What is the extreme f(x)=y^4-4y^3+2x^2+8xy ?

The extreme f(x)=y^4-4y^3+2x^2+8xy is Saddle(0,0),Minimum(2,-1),Minimum(-8,4)

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extreme f(x)=y^4-4y^3+2x^2+8xy

extreme points

f (x) = y^{4} - 4 y^{3} + 2 x^{2} + 8 x y

Saddle (0, 0), Minimum (2, - 1), Minimum (- 8, 4)

Solution steps

Find the critical points:

(0, 0), (2, - 1), (- 8, 4)

\frac{\partial ^{2} f}{\partial x ^{2}} = 12 y^{2} - 24 y

D (x, y) = 4 (12 y^{2} - 24 y) - 64

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(2, - 1) :

Minimum

Check critical point

(- 8, 4) :

Minimum

Saddle (0, 0), Minimum (2, - 1), Minimum (- 8, 4)

f (x) = x^{2} e^{- 9 x}

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

y = \frac{ln ( x )}{x}

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

f (x) = 4 x^{5} - 8 x^{4} - 5 x^{3} + 10 x^{2} + x - 2

What is the extreme f(x)=y^4-4y^3+2x^2+8xy ?
The extreme f(x)=y^4-4y^3+2x^2+8xy is Saddle(0,0),Minimum(2,-1),Minimum(-8,4)