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

The extreme f(x,y)=y^3-yx-2xy^2+2x^2 is Saddle(0,0),Minimum(3/32 , 1/4),Saddle(1/4 , 1/2)

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

extreme points

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

Saddle (0, 0), Minimum (\frac{3}{32}, \frac{1}{4}), Saddle (\frac{1}{4}, \frac{1}{2})

Solution steps

Find the critical points:

(0, 0), (\frac{3}{32}, \frac{1}{4}), (\frac{1}{4}, \frac{1}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y - 4 x

D (x, y) = - 16 y^{2} + 16 y - 16 x - 1

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{3}{32}, \frac{1}{4}) :

Minimum

Check critical point

(\frac{1}{4}, \frac{1}{2}) :

Saddle

Saddle (0, 0), Minimum (\frac{3}{32}, \frac{1}{4}), Saddle (\frac{1}{4}, \frac{1}{2})

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

f (x) = \frac{1}{x ^{3} + 3 x ^{2} + 5 x + 1}

y = 32 cos (10 x) + 25

f (x) = - 2 x^{2} + 4

\frac{1 - 2 x}{x ^{2} - x}

What is the extreme f(x,y)=y^3-yx-2xy^2+2x^2 ?
The extreme f(x,y)=y^3-yx-2xy^2+2x^2 is Saddle(0,0),Minimum(3/32 , 1/4),Saddle(1/4 , 1/2)