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

The extreme f(x)=2x^3+3xy+2y^3 is Saddle(0,0),Maximum(-1/2 ,-1/2)

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

extreme points

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

Saddle (0, 0), Maximum (- \frac{1}{2}, - \frac{1}{2})

Solution steps

Find the critical points:

(0, 0), (- \frac{1}{2}, - \frac{1}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = 12 y

D (x, y) = 144 x y - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- \frac{1}{2}, - \frac{1}{2}) :

Maximum

Saddle (0, 0), Maximum (- \frac{1}{2}, - \frac{1}{2})

f (x) = - 5 cos (x - 3 π)

f (x) = \frac{- 5}{x ^{2} - 4}

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

f (x) = - 3 x^{4} - 7 x^{3} + 23

f (x) = x 6 - x

What is the extreme f(x)=2x^3+3xy+2y^3 ?
The extreme f(x)=2x^3+3xy+2y^3 is Saddle(0,0),Maximum(-1/2 ,-1/2)