What is the extreme f(x)=12x-x^3-y(y^2+3y-9)+4 ?

The extreme f(x)=12x-x^3-y(y^2+3y-9)+4 is Saddle(-2,1),Minimum(-2,-3),Maximum(2,1),Saddle(2,-3)

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extreme f(x)=12x-x^3-y(y^2+3y-9)+4

extreme points

f (x) = 12 x - x^{3} - y (y^{2} + 3 y - 9) + 4

Saddle (- 2, 1), Minimum (- 2, - 3), Maximum (2, 1), Saddle (2, - 3)

Solution steps

Find the critical points:

(- 2, 1), (- 2, - 3), (2, 1), (2, - 3)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 3 (2 y + 2)

D (x, y) = 18 x (2 y + 2)

Check the sign of at each critical point

Check critical point

(- 2, 1) :

Saddle

Check critical point

(- 2, - 3) :

Minimum

Check critical point

(2, 1) :

Maximum

Check critical point

(2, - 3) :

Saddle

Saddle (- 2, 1), Minimum (- 2, - 3), Maximum (2, 1), Saddle (2, - 3)

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

f (x) = 64 - x^{2}

f (x, y) = x y - 4 y - 16 x + 64

y = cos (x)

2 x^{2} + 24 x - 6

What is the extreme f(x)=12x-x^3-y(y^2+3y-9)+4 ?
The extreme f(x)=12x-x^3-y(y^2+3y-9)+4 is Saddle(-2,1),Minimum(-2,-3),Maximum(2,1),Saddle(2,-3)