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extreme f(x,y)=x^4-2x^2+y^3-3y

Solution

extreme points

f (x, y) = x^{4} - 2 x^{2} + y^{3} - 3 y

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y

D (x, y) = 6 y (12 x^{2} - 4)

Check the sign of at each critical point

Check critical point

(0, 1) :

Saddle

Check critical point

(0, - 1) :

Maximum

Check critical point

(- 1, 1) :

Minimum

Check critical point

(- 1, - 1) :

Saddle

Check critical point

(1, 1) :

Minimum

Check critical point

(1, - 1) :

Saddle

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

ln (x^{4} + 27)

f (x, y) = \frac{2 y - x}{( x - 1 ) ^{2} + ( y - 2 ) ^{2}}

f (x, y) = x - 1 y - 1

x (x + 2)^{3}

f (x, y) = 8 x^{3} + 2 x y - 3 x^{2} + y^{2} + 1