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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

\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

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

Maximum

Check critical point

(0, 0) :

Saddle

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

f (x) = f (x) = 3 x^{4} - 8 x^{3} - 6 x^{2} + 24 x - 3

- 2.75 x^{2} + 6.05 x + 270

\frac{d}{d x} (x^{2} + x^{3} y^{2})

f (x) = x^{3} - 18 x y + y^{3}

2 x - 13

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