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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

Maximum

Check critical point

(0, 1) :

Maximum

Check critical point

(- 1, 0) :

Minimum

Check critical point

(- 1, - 1) :

Saddle

Check critical point

(- 1, 1) :

Saddle

Check critical point

(1, 0) :

Minimum

Check critical point

(1, - 1) :

Saddle

Check critical point

(1, 1) :

Saddle

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

f (x) = \frac{x}{8 x - 3}, 8 \leq x \leq 13

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

f (x) = \frac{x}{( x - 7 ) ^{2}}

y = \frac{e ^{x}}{x ^{2} - 3 x + 2}

g (t) = t^{4} + t^{3} + t^{2} + 1