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) = x^{2} - 4 x + 2

x^{2} - 4 x + 5

f (x) = e^{- 3.5 x^{2}}

f (x, y) = 8 - 2 x^{2} - 2 y^{2}

f (x, y) = 4 y x

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)