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

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

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 144 x^{2} y^{2} - 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

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

5 x + 1 + \frac{2}{x - 2}

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

y = (x^{2} - 3) e^{x}

f (x) = x^{3} - 3 x - 5

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