What is the extreme f(x,y)=y^3x^5+yx ?

The extreme f(x,y)=y^3x^5+yx is Saddle(0,0)

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extreme f(x,y)=y^3x^5+yx

extreme points

f (x, y) = y^{3} x^{5} + y x

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 x^{5} y

D (x, y) = - 105 x^{8} y^{4} - 30 x^{4} y^{2} - 1

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

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

2 x^{3} - 3 x^{2} - 12 x + 8

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

g (x, y) = ln (x^{2} + y^{2} - 1)

f (x) = x^{3} - 3 x - 1

What is the extreme f(x,y)=y^3x^5+yx ?
The extreme f(x,y)=y^3x^5+yx is Saddle(0,0)