What is the extreme f(x,y)=x^2y+2xy^2-8x+4 ?

The extreme f(x,y)=x^2y+2xy^2-8x+4 is Saddle(0,2),Saddle(0,-2)

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extreme f(x,y)=x^2y+2xy^2-8x+4

extreme points

f (x, y) = x^{2} y + 2 x y^{2} - 8 x + 4

Saddle (0, 2), Saddle (0, - 2)

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = 4 x

D (x, y) = - 4 x^{2} - 8 x y - 16 y^{2}

Check the sign of at each critical point

Check critical point

(0, 2) :

Saddle

Check critical point

(0, - 2) :

Saddle

Saddle (0, 2), Saddle (0, - 2)

f (x) = x^{4} - 8 x^{2} + 5

f (x) = 4 x^{3} - 12 x^{2}

f (x, y) = \frac{1}{x + y}

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

f (x, y) = x - 9 y + 17

What is the extreme f(x,y)=x^2y+2xy^2-8x+4 ?
The extreme f(x,y)=x^2y+2xy^2-8x+4 is Saddle(0,2),Saddle(0,-2)