What is the extreme f(x,y)=(6xy)/(x^2+y^2+1) ?

The extreme f(x,y)=(6xy)/(x^2+y^2+1) is None

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extreme f(x,y)=(6xy)/(x^2+y^2+1)

extreme points

f (x, y) = \frac{6 x y}{x ^{2} + y ^{2} + 1}

None

Solution steps

Find the critical points:

None

\frac{6 x y}{x ^{2} + y ^{2} + 1}

has no critical points, therefore there are no extreme points

None

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

f (x) = - x^{2} + 8, - 2 \leq x \leq 4

f (x) = - x^{4} + 4 x^{3} + 9 x - 6

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

f (x, y) = - 3 x^{2} - 6 y^{2} - 6 x y + 66 x + 102 y

What is the extreme f(x,y)=(6xy)/(x^2+y^2+1) ?
The extreme f(x,y)=(6xy)/(x^2+y^2+1) is None