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

The extreme f(x)=x^3+y^3-3x^2-9y^2-9x is Saddle(-1,6),Maximum(-1,0),Minimum(3,6),Saddle(3,0)

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

extreme points

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

Saddle (- 1, 6), Maximum (- 1, 0), Minimum (3, 6), Saddle (3, 0)

Solution steps

Find the critical points:

(- 1, 6), (- 1, 0), (3, 6), (3, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y - 18

D (x, y) = (6 x - 6) (6 y - 18)

Check the sign of at each critical point

Check critical point

(- 1, 6) :

Saddle

Check critical point

(- 1, 0) :

Maximum

Check critical point

(3, 6) :

Minimum

Check critical point

(3, 0) :

Saddle

Saddle (- 1, 6), Maximum (- 1, 0), Minimum (3, 6), Saddle (3, 0)

f (x, y) = 2 x^{2} - 8 x + y^{2} - 8 y + 6

f (x) = x e^{- 4 x}, 0 \leq x \leq 2

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

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

x^{\frac{1}{5}} (x + 6)

What is the extreme f(x)=x^3+y^3-3x^2-9y^2-9x ?
The extreme f(x)=x^3+y^3-3x^2-9y^2-9x is Saddle(-1,6),Maximum(-1,0),Minimum(3,6),Saddle(3,0)