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

The extreme f(x,y)=(x-y)^2-x^4-y is Saddle(-1/(\sqrt[3]{4)},-(-\sqrt[3]{4}+2)/(2\sqrt[3]{4)})

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

extreme points

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

Solution steps

Find the critical points:

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

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

Check the sign of at each critical point

Check critical point

Saddle

f (x) = x^{3} - x^{2} - x + 9, - 1 \leq x \leq 2

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

r - s - 1

w (x, y) = x e^{2} y + e - y

f (x) = - 2 x^{2} + 64 x - 80

What is the extreme f(x,y)=(x-y)^2-x^4-y ?
The extreme f(x,y)=(x-y)^2-x^4-y is Saddle(-1/(\sqrt[3]{4)},-(-\sqrt[3]{4}+2)/(2\sqrt[3]{4)})