What is the extreme f(x)=(xy+6)/(3sqrt(x^2-y)) ?

The extreme f(x)=(xy+6)/(3sqrt(x^2-y)) is Saddle(-(3^{1/3})/(2^{1/3)},-\sqrt[3]{18})

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extreme f(x)=(xy+6)/(3sqrt(x^2-y))

extreme points

f (x) = \frac{x y + 6}{3 x ^{2} - y}

Solution steps

Find the critical points:

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{4 x ^{3} - x y + 18}{12 ( x ^{2} - y ) ^{\frac{5}{2}}}

D (x, y) = \frac{48 x ^{5} - 4 x ^{4} y ^{2} - 132 x ^{3} y + 5 x ^{2} y ^{3} - 108 x ^{2} + 84 x y ^{2} - y ^{4} + 108 y}{36 ( x ^{2} - y ) ^{5}}

Check the sign of at each critical point

Check critical point

Saddle

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

f (x) = \frac{1}{2} x^{4} - x^{2} - 1

f (x) = \frac{x - 2}{x ^{2} - 6 x + 9}

f (x) = 3 + 9 x^{2} - 6 x^{3}

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

What is the extreme f(x)=(xy+6)/(3sqrt(x^2-y)) ?
The extreme f(x)=(xy+6)/(3sqrt(x^2-y)) is Saddle(-(3^{1/3})/(2^{1/3)},-\sqrt[3]{18})