What is the extreme f(x,y)=x^3+3*x*y-y^2 ?

The extreme f(x,y)=x^3+3*x*y-y^2 is Saddle(0,0),Maximum(-3/2 ,-9/4)

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extreme f(x,y)=x^3+3xy-y^2

extreme points

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

Saddle (0, 0), Maximum (- \frac{3}{2}, - \frac{9}{4})

Solution steps

Find the critical points:

(0, 0), (- \frac{3}{2}, - \frac{9}{4})

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

D (x, y) = - 12 x - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- \frac{3}{2}, - \frac{9}{4}) :

Maximum

Saddle (0, 0), Maximum (- \frac{3}{2}, - \frac{9}{4})

f (x) = 4 ln (2 x e^{- x})

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

f (x) = \frac{y - 3}{y ^{2} - 3 y + 9}

f (x) = (16 - x) (100 x + 1300)

f (x) = \frac{1}{3 x ^{\frac{2}{3}}}

What is the extreme f(x,y)=x^3+3*x*y-y^2 ?
The extreme f(x,y)=x^3+3*x*y-y^2 is Saddle(0,0),Maximum(-3/2 ,-9/4)