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

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

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

extreme points

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

Minimum (0, 0), Saddle (\frac{1}{3}, \frac{2}{3}), Saddle (\frac{1}{3}, - \frac{2}{3})

Solution steps

Find the critical points:

(0, 0), (\frac{1}{3}, \frac{2}{3}), (\frac{1}{3}, - \frac{2}{3})

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(\frac{1}{3}, \frac{2}{3}) :

Saddle

Check critical point

(\frac{1}{3}, - \frac{2}{3}) :

Saddle

Minimum (0, 0), Saddle (\frac{1}{3}, \frac{2}{3}), Saddle (\frac{1}{3}, - \frac{2}{3})

f (x) = x^{3} + 5 x^{2} - 8 x + 1

f (x) = 7 x^{\frac{8}{3}}

\frac{6}{x} + \frac{3}{y} + 14 x + 5 y

y = x - sin (- x), 0 \leq x \leq 2 π

x^{2} (2 - x)^{3}

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