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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = 36 y^{2} - 36 x^{2}

Check the sign of at each critical point

Check critical point

(0, 1) :

Minimum

Check critical point

(0, - 1) :

Maximum

Check critical point

(1, 0) :

Saddle

Check critical point

(- 1, 0) :

Saddle

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

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

f (x) = 2 (5 x)^{x}

f (x, y) = 400 - 16 x^{2} - 81 y^{2}

f (x) = ln (7 + x^{2})

y = - x^{3} + \frac{3}{2} x^{2} + 18 x

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