What is the extreme f(x)=y^3-x^3-2xy+6 ?

The extreme f(x)=y^3-x^3-2xy+6 is Saddle(0,0),Maximum(2/3 ,-2/3)

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extreme f(x)=y^3-x^3-2xy+6

extreme points

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

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y

D (x, y) = - 36 x y - 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

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

Maximum

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

f (x) = 7 θ - 9 sin (θ)

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

f (x) = (\frac{1}{9} x^{2} + y^{2}) e^{\frac{- x}{3} + y}

f (x) = 16 x^{3} (x + 1)^{2}

2 x^{\frac{3}{2}} - 9 x + 6

What is the extreme f(x)=y^3-x^3-2xy+6 ?
The extreme f(x)=y^3-x^3-2xy+6 is Saddle(0,0),Maximum(2/3 ,-2/3)