What is the extreme f(x,y)=xy^2+12x^3y-xy ?

The extreme f(x,y)=xy^2+12x^3y-xy is

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extreme f(x,y)=xy^2+12x^3y-xy

extreme points

f (x, y) = x y^{2} + 12 x^{3} y - x y

Minimum (\frac{1}{2 15}, \frac{2}{5}), Maximum (- \frac{1}{2 15}, \frac{2}{5 15} 15), Saddle (\frac{1}{2 3}, 0), Saddle (- \frac{1}{2 3}, 0)

Solution steps

Find the critical points:

(\frac{1}{2 15}, \frac{2}{5}), (- \frac{1}{2 15}, \frac{2}{5 15} 15), (\frac{1}{2 3}, 0), (- \frac{1}{2 3}, 0)

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

D (x, y) = - 1296 x^{4} + 72 x^{2} - 4 y^{2} + 4 y - 1

Check the sign of at each critical point

Check critical point

(\frac{1}{2 15}, \frac{2}{5}) :

Minimum

Check critical point

(- \frac{1}{2 15}, \frac{2}{5 15} 15) :

Maximum

Check critical point

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

Saddle

Check critical point

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

Saddle

Minimum (\frac{1}{2 15}, \frac{2}{5}), Maximum (- \frac{1}{2 15}, \frac{2}{5 15} 15), Saddle (\frac{1}{2 3}, 0), Saddle (- \frac{1}{2 3}, 0)

f (x) = \frac{2 x - 4}{x - 1}

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

y = x^{2} - 4 x - 5

x^{3} - 6 x^{2} + 9 x + 9

f (x, y) = 16 - x^{2} - y^{2}

What is the extreme f(x,y)=xy^2+12x^3y-xy ?
The extreme f(x,y)=xy^2+12x^3y-xy is