What is the extreme f(x,y)=x^3+y^3-6y^2-3x+9 ?

The extreme f(x,y)=x^3+y^3-6y^2-3x+9 is Saddle(-1,4),Maximum(-1,0),Minimum(1,4),Saddle(1,0)

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extreme f(x,y)=x^3+y^3-6y^2-3x+9

extreme points

f (x, y) = x^{3} + y^{3} - 6 y^{2} - 3 x + 9

Saddle (- 1, 4), Maximum (- 1, 0), Minimum (1, 4), Saddle (1, 0)

Solution steps

Find the critical points:

(- 1, 4), (- 1, 0), (1, 4), (1, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y - 12

D (x, y) = 6 x (6 y - 12)

Check the sign of at each critical point

Check critical point

(- 1, 4) :

Saddle

Check critical point

(- 1, 0) :

Maximum

Check critical point

(1, 4) :

Minimum

Check critical point

(1, 0) :

Saddle

Saddle (- 1, 4), Maximum (- 1, 0), Minimum (1, 4), Saddle (1, 0)

y = x^{3} - 3 x + 5

e^{- \frac{2}{x}}

f (x) = x - ln (x), \frac{1}{2} \leq x \leq 2

f (x, y) = 9 x^{2} + 18 x y + y^{3}

f (x) = 2 x^{2} + 3 x - 5

What is the extreme f(x,y)=x^3+y^3-6y^2-3x+9 ?
The extreme f(x,y)=x^3+y^3-6y^2-3x+9 is Saddle(-1,4),Maximum(-1,0),Minimum(1,4),Saddle(1,0)