What is the extreme f(x)=(x^2+y^2)^2=2*(x^2-y^2) ?

The extreme f(x)=(x^2+y^2)^2=2*(x^2-y^2) is Saddle(0,0)

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extreme f(x)=(x^2+y^2)^2=2*(x^2-y^2)

extreme points

f (x) = (x^{2} + y^{2})^{2} = 2 \cdot (x^{2} - y^{2})

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 4

D (x, y) = - 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

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

f (x) = x^{\frac{2}{3}} (x - 2), - 2 \leq x \leq 2

f (x) = x^{2} - 2 x - lo g_{3} (y)

f (x) = 7 - (y) e^{- 0.45 x}

f (x) = x^{3} - 36 x^{2}, - 12 \leq x \leq 36

What is the extreme f(x)=(x^2+y^2)^2=2*(x^2-y^2) ?
The extreme f(x)=(x^2+y^2)^2=2*(x^2-y^2) is Saddle(0,0)