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

The extreme f(x)=3xy^2-2x+y is Saddle(1/(2sqrt(6)),-sqrt(2/3)),Saddle(-1/(2sqrt(6)),sqrt(2/3))

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

extreme points

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

Saddle (\frac{1}{2 6}, - \frac{2}{3}), Saddle (- \frac{1}{2 6}, \frac{2}{3})

Solution steps

Find the critical points:

(\frac{1}{2 6}, - \frac{2}{3}), (- \frac{1}{2 6}, \frac{2}{3})

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

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

Check the sign of at each critical point

Check critical point

(\frac{1}{2 6}, - \frac{2}{3}) :

Saddle

Check critical point

(- \frac{1}{2 6}, \frac{2}{3}) :

Saddle

Saddle (\frac{1}{2 6}, - \frac{2}{3}), Saddle (- \frac{1}{2 6}, \frac{2}{3})

x^{2} - y^{2} - 2 x + 4 y + 6

f (x) = ln (4 x^{2} + 18 y^{2} - 36)

f (x) = x^{3} + y^{3} - 15 x y

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

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

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