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

The extreme f(x,y)=(x^4-y^4)/(x^2-y^2) is Minimum(0,0)

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

y = - 2 x^{3} + 6 x^{2} - 3

f (x) = x ln (x^{2} - y)

y = 3 x^{5} - 20 x^{3}

F (x, y) = 5 x y^{3} - y^{4} + \frac{x ^{2}}{2}

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

What is the extreme f(x,y)=(x^4-y^4)/(x^2-y^2) ?
The extreme f(x,y)=(x^4-y^4)/(x^2-y^2) is Minimum(0,0)