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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = (- 6 x + 6)^{2} - 36 y^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(2, 0) :

Maximum

Check critical point

(1, 1) :

Saddle

Check critical point

(1, - 1) :

Saddle

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

f (x) = \frac{2 e ^{x}}{x ^{5}}, [0, \infty]

5 x^{3} e^{- x}

f (x) = x - 2 sin (x), (0, 2 π)

f (x) = \frac{- 1}{5 x ^{2} + 25 x + 30}

f (x) = \frac{5 x}{x ^{2} - 1}

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