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

Solution

extreme points

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

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

Solution steps

Find the critical points:

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

Check the sign of at each critical point

Check critical point

(- 1, 0) :

Saddle

Check critical point

(- 1, - 1) :

Minimum

Check critical point

(- 1, 1) :

Minimum

Check critical point

(1, 0) :

Maximum

Check critical point

(1, - 1) :

Saddle

Check critical point

(1, 1) :

Saddle

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

f (x) = x^{4} - 2 x^{3} + x + 1

f (x) = 3 x^{3} + 16 x

f (x) = x - 2 sin (x)

f (x) = x 2 - x^{2}