What is the extreme f(x,y)=y^3+3x^2y-6xy+1 ?

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

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extreme f(x,y)=y^3+3x^2y-6xy+1

extreme points

f (x, y) = y^{3} + 3 x^{2} y - 6 x y + 1

Saddle (2, 0), Saddle (0, 0), Minimum (1, 1), Maximum (1, - 1)

Solution steps

Find the critical points:

(2, 0), (0, 0), (1, 1), (1, - 1)

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

D (x, y) = 36 y^{2} - 36 x^{2} + 72 x - 36

Check the sign of at each critical point

Check critical point

(2, 0) :

Saddle

Check critical point

(0, 0) :

Saddle

Check critical point

(1, 1) :

Minimum

Check critical point

(1, - 1) :

Maximum

Saddle (2, 0), Saddle (0, 0), Minimum (1, 1), Maximum (1, - 1)

f (x) = \frac{1}{4} (x - 8)^{4} + \frac{17}{3} (x - 8)^{3}

f (x, y) = \frac{x}{∣ y ∣}

f (x) = x^{2} - 10 x + 21

f (x, y) = e^{x} (x^{2} + y^{2})

f (x, y) = x^{3} - 2 y^{2} - 6 x y + 4

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