What is the extreme f(x)=x^3-8y^3+4xy-8 ?

The extreme f(x)=x^3-8y^3+4xy-8 is Saddle(0,0),Minimum(2/3 ,-1/3)

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extreme f(x)=x^3-8y^3+4xy-8

extreme points

f (x) = x^{3} - 8 y^{3} + 4 x y - 8

Saddle (0, 0), Minimum (\frac{2}{3}, - \frac{1}{3})

Solution steps

Find the critical points:

(0, 0), (\frac{2}{3}, - \frac{1}{3})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 48 y

D (x, y) = - 288 x y - 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{2}{3}, - \frac{1}{3}) :

Minimum

Saddle (0, 0), Minimum (\frac{2}{3}, - \frac{1}{3})

7 + 12 x - x^{3}

f (x) = x + e^{- 3 x}, - 3 \leq x \leq 4

f (x) = x e^{- e}, 0 \leq x \leq 2

f (x) = 4 x^{2} - 16 x

2 - 2 x^{2}

What is the extreme f(x)=x^3-8y^3+4xy-8 ?
The extreme f(x)=x^3-8y^3+4xy-8 is Saddle(0,0),Minimum(2/3 ,-1/3)