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

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

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

extreme points

f (x) = 3 x y + x^{3} - 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) = - 2 x^{4} - 24 x^{3} + 12

f (x) = 5 x^{2} + 4 x

f (x, y) = - \frac{1}{4} x^{4} + \frac{2}{3} x^{3} + 4 x y - y^{2}

f (x, y) = x y (10 - 2 x - 3 y)

P (b, s) = b + 2 s

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