What is the extreme f(x)=x^2y+y^3-12y ?

The extreme f(x)=x^2y+y^3-12y is Minimum(0,2),Maximum(0,-2),Saddle(2sqrt(3),0),Saddle(-2sqrt(3),0)

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extreme f(x)=x^2y+y^3-12y

extreme points

f (x) = x^{2} y + y^{3} - 12 y

Minimum (0, 2), Maximum (0, - 2), Saddle (23, 0), Saddle (- 23, 0)

Solution steps

Find the critical points:

(0, 2), (0, - 2), (23, 0), (- 23, 0)

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

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

Check the sign of at each critical point

Check critical point

(0, 2) :

Minimum

Check critical point

(0, - 2) :

Maximum

Check critical point

(23, 0) :

Saddle

Check critical point

(- 23, 0) :

Saddle

Minimum (0, 2), Maximum (0, - 2), Saddle (23, 0), Saddle (- 23, 0)

\frac{e ^{x}}{x + 1}

f (x) = x^{2} - 2 x - 3, 0 \leq x \leq 2

f (x) = x (14 - 2 x) (15 - x)

\frac{1}{4} x^{4} - \frac{2}{3} x^{3} - \frac{1}{2} x^{2} + 2 x - 1

f (x) = 2 x^{2} + 3 y^{2} - 7

What is the extreme f(x)=x^2y+y^3-12y ?
The extreme f(x)=x^2y+y^3-12y is Minimum(0,2),Maximum(0,-2),Saddle(2sqrt(3),0),Saddle(-2sqrt(3),0)