0) :

Maximum

Check critical point

(1, 0) :

Maximum

Saddle (0, 0), Maximum (- 1, 0), Maximum (1, 0), Saddle (0, 0), Maximum (- 1, 0), Maximum (1, 0), Saddle (0, 0), Maximum (- 1, 0), Maximum (1, 0)

f (x) = x - ln (x), 0 \leq x \leq 2

f (x) = x^{3} - 24 ln (x) + 5

f (x) = 16 x + 19 x^{\frac{16}{19}}

f (t) = t^{2} + \frac{1}{t}

y = \frac{1}{3} x^{3} + 2 x^{2} + 3 x

What is the extreme 2x^2-2y^2-x^4-2x^2y^2-y^4 ?
The extreme 2x^2-2y^2-x^4-2x^2y^2-y^4 is