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

The extreme f(x)=(2xy)/(x^2+y^2-4) is None

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

extreme points

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

None

Solution steps

Find the critical points:

None

\frac{2 x y}{x ^{2} + y ^{2} - 4}

has no critical points, therefore there are no extreme points

None

f (x, y) = 2 x^{2} + 3 x y + 4 y^{2} - 7 x - 11 y

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

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

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

y = x^{3} - 3 x + 1

What is the extreme f(x)=(2xy)/(x^2+y^2-4) ?
The extreme f(x)=(2xy)/(x^2+y^2-4) is None